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Problem Books in Mathematics Series Editor: P.R. Halmos
Unsolved Problems in Intuitive Mathematics, Volume I: Unsolved Problems in Number Theory by Richard K. Guy 1981. xviii, 161 pages. 17 illus.
Theorems and Problems in Functional Analysis by A.A. Kirillov and A.D. Gvishiani 1982. ix, 347 pages. 6 illus. Problems in Analysis by Bernard Gelbaum 1982. vii, 228 pages. 9 illus.
A Problem Seminar by Donald J. Newman 1982. viii, 113 pages.
Problem-Solving Through Problems by Loren C. Larson 1983. xi, 344 pages. 104 illus.
Demography Through Problems by N. Keyfitz and J.A. Beekman 1984. viii, 141 pages. 22 illus.
Problem Book for First Year Calculus by George W. B/uman 1984. xvi. 384 pages. 384 illus. Exercises in Integration by Claude George 1984. x. 550 pages. 6 illus. Exercises in Number Theory by D.P. Parent 1984. x. 541 pages.
Problems in Geometry by Marcel Berger et al. 1984. viii. 266 pages. 244 illus.
Claude George
Exercises in Integration
With 6 Illustrations
Springer-Verlag New York Berlin Heidelberg Tokyo
Translator J.M. Cole
Claude George University de Nancy I UER Sciences Mathematiques Boite Postale 239 54506 Vandoeuvre les Nancy Cedex France
17 St. Mary's Mount Leybum, North Yorkshire DL8 5JB U.K.
Editor Paul R. Halmos Department of Mathematics Indiana University Bloomington, IN 47405 U.S.A.
AMS Classifications: OOA07, 26-01, 28-01
Library of Congress Cataloging in Publication Data George, Claude. Exercises in integration. (Problem books in mathematics) Translation of: Exercices et problemes d`int6gration. Bibliography: p. Includes indexes.
1. Integrals, Generalized-Problems, exercises, etc. I. Title. II. Series. QA312.G39513
Printed and bound by R.R. Donnelley & Sons, Harrisonburg, Virginia. Printed in the United States of America.
987654321 ISBN 0-387-96060-0 Springer-Verlag New York Berlin Heidelberg Tokyo ISBN 3-540-96060-0 Springer-Verlag Berlin Heidelberg New York Tokyo
Introduction
Having taught the theory of integration for several years at the University of Nancy I, then at the Ecole des Mines of the same city, I had followed the custom of the times of writing up detailed solutions of exercises and problems, which I used to distribute to the students every week.
Some colleagues who had had
occasion to use these solutions have persuaded me that this work would be interesting to many students, teachers and researchers. The majority of these exercises are at the master's level; to them I have added a number directed to those who would wish to tackle greater difficulties or complete their knowledge on various points of the theory (third year students, diploma of education students, researchers, etc.).
This book, I hope, will render to students the services that this kind of book brings them in general, with the reservation that can always be made in this case: that certain of them will be tempted to look at the solution to the exercises which are put to.them without any personal effort.
There is hardly any need to
emphasize that such a use of this book would be no benefit. the other hand, the student who
On
after having worked seriously
upon a problem, seeks some pointers from the solution, or compares it with his own, will be using this work in the optimal way.
V
INTRODUCTION
vi
Teachers will find this book to be an important, if not exhaustive, list of exercises, certain of which are more or less standard, and others which may seem new. I have also noted (and this is what led me to edit these sheets)
that from one year to another one sometimes forgets the solution of an exercise and that one has to lose precious time in redisThis is particularly true for those solutions of
covering it.
which one remembers the heuristic form but of which the writing up is delicate if one wishes to be clear and precise at the same time.
Now, if one requires, quite rightly, that students
write their homework up correctly, then it is befitting to submit impeccable corrections to them, where the notations are judiciously chosen, phrases of the kind "it is clear that ... " used wittingly, and where the telegraphic style gives way to conciseness.
It is often the incorporation of these corrections which
demands the most work; I have therefore striven to take pains with the preparation of the proposed solutions, always remaining persuaded that perfection in this domain is never attained.
If
this book encourages those who have to present (either orally or in writing) correct versions of problems to improve the version they submit, the object I have set myself will be partly realised. In this book researchers will find some results that are not always treated in courses on integration; they are either properties whose use is not as universal as those which usually appear and which are therefore found scattered about in appendices in various works, or are results that correspond to some technical lemmas which I have picked up in recent articles on a variety of subjects: group theory, differential games, control theory, probability, etc.,
...
.
In presenting such a work it is just as well to make explicit those points of the theory that are assumed to be known.
This is
the object of the brief outline which precedes the eleven chapters of exercises.
INTRODUCTION
vii
In view of the origin of this book, it is evident that I took as a reference point the course that I gave at the time.
After
having taught abstract measure theory one year, I opted the next for a course expounding only the Lebesgue integral.
This is not
the place to discuss the advantages and inconveniences of each of the two points of view for the first year of a master's programme. I will say only that I have always considered the course that I gave to be more a course in analysis in which it is decided to use the Lebesgue integral than as a dogmatic exposition of a particular theory of integration.
The choice of exercise reflects
this attitude, especially in the emphasis given to trigonometric series, thereby paying the hommage due to the theory which is the starting point of the works of Cantor, Jordan, Peano, Borel, and Lebesgue.
From this it results that, except for the seven exer-
cises of Chapter 2 concerning a-algebras, all the others deal with Lebesgue measure on]Rr.
The advantage that has to be conceded to
this point of view is that it avoids the vocabulary of abstract measure theory, which constitutes an artificial obstacle for those readers who might not yet be well versed in this theory.
As for
students who might have followed a more sophisticated course, I can assure them that by substituting du for dx and u(E) fore meas(E) they will essentially rediscover the problems as they are commonly put to them, except for pathological examples about measures that are not a-finite and the applications of the RadonNikodym Theorem.
Furthermore, on this latter point the more per-
spicacious amongst them will not fail to see that the chapter treating the relationships between differentiation and integration is not foreign to this theorem.
Truthfully, there is another
point that is not tackled in this book, namely the matter of Fourier transforms of finite positive measures and Stone's Theorem, which to my mind is better suited to a course on probability. As was mentioned above, numerous exercises are devoted to trigonometric series, which provides an important set of applications
Viii
INTRODUCTION
of Lebesgue's theory.
This has led me to include some exercises
on series, summation processes, and trigonometric polynomials. Other exercises use the theory of holomorphic functions.
In
particular, some results of the PhrUgmen-Lindeltff type arise on
two occasions; in each instance I have given its proof under the hypotheses that appear in the exercise.
Quite generally, I have
included in the solutions, or in an appendix to them, the proofs of certain points of analyis, topology, or algebra which students may not know.
I have chosen to make each solution follow immediately after the corresponding problem.
The other method, which consists of
regrouping the former in a second part of the work, seemed to me (from memories I have retained from my student days) much less
manageable, especially when the problem is long, for it then becomes necessary to return often to the back of the book in order to follow the solution. I find it difficult to cite the origin of these exercises.
Many are part of a common pool of knowledge, handed down, one might say, in the public domain.
Others are drawn from different
classic works where they are proposed without proof or followed by more or less summary indications (in this respect it is interesting to note that in forcing oneself to write down the solutions one discovers a certain number of errors -just as many in the questions as in the suggestions offered).
Certain of the exer-
cises in this book were communicated to me orally by colleagues; I would thank them for their help here.
Lastly, others are, as
I have already said, lemmas found here and there, and which I have sometimes adapted.
DEFINITION: A family A of subsets of a set X is called a a-ALGEBRA ("sigma algebra") if 0 e A, and if A is closed under complementation and countable union.
From this it follows that the set X itself belongs to the a-algebra A, and that the a-algebra A is closed under countable intersecFor two sets A,B e A let us denote A - B = {x:x a A,x $ B};
tion.
then we have (A - B)e A. The smallest a-algebra containing the open sets of ]R a-algebra of BOREL SETS of ]R
;
is the
this a-algebra is also the small-
est a-algebra which contains the closed (resp. open) rectangles
of]R
.
DEFINITION: A (positive) MEASURE on a a -algebra A is a mapping u of A into [0,oo] such that if E is the disjoint union of a sequence of sets En e A, then u(E) = I u(En).
It follows that u(o) = 0, and then, if E is the union (not necessarily disjoint) of the sets En, U(E) 5 L u(En).
An equi-
valent definition is the following: If E is the union of a fin-
1
CHAPTER 0: OUTLINE
2
ite number of Ei's, each of which is in A and which are pairwise disjoint, then p(E) = p(E1) +
+ p(Ep); and furthermore p(A) _
limu(An) when A is the union of an increasing sequence of sets An of A.
If p is a measure and A is the intersection of a decreas-
ing sequence of sets An e A and if u(A1) < -, then p(A) = limp(An) There exists one and only one positive measure v on the a-algebra of Borel sets of ]R
such that, for every rectangle P, its
measure v(P) is equal to the volume of P.
DEFINITION: A set E of]R is called a NEGLIGEABLE SET if there exists a Borel set A such that E C A and v(A) = 0.
This definition is equivalent to the existence, for every e> 0, of a sequence of rectangles covering E, the sum of the volumes of which is less than c.
A countable union of negligeable sets is
negligeable, and every affine sub-manifold of iRp that is of dimension less that p is negligeable. DEFINITION: A set of ]R
is called a LEBESGUE MEASURABLE SET (or
simply a MEASURABLE SET) if it belong to the smallest a-algebra containing the Borel sets and negligeable sets of]R1. In order that E C ]R
be measurable it is necessary and suffic-
ient that there exist the Borel sets A and B such that A C E C B and v(B - A) = 0; upon then setting meas(E).= v(A) one unambiguously defines a positive measure on the a-algebra of Lebesgue measurable sets of ]Rp.
SURE ON I(.
This measure is called the LEBESGUE MEA-
A set is negligeable if it is measurable and of
(Lebesgue) measure zero.
This is why one also uses the expres-
sion SET OF MEASURE ZERO to denote a negligeable set.
The Lebesgue measure is invariant under translation as well as under unimodular linear transformations (i.e., those with determinant equal to ±1).
A homothety of ratio A multiplies the
Lebesgue measure by JAJp (where p is the dimension of the space).
OF THE COURSE
3
DEFINITION: If A is a o-algebra of subsets of X and B is a Q algebra of subsets of Y, a mapping f:X + Y is said to be an A B-MEAS-URABLE MAPPING if f 1(B) e A for a Z When Z B e Y= B.
Ilzp and B is
the a-algebra of Borel sets, one says, simply, that f is an A-MEASURABLE MAPPING.
In this case the definition is equivalent to re-
quiring f_1(V) eA for every open set V of Y.
Furthermore, when
X = n2q, the mapping f is said to be a BOREL MAPPING or a LEBESGUE-
MEASURABLE MAPPING according as A is the a-algebra of Borel sets or the Lebesgue-measurable sets of X. If f:X -;Ill, in order that f be A-measurable it is sufficient
that (f < a) = {x:f(x) < a} e A for all a em (and even for a e
This condition is taken as the definition of the A-measurability of an ARITHMETIC FUNCTION, that is to say of a mapping of X into [-co,+m] =3-R.
If f is an A-measurable mapping of X into Iltp and g
a Borel mapping of Ilzp into zzq, then gof is A-measurable. note that every continuous mapping of Ilzp into Ilzq is Borel.
Let us If
(fn) is a sequence of A-measurable arithmetic functions, the functions supfn,inffn,limsupfn,liminffn are also A-measurable. DEFINITION: A function is called a SIMPLE FUNCTION (with respect to the a-algebra A) if it is a linear combination of characteristic functions of sets of the a-algebra A.
For every A-measurable positive arithmetic function f there exists an increasing sequence of positive simple functions which converges to f at every point of X.
DEFINITION: A property holding on the points of a set A of7Rp is said to be true ALMOST EVERYWHERE ON A SET A if the set of points
of A for which this property is not satisfied has measure zero.
If f and g are two mappings from z
into zzq (or m) such that
f is measurable and f = g almost everywhere, then g is measurable.
CHAPTER 0: OUTLINE
4
This allows the notion of measurability to be extended to functions that are defined only almost everywhere.
DEFINITION: A function defined on]RP is called a STEP FUNCTION if it is a linear combination of characteristic functions of rect-
angles of]RP. Every measurable arithmetic function on]R
is the limit almost
everywhere of a sequence of step functions.
THEOREM: (Regularity of the Lebesgue Measure): For every measurable set E ofiRP one has: sup{meas(K):K compact K C E}; meas(E) = inf{meas(V):V open V D E}.
THEOREM: (Egoroff): Let X be a measurable set of]R
such that
meas(X) < co and (fn) a sequence of measurable functions such that
fn - f almost everywhere on X.
For every e > 0 there exists a
measurable set A C X such that: (i): meas(X - A) < ci (ii): fn -> f uniformly on A.
0.2
INTEGRATION OF MEASURABLE POSITIVE FUNCTIONS
NOTATION: If cp is a simple function on Min that takes the positive
values a1,...,ap on the (disjoint) measurable sets Al....)Ap, we set
I
n
q,(x)dx =
aimeas(Ai),
9 _
i=1
OF THE COURSE
5
with the convention that a.(+-)
or 0 according as a > 0 or
a = 0. DEFINITION. With the above notation, if f is a positive measurable arithmetic function on ]Rn there exists an increasing sequence ((pi) of positive simple functions which tends towards f One then sets:
at every point.
f(x)dx = if = limf
J
. .
This element of [0,+-]=]K +, which does not depend upon the sequence
(Ti) selected, is called the (LEBESGUE) INTEGRAL of f on7Rn.
This (Lebesgue) integral possesses the following properties (where f and g denote measurable positive arithmetic functions): PROPERTY (1): If f = g almost everywhere, then if = Jg;
PROPERTY (2): Jf = 0 if and only if f = 0 almost everywhere;
PROPERTY (3): if < - implies f < m almost everywhere;
PROPERTY (4): f 4 g almost everywhere implies if
PROPERTY (5):
PROPERTY (6)
1<
Jg;
J(f + g) = if + fg;
If A e]-R+., then JAf = AJf.
One can prove the following two fundamental results:
THEOREM: (Lebesgue's Monotonic Convergence): If (fn) is an increasing sequence of measurable positive arithmetic functions, then
Jlimf
n
n
=
limif-' n
CHAPTER 0: OUTLINE
6
LEMMA: (Fatou): If (fn) is a sequence of positive arithmetic functions, then
Jliminffn " liminf Ifn. n n
These two essential properties of the Lebesgue integral are equivalent to the following statements: Let (fn) be a sequence of positive arithmetic functions:
(a):
f(I fn) = E (Jfn);
(b):
If fn -* f almost everywhere, and if there exists A eIt+
such that
Jfn 5 A
for all n,
then
Jf , A. Property (1) allows the definition of the integral to be extended to measurable arithmetic functions that are defined only almost everywhere. And lastly:
NOTATION: If f is a measurable positive arithmetic function, and if E is a measurable set of ]R , we set
fEf(x)dx=JE =fiv, where 1lE denotes the CHARACTERISTIC FUNCTION of E.
The mapping E + J f defines a positive measure on the 'a-algebra
E
OF THE COURSE
7
of (Lebesgue) measurable sets.
0.3
INTEGRATION OF COMPLEX MEASURABLE FUNCTIONS
NOTATION: For every real function u, we set: u+ = sup(u,O) _ I(luI + u), u- = sup(-u,0) _ i(lul - u), so that
u = U+ - u_,
= u+ + u-,
lu l
u+u- = 0.
Let f be a complex measurable function on7Rn, and let u and v be its real and imaginary parts.
The function f is measurable
if and only if u+U_,v+,v- are measurable. DEFINITION: f is (LEBESGUE) INTEGRABLE ON]R if it is measurable
and if
11fl < The INTEGRAL is then defined by setting
Jf
=
Ju+
- Ju
+
iJv -
iJv,
(which has a meaning, because u+,u_,v+,v_ are majorised by lfl).
The Lebesgue integral possesses Properties (1) and (5) of the preceding Section; it also possesses Property (4) when f and g are real, as well as Property (6) with A e T.
The sum, and the
pointwise maximum and minimum of a finite number of integrable
CHAPTER 0: OUTLINE
8
functions are integrable.
If f is integrable, then
ff1
The two essential properties of the Lebesgue integral are the following:
THEOREM: (Lebesgue's Dominated Convergence): Let (f ) be a sequence n of integrable functions that converges to f almost everywhere. If there exists a measurable positve arithmetic function g such that
lfnl 5 g for all n,
and Jg < W,
then f is integrable, and
Jf=
lim n
Jfn.
THEOREM: (Term by Term Integration of Series of Functions): If (un) is a sequence of integrable functions such that
I
< JUj
n
then the series u(x) = I un(x) n
is absolutely convergent for almost all x, the function u (which is defined almost everywhere) is integrable, and
Ju
=
n
fUn
OF THE COURSE
9
If f is integrable and if E is a measurable set, we again set
JE =
JLEf
(the integral of f over E).
In fact this integral depends only
upon the values of f on E and can be defined whenever f itself is only defined on the set E; it suffices, for example, that the function obtained by extending f by zero outside E be integrable over]R".
In this case one says that f is INTEGRABLE OVER E.
All
the properties of the Lebesgue integral over3Ru extend to this Furthermore, if f is integrable over E then it is integrable
case.
over every measurable set contained in E, and if E is the disjoint
union of a sequence (En) of measurable sets,
f
fE
L JE f. = n n
Similarly, if E is the union of an increasing sequence (En) of measurable sets,
J f = l nimfE f. E
n
This formula is still valid if E is the intersection of a decreasing sequence (E ) of measurable sets and if f is integrable n over E1.
In the case of integration on ]R, it is convention to write
when I is an interval with endpoints a and b(-o 4 a 4 b 4+-), and f is either a measurable positive arithmetic function on I, or a complex integrable function on this interval.
When a > b
10
CHAPTER 0: OUTLINE
we write
when f is integrable over [a,b], and a "Chasles' Formula" can then If -- < a < b < +-, then f is Lebesgue-integrable
be written.
over [a,b] whenever it is Riemann-integrable over this interval, and the two integrals are equal.
However, when the interval is
infinite the Lebesgue integral only constitutes an extension of the notion of absolutely convergent (generalized) Riemann integral. The GENERALIZED (or SEMI-CONVERGENT) LEBESGUE INTEGRALS can be defined in the following way.
For example, let us assume that f
is (Lebesgue) integrable over every interval [0,M], 0 : M < one then sets
f J_
M M}oj0
f
when this limit exists.
In this respect let us note the follow-
ing Proposition:
PROPOSITION: (Second Mean Value Formula): If f is a decreasing positive function on [a,b], and g an integrable function on this interval, then
IJbfgl : f(a)
a
sup IJxgl. a,
a
To close this Section let us indicate that if f is a mapping from3RP into3Rq of which the q coordinates are integrable, the integral of f is the element 0f]
of which each component is
equal to the integral of the corresponding component of f.
Every-
thing that has been said above remains valid when the absolute
OF THE COURSE
11
value is replaced by a norm on U
0.4
.
FUBINI'S THEOREM
THEOREM: (Fubini) : Let X = Iltp, y = ]R jjXxf (x,y)dxdy
then the formula
fXdxjf (x,y)dy =
=
dyfXf(x,y)dx fy
is valid in each of the following two cases: (1): f is a measurable positive arithmetic function on X x Y; (2): f is an integrable function over X x Y.
Amongst other things the validity of the formula means (according to the case) that for almost all x e X the function y Fa f(x,y)
is measurable (resp., integrable) on Y, and that the function x '
J f(x,y)dy, defined almost everywhere on X, is measurable
X (resp., integrable) on X.
In particular, in order to have the rule of "interchangeability of the order of integration" it suffices to be assured, when f is a measurable complex function on Xx Y, that
JJ1If(x,y)ldy
<
Whenever f is positive and measurable this interchange is always legitimate.
Certain proofs of Fubini's Theorem use the following Lemma, which is interesting in its own right:
LEMMA: In order that a set have measure zero it is necessary and sufficient that there exist an increasing sequence (qn) of positive step functions such that
sup 9n n
<
lima (x) n
n
if x e E.
12
CHAPTER 0: OUTLINE In this statement one can replace "step functions" by "compact-
ly supported continuous functions".
CHANGE OF VARIABLES
0.5
Let V be an open set of ]R
and u a diffeomorphism of V onto
u(V), that is to say a bijection of V onto u(V) such that u and u-1 are continuously differentiable.
The JACOBIAN of u is denoted
J(u).
PROPOSITION: The formula
Ju(V)
f(x)dx = J f(u( x))IJ(x)ldx V
is valid in each of the following two cases: (1): f is a measurable positive arithmetic function on u(V); (2): f is an integrable function over V.
Amongst other things the validity of the formula means that fou is measurable (resp. (fou)IJI is integrable) on V.
Spherical Coordinates in ]R
e with the half-hyperplane defined by x1 < 0, x2 = 0 removed, possesses a proper parametric representation:
x1 = rsinen_2
sin82sino1coscp,
x2 = rsinOn_2
sinO2sin81sin9,
x 3 = rsin n_2
sin8 2 cos 1,
x4 = rsin4n_2
coso2,
xn = rcos8n_2,
OF THE COURSE
13
where r > 0, 0 < of < it,
ITI < R.
The formula for the change of
variables in this case comes down to replacing the xi by their expression as a function of r, the oils, and cp, and dx = dx1dx2...
dxn by rn-lsinn-ton-2
... sin2o2sinoldrdSn-2...de2doldq.
Let us also recall that the volume of the ball x2 +
+X
2
n
<, 1
is
n/2 v_
n
r 2 + 1
n
1
0.6
THE LP SPACES In what follows we make the convention of setting 1/co = 0,
a.- = 0 or - according as a = 0 or a > 0, and coo = m if a > 0. HOLDER'S INEQUALITY: Let 1 < p,q < - be such that p1
+q =1,
and let f and g be measurable positive arithmetic functions on
31n; then:
Jfg ` [JfPJ l/p [Jgq/ 1/q MINKOWSKI'S INEQUALITY: Let 1 S p < c, and let f and g be measurable positive arithmetic functions on ]R", then
If (f + g)PJ 1/p
[JgPJ'/P.
[JfPJ <
1/p
+
When p = q = 2, Holder's Inequality is known as the (CAUCHY-
CHAPTER 0: OUTLINE
14
SCHWARZ INEQUALITY.
For 0 < p < 1 one still has an inequality
which is obtained by replacing 4 with : in Minkowski's Inequality (it is sometimes called MINKOWSKI'S SECOND INEQUALITY). In order to have equality in Hblder's Inequality it is neces-
sary and sufficient that there exist a 3 0 such that fP = agq almost everywhere.
For Minkowski's Inequality when 1 < p < -, equal-
ity is only obtained if f = ag almost everywhere. NOTATION: For every measurable function f on IItn one sets
llfll
where 0
= (Jlfll1/p
Also, there exists M e-
such that lfl 4 M almost everywhere,
The element M
and such that this does not hold for any M' < M.
is called the ESSENTIAL SUPREMUM Of lfl and one defines
lifilm = ess sup If I
.
DEFINITION: With the preceding notations, for 0 < p << - one de-
fines the following functional spaces:
Lp(]Rn)
{ f: f is measurable on IItn; ll f llp < -).
These spaces are vector subspaces of the space of functions
on] . When 1 : p
the mapping f i+ llflip is a semi-norm on LP (where
we write LP for LP(]R )); the kernel of this semi-norm coincides
with the vector subspace N of functions zero almost everywhere, so that LP/N is canonically provided with a norm.
In this work we
shall make not distinction between Lp and LP/N, which amounts to identifying two functions that are equal almost everywhere.
Let
us point out that certain authors denote by XP that which we denote by LP, and they keep the latter notation for .P/N.
Having
taken into account the convention we have just indicated, in the
OF THE COURSE
15
remainder we shall consider f ' If 11
as a norm on LP (1,
THEOREM: Let 1 5 p 5 00 and let (un) be a sequence of elements of LP such that
G iiunilp<
n
Then for almost all x the series
u(x) = I u(x) n
is absolutely convergent; the function u thus defined almost everywhere belongs to LP and one has
in the sense of convergence in the norm of L. COROLLARY: For 1 < p 5 00 the spaces LP are complete.
if f
n
-> f in LP there exists a sub-sequence fn
Furthermore,
such that f
n
.
-r f
almost everywhere. It will be noted that if fn -> f in L , then fn -> f almost everywhere.
f in LP, 1
This property may fail if fn -
When f e LP, g e Lq, 1 : p,q <
integrable and
Jfgl
6
IIfIIpIIgIIq
= 1, the function fg is
+
p
p < 00.
Q
(Holder's Inequality).
The spaces LP (1 S p 5 00) are BANACH SPACES; L2 is even a Hil-
bert space if the scalar product is defined by
=
Jf0
2(fig)
f,g e L.
16
CHAPTER 0: OUTLINE For every measurable set E of Rn one similarly defines the
spaces Lp(E) by everywhere replacing the integrals taken over Rn by integrals over E.
The space Lp(E) is identified with the
closed vector subspace of Lp(IZn) formed of the functions that vanish outside E.
When meas(E) < W one has the inclusions
Lq(E) C Lp(E) if 0 < p < q < -. DENSITY THEOREM NO. 1: Let E be a measurable set of Rn; then the
characteristic functions of the measurable sets A C E are total
forLp(E) for 1SpSW. DENSITY THEOREM NO. 2: Let V be an open set of Rn, then the characteristic functions of the rectangles P such that P C V are total in LP(V) for 1 4 p <
The set of continuous functions compactly
supported in V are dense in Lp(V), 1 5 p < Let us recall that a set A of a metric space E is said to be DENSE in E if A = E; if E is a normed space, A is said to be TOTAL in E if the linear combinations of elements of A are dense in E.
0.7
CONVOLUTION
DEFINITION: Two measurable functions f,g on Rn are said to be CONVOLVAELE if the function y H f(x - y)g(y) is integrable for almost all x; in this latter case one can define almost everyWhere the CONVOLUTION PRODUCT OF TWO FUNCTIONS f and g by
(f*g)(x) = Jf(x - y)g(y)dy.
We have:
f*g = g*f. If f is convolvable with g a n d h, then it is with Ag + uh (A,1,
e ¢), and
OF THE COURSE
17
f*(Ag + uh) = A(f*g) '+ u(f*h). If IfI
1, fl,
IgI
1, gl, f and g are convolvable whenever fl and
5l are; from this it follows that if f = u + iv, g = r7+ is (u,v, r,s real), f and g are convolvable if and only if each of the functions u+,u_,v+,v_ is convolvable with each of the functions r+,r_,s+,s_.
Whenever f and g are convolvable, and are zero respectively outside measurable sets A and B, f*g is zero outside A + B.
In
particular, if f and g are compactly supported, f*g is also compactly supported. In addition to the spaces Lp (i.e., Lp(]R )) introduced in the
preceding Section it is useful to define the following functional spaces:
DEFINITION: If 1 < p < m we define LPoc to be the vector space of measurable functions f such that for every compact set K of Rn one has [JKIf1i'/p
HAIp,k =
<
_-
The functions belonging to LPoc are said to be LOCALLY p-INTEGRABLE. If p
Lloc is the space of LOCALLY BOUNDED MEASURABLE FUNCTIONS,
that is to say that for every compact set K one has
Ilfi,
K = ess sup If (X) I
<
xeK
One can be satisfied with looking at compact sets of the type KN = {x:Ixl <, N}, where N 3 1 is an integer and x -> IxI a norm on
RR,
We then set
IIfIIp,K
=
n
and the formula
IIfIIp,N,
18
CHAPTER 0: OUTLINE
IIf - gIIP,N
W
d(f,g) =
2N=1 1
1 +
If - gIIp,N
defines a metric on LPoc (with the usual condition of identifying two functions; which are equal almost everywhere).
This metric is
compatible with the vector space structure on LPoc, fi ; f if and only if
If - fiIIP,K
0 for every compact set K, and LPoc is com-
plete in this metric.
DEFINITION: The space of p-INTEGRABLE COMPACTLY SUPPORTED MEASURABLE FUNCTIONS is written LP for 1 < p 4
it is a vector sub-
We say that fi - f in LP if fi I fin LP and if,
space of LP.
furthermore, there exists A > 0 such that for all i one has fi(x) = 0 whenever IxI > A.
This latter condition is expressed by say-
ing that the fi have their SUPPORT contained in a fixed compact set.
DEFINITION: The SPACE OF k-FOLD CONTINUOUSLY DIFFERENTIABLE FUNCn TIONS on ]R is written Ek (0 s k s co). (For k = 0, E0 is defined to be the space of continuous functions on 3zn; in this case we
may also write C instead of E0). For every n-tuple s = (s1,...,sn) of integers greater than or equal to zero one sets
Isl = s1 + .. + sn,
and we define
Ds
D
f=
ISI
,f
feEk,
IsI
<<
k.
axs i...axnn '
1
If 0 S k < -, for every integer N 3 1 and f e Ek we set:
pN(f) = sup{IDsf(x)I:Isj : k,lxl 4 N),
OF THE COURSE
19
and ifk= -, pN(f) = sup{IDsf(x)I:Isj < N,jxj 4< N}.
With these notations the formula W
pN(f - g)
-N 2-
d(f,g) =
1 +
N11
pN
f -
g
defines a metric on Ek compatible with the vector space structure, and for which Ek is complete.
Furthermore, fi - f in Ek if for s
every s such that Isl < k one has limDsf. = D f uniformly on every
compact set of ]R' . DEFINITION: The vector space C is the vector SPACE OF BOUNDED CONTINUOUS FUNCTIONS on ]R
provided with the NORM
IIfL = suplf(x)I It is a Banach space.
DEFINITION: The closed vector subspace of CW formed of UNIFORMLY CONTINUOUS BOUNDED FUNCTIONS on ]R' is denoted UC`°.
DEFINITION: The space of CONTINUOUS FUNCTIONS of ]R" WHICH TEND TO ZERO AT INFINITY is denoted CO.
It is a closed vector subspace
ofUC . DEFINITION: The vector subspace of Ek consisting of COMPACTLY SUPPORTED k-FOLD CONTINUOUSLY DIFFERENTIABLE FUNCTIONS is written Dk (0 5 k
co).
If k = 0 the,space D0, that is to say the space
of compactly supported continuous functions, is also written K. We shall say that fi -> f in Dk if for all s (Isl s k) one has
limDsfi = Dsf uniformly on ]n and if the fi have their support i
contained in a fixed compact set.
20
CHAPTER 0: OUTLINE If E and F are two of the spaces that have just been defined
we shall write E - F if E C F and if fi - f in E implies that
fi -> f in F.
It is clear that E - F and F -> G implies E -* G.
We have the following diagram, where 0 < k < - and 1 < p <
E' --a
Fk
-p C - + L
loc
-' LPloc - Lbe
T C
t LW
Lp
L1
D"->Dk--* K ->LWc
c
c
UCW
t
It will be noted that D
-> E for any space E from the table
above, and also E -> Lloc for every E; note also that the Lp spaces
are not mutually comparable. NOTATION: For every function f and every a e Rn, we define the
TRANSLATION OF f BY a by f.(x) = f(x - a).
Sometimes the notation Taf is used to designate fa.
If f and g
are convolvable, then
(f*g)a = fa,:g = ff:ga. DEFINITION: If E is one of the function spaces defined above, one says that E is INVARIANT UNDER TRANSLATION if f e E implies that
fa e E for every a eRn. Furthermore, if ai -> a implies that f,'. -f I in E, one says that the TRANSLATIONS OPERATE CONTINUOUSLY ON E.
OF THE COURSE
21
Cm,UC, CO,Ek and
THEOREM: (1): The spaces LP,LlOC,LC (1 5 p are invariant under translation;
Dk (0 5 k
(ii): The translations operate continuously on all these spaces,
except upon LW,L
loc
,L and Cam. c
NOTATION-DEFINITION: If F,GH are three of the function spaces
defined above, the notation F::G C H expresses that if f e F, g e G, then f and g are convolvable and f*g e H,
Furthermore, if fixgi -)-
fs:g in H whenever fi + f in F and gi -*-g in G, one writes F::G C H (continuously).
In the case where F = G = H = A it is said that
A is a CONVOLUTION ALGEBRA.
Lastly, if A is a convolution algebra
and A*E C E, one says that A OPERATES IN E; A is said to OPERATE CONTINUOUSLY IN E if A^E C E continuously.
THEOREM: (i): L1*L C UCW (continuously); furthermore,
IIffgII < (ii): Lp*Lq C C0 (continiously) if 1 < p,q < -, p + q = 1; furthermore,
IIf*gL s IIfIIpIIgIIq; (iii): L1s:Lp C LP (continuously) if 1 , p , m; furthermore,
11f* g11
, IIfII1IIgIIp.
In particular, L1
(iv): L
1
is a convolution algebra
operates continuously in Lp (1 4 p ,
(v). Dk*Lloc C Ek (continuously), 0 < k
(vi): L1 it a convolution algebra;
oo),Cm,UC
and CO;
CHAPTER 0: OUTLINE
22
(vii): L1 operates continuously in Lioc,L0 (1 5 p
W) Ek and
k D;
(viii): Dk is a convolution algebra.
Lastly, we have the formula
DS(f*g) = ff:Dsg,
REMARK: Since D -> L1 3 L1 (continuously), the convolution alge-
bras D and L1 operate continuously in all the function spaces which have previously been defined.
NOTATION-PROPOSITION: For every function f defined on1Rn we set: f(x) = f(-x),
.'(x) = f(-x).
When f and g are convoZvable, so are fand I (resp. f and g), and
(f*g)" = f*.,
(f g)- = f"g.
Furthermore : = 1, we set
NOTATION-PROPOSITION: If g e LP, h e Lq, p + 4
(g,h) = Jgh = (g*)(O) (g1h) = Jgh = (g*h)(O).
When f eL1, geLP, heLq, p + q = 1,
(f*g,h) - (g,f*h),
(f,eg1h) = (gl?*h).
OF THE COURSE
23
NOTATION: We denote by LPQ1r) the set of measurable functions on iR
that have period 2nand are such that
If 11p =
[
p
J2nIfIP)1/P 0
IIfII,, = ess sup I f(x) I
< Co,
05x62n
The set of k-fold continuously differentiable functions on ]R with period 2n is denoted Ek(W) (here 0 < k 4 Co, and E0(i') is
also denoted C(ur)). DEFINITION: The CONVOLUTION PRODUCT OF TWO MEASURABLE FUNCTIONS f and g WITH PERIOD 21E is defined by the formula: 2n
f(x - y)g(y)dy.
(f*g)(x) = 2nJ 0
Defining fi -> f in Ek(IF) to mean that f(S) f(S) uniformly for every integer s with 0 4 s < k, the following hold: LP(a)&rLq(a) C C(a)
(continuously), if
L1(a)*LP(a) C LP(Ir)
(continuously),
L'(T.')*E (a) C Ek(a)
(continuously),
= 1,
+
p
4
as do the inequalities:
IIf* IIC 5 IIfIIpIIgIIq, if feLP(a), geLcl(r), p +
= 1, q
IIf*IIp -1 IIfIIlIIgIIp,
if feL'(l), geLP(a).
By defining J`,J as above, and
24
CHAPTER 0: OUTLINE 21
(f,9) = 2nJ
2n f9,
f9
(.fig) = f1J 0
0
one obtains the same formulae as above.
Let us note that in the
formulae defining f*g,(f,g) and (fig), one can replace the range of integration (0,2n) by any interval of length 2n.
Clearly one can consider functions having an arbitrary period T > 0, it is then just a matter of replacing it by T/2 everywhere.
0.8
REGULARISATION OF FUNCTIONS
DEFINITION: One calls an approximate identity in L1 every sequence (rpi) of integrable functions that satisfies the following conditions:
(i): There exists a constant M such that 11,Pi11l < M for all i; (ii): lim J'Pi = 1; i
(iii): For every a > 0,
limJ i
m = 0, JxJ>,a i
An approximate identity (ml) is said to be compact if all the functions cp. vanish outside the same compact set of]Rn. i
In L1 there exist compact approximate identities consisting of
functions belonging to V ; these are called REGULARISING SEQUENCES. THEOREM: Let (rp
be an approximate identity in L1.
If E is one
of the spaces LP (1 < k < .o) or UC , then f o r every function f e E
one has 9 i *f -> f in E. If the approximate identity (W.) is compact this property extends to the spaces LPoc,Lp (1 < p < o), Ek,Vk (0 < k From this one deduces the following corollaries:
THEOREM: (Density); D
!1 < p < c ), Ek, Vk.
is dense in each of the spaces Lp;LPoc, LP c
OF THE COURSE
25
LEMMA: (Calculus of Variations): If f e L10c is such that Jf9 = 0 for any rpe D , then f = 0 almost everywhere.
DEFINITION: An approximate identity in L1(1r) is a sequence (p.) of integrable functions with period 27t, such that
(i):II(piII1,M; (
(ii) : lim 2l -71 i = 1; (iii): For all a, 0 < a < n, liml
i
p.(x)dx = 0.
a
If E denotes one of the spaces LP('T)
(1 < p < m) or Ek(a),
(0 , k < m), then Ti*f -> f in E for every function f e E.
In
L'(T.') there exist approximate identities consisting of functions
belonging to Em(u); from this it follows that Em(r) is dense in each of the spaces LP(,r), 1 < p < m and Ek(a).
0.9
FOURIER TRANSFORMATIONS
DEFINITION: For every f e L1 (= L1(R)) we give the name FOURIER
TRANSFORM of f to the function
E f(y)
tm
e-2%ixy f(x)dx,
=- J-W
y e3R.
The symbol f is also employed to designate Ff.
F f(y) _ (tme21Eixyf(x)dx,
y e 3R,
m
is called the ADJOINT FOURIER TRANSFORM of f.
Upon setting
The function
26
CHAPTER 0: OUTLINE
e (x) =
e2itiax
a eat, x eat.
a
one obtains the following Formulas, where f and g belong to L1:
(0 9.1)
f'':ea = ?(a)ea$
F(f) = Tf-,
F(f) = F(f),
(0 9.2)
F(.f) = (Ff)-,
F(f) = (Ff)-,
(0.9.3)
F(f*g) _ Ff.Fg,
F(f*g) _ f. Fg.,
(0.9.4)
F(fa)
F(fa) = eaF(f),
(0.9.5)
F(eaf) _ (ff)_a,
(0.9.6)
e-aF(f),
F(eaf) = (Ff)a,
(Fflg) = (flpg),
(0.9.7)
f::g = F(Ff.g).
(0.9.8)
Furthermore, if f is piecewise continuously differentiable, and if f and f' belong to L1, then f'(y) = 2uiyf(y).
Whenever f and x } g(x) = xf(x) belong to L1, f is continuously differentiable and
g(y) =
2ni
f' (y).
THEOREM: (Riemann-Lebesgue): F is a linear mapping of L1 into C0
such that
I j Ff l l
11f111-
THEOREM: (Fourier's Inversion): If f and Ff belong to L1, then f = F(Ff) almost everywhere. almost everywhere.
In particular, if Ff = 0, then f = 0
OF THE COURSE
27
THEOREM: (Fourier-Plancherel) : If f e L1 f1 L2, then
11 Ff 11
2=
11f112-
It follows from the Fourier and Fourier-Plancherel Theorems that the restriction of F and F to L1f1L2 extend in an unique way into two isometries of L2 onto itself, which are inverses of each other; these will still be denoted F and F.
If f e L2 and
f = Ff, then M
f (Y)
JM
e-2nixy f(x)dx,
M
f(x) = lim
e
2nixy-f(y)db,
M}-m-M
in the sense of convergence in L2.
Formulas (2),(3),(5),(6),(7)
are still valid for f,g e L2; Formula (4) generalises to the case where f e L1 and g e L2; and Formula (8) can be replaced by
F(fg) = F f*Fg whenever f,g e
L2.
Lastly, let us note that if f e L1 and Ff e L2,
then f e L2 0.10
INTEGRATION AND DIFFERENTIATION
NOTATION: We provide]R with the norm lxl
-- Max(lx1l,...,lxnl).
By the letter C we denote a CUBE containing the origin, that is to say, a set such that 0 e C and C = {x:!x - xol 4 a}, where x0 ein and a > 0.
The number 2a is called the diameter of the
cube C and it will be denoted by 6(C). THEOREM: (Lebesgue): Let f e L1 There exists a negZigeable loc set N of 3Rn such that if x $ N then for every complex number a one
has Zi?n 1 C mess 6( )#O
J Clf(x
+ t) - aldt = If(x) - a
l
CHAPTER 0: OUTLINE
28
In particular, for almost all x,
f(x) =
lim
d(C)+0 mess C
J
f(x + t)dt. C
DEFINITION: A function f is a FUNCTION Of BOUNDED VARIATION ON AN INTERVAL [a,b] if n-1 V(f;a,b) = sup A
If(xi+1) - f(xi)I <
E
i=0
the supremum being taken over all decompositions A
_ (a = x0 < x1< ... < xn = b)
of the interval [a,b].
If f is of bounded variation on [a,b] and on [b,c], then it is on [a,c], and V(f;a,c) = V(f;a,b) + V(f;b,c).
THEOREM: (Jordan): A real function is of bounded variation on [a,b] if and only if i-t is the difference of two increasing functions on [a,b].
A complex function is of bounded variation if its real and imaginary parts are.
Every function f with bounded variation is
regular, and upon setting V(x) = V(f;a,x), a << x
b, one has
V(x + 0) - V(x) = If(x + 0) - f(x)I, V(x) - V(x - 0) = If(x) - f(x - 0)I) at every point where these symbols have a meaning.
Therefore, in
Jordan's Theorem, if f is real, continuous, and of bounded vari-
OF THE COURSE
29
ation on [a,b], then on this interval it is the difference of two continuous increasing functions.
Jordan's Theorem extends to
real functions defined on an open interval I and with bounded variation on every compact interval contained in I. DEFINITION: A function f is said to be a FUNCTION WITH BOUNDED VARIATION on]R if V(f;O,x) and V(f;-x,O) have finite limits when
x
the sum of these limits is called the VARIATION OF THE
FUNCTION f on u . THEOREM: (Lebesgue): Every function with bounded variation on [a,b] is differentiable almost everywhere, its derivative is integrable, and b
a
If'(x)Idx
V(f;a,b).
DEFINITION: A function f is called ABSOLUTELY CONTINUOUS on [a,b] if, for all c > 0, there exists 6 > 0 such that for every finite sequence of mutually disjoint sub-intervals ]ai,si[ of [a,b] one has
i
If(si) - f(ai)I < e
whenever
i
(R. - ai) < 6.
THEOREM: (Lebesgue): If f is absolutely continuous on [a,b] f is differentiable almost everywhere, f' is integrable and
b
f(b) - f(a) = J.f'(x)dx, a
b V(f;a,b) = J If'(x)Idx. a
Conversely, if F is integrable on [a,b], and if
f(x) = JF(t)dt, x a
a < x < b,
CHAPTER 0: OUTLINE
30
then f is absolutely continuous and f' = F almost everywhere. The theory of differentiation makes use of the two following Lemmas, which are interesting in their own right. LEMMA (1): Let K be a compact set of atn covered by a family of open cubes.
From this family there can be chosen a finite sequence
C1.$---.$Cp of mutually disjoint cubes such that
meas(K) < 3n
meas(Cp). k=1
LEMMA (2): (The Setting Sun Lemma): Let f be a real continuous function on [a,b]., E the set of points x of this interval for
which there exists a y such that x < y < b and f(x) < f(y).
Then
the set E is the disjoint union of a sequence of intervals with end points an < bn such that f(an)
0.11
f(bn)
TRIGONOMETRIC SERIES If (un)nea is a sequence of complex numbers indexed by some
positive or negative integers, one sets
W
+W
-
un = u0 +
I
N (un + u_n) = lim
E
un,
N- n=-N
n=1
when this last limit exists.
luni < - one also has
When I00
u
n
=lim N-
M E
n=-N
u.n
M-
A trigonometric series is a formal series of the type
OF THE COURSE
tm c e n
31
m
a
inx
=
2
(ancosnx + bnsinnx),
+
n=1
where, upon agreeing to set b0 = 0,
an = cn + c-n, b
n
C
= i(c n - c -n ),
- ib ), n = Y(a n n
n > 0.
c-n = '(an + ibn),
In what follows it is assumed that f e L1(7r) (L1(w) has been
defined in Section 0.7).
For every n e 2z we set
(2n e-inxf(x)dx.
?(n) = I
0
The ?(n) are called the FOURIER COEFFICIENTS of f, and the formal series
inx
is the FOURIER SERIES of f.
We use the notation
L oneinx
f(x) ti
-W
to indicate that the second member is the Fourier series of f, that is to say that cn = ?(n) for every n ez.
It will be noted
that:
12n r
an
f(x)cosnxdx,
n 0
bn =
1 n
2n
f(x)sinnxdx.
On setting en(x) = einx we have the following FORMULAS
32
CHAPTER 0: OUTLINE
.'(n) _ (f l en), fsaen = f(n)en,
f(n) = ?(n)
and f(n) = ?(-n),
f*g(n) = f(n)g(n),
P (n) = in(n)
if f is absolutely continuous.
For the explicit calculation of the Fourier series of a function f the following result, which generalises (5) in the Formulas above, is useful:
PROPOSITION: If f is a pieeewise continuously differentiable function, that is to say, if there exist points.
-u 4 al < a2 < ... < a p
it
<
such that f coincides on each open interval ]as,as+l[' 1 5 s S p and ap+l = al + 21E, with the restriction of a function fs contin-
uously differentiable on the closed interval [as,as+1]' upon denoting by [f'] a function with period 27[ that coincides with f'
s
on each ]as'as+1[, one has:
ina
{f(as + o) - f(as - o)}e
in?(n) = [f'](n) + 2n s=1
EXAMPLE 1.
f(x)
If f(x) = x when -n < x <
,, 2
n=1
(-1)n+1 sinnx n
it,
S.
OF THE COURSE EXAMPLE 2.
33
If f(x) = -1 when -n < x < 0 and f(x) = 1 when 0 < x
< it,
f(x) 4 4
sin(2n + 1)x
2n+1
nn=
EXAMPLE 3.
If f(x) = it - Ixi when -n
f(x)ti'-`+4 2 it
x < it,
cos(2n + 1)x n=0
On + 1)2
THEOREM: (Riemann-Lebesgue):
lim ?(n) = 0.
Ini-'°° THEOREM: (Fourier): If f(n) = 0 for all n ea, then f = 0 almost everywhere.
If f is p-fold continuously differentiable, f(n) = o(Inl-P); if f has bounded variation on [0,2n], f(n) = 0(1/Inj).
Converse-
ly, if for an integer p one has I InLPIf(n)I < =, then f coincides almost everywhere with a p-fold continuously differentiable function.
NOTATION: We set:
N SN(f;x) =
I
f(n)einx
(FOURIER SUMS),
n=-N
aN (f x)= 3
s0(f;x) + ... + SN(f;x)
N+1
(FEJER SUMS),
then:
SN(f;x) = (f*DN)(x), with
aN(f;x) = (f'N)(x),
34
CHAPTER 0: OUTLINE
DN (x) = sin(N + Z)x
(DIRICHLET'S KERNEL),
sin Zx
FN(x)
1(sisN
(FEJER'S KERNEL).
+ 1)jx inrx
N +
THEOREM: (Localisation): If f(x + 0) and f(x - 0) exist at a point
x, we set
cpx(t) = f(x + t) + f(x - t) - f(x + 0) - f(x - 0). For every 0
d < it one has:
sN(f;x) - '-z(f(x + 0) + f(x - 0))
1
d
(X(t)
sin(Nt+ zt
dt + r, (X,6).,
IO
with lime N(x,5) = 0. N-
(0.11.6)
Furthermore, if f is continuous on I = ]a,b[, the convergence in (0.11.6) is uniform for x belonging to a compact set contained in I.
THEOREM: (Jordan-Dirichlet): If f is of bounded variation on every compact interval contained in an open interval I, its Fourier series converges at every point x of I to 2(f(x + 0) + f(x - 0)). Furthermore, if f is continuous on I, its Fourier series converges uniformly towards f on every compact set of I. If f is of bounded variation on [0,2n] there exists a constant A such that IsN(f;x)I < A for every integer N >, 0 and for all
x e]R.
OF THE COURSE
35
THEOREM: If one of the functions f or g is of bounded variation
on [0,27E], then:
1r2n 2n
f(x)g(x)dx =
1
f(n)g(-n)
n=-W
0
.(n)(2neinxg(x)dx.
=
E
n=-0
0
In other words: In order to integrate f with respect to g(x)dx one can integrate its Fourier series term by term. THEOREM: (Fejer): If f(x + 0) and f(x - 0) exist at a point x, then lima N(f;x) = j(f(x + 0) + f(x - 0)).
N Whenever f is continuous on a compact interval, one has aN(f;x)-> f(x) uniformly on this interval.
The Fejer kernels form an approximate identity, so that if
f e LP(T), 1 < p <
then aN(f) -> f in LP(a).
THEOREM: (Fejer-Lebesgue): 1imoN(f;x) = f(x)
(0.11.7)
at every point x such that
lim LJhlf(x + t) - f(x)ldt = 0. 0 h+0
In particular, Equation (0.11.7) is true for almost all x. THEOREM: (Plancherel): In order that f e L2('T) it is necessary and
sufficient that
CHAPTER 0: OUTLINE OF THE COURSE
36
E
If(n)I2
<
and in this case 27E
if(x)I2dx
2n0
=
L
I?(n)I2.
NOTE
In the statements of the exercises in the following chapter,
the functions considered are always complex valued when no further indication is given.
Similarly, unless otherwise indicated, the
sets considered are measurable sets of ]R
.
CHAPTER 1
Measurable Sets
Let E1,...,E n be a finite sequence of sets of fin-
EXERCISE 1.1: ite measure.
For every integer p (1 . p _< n) set
meas(E. P
(a):
Z1<
< p
1P
11
Show that n
meas(E1 U
U En) =
(-1)p-1aP
(POINCARE'S FORMULA).
1
p=1
(b):
For every integer s (1 s s < n) we denote by GS the
set of points which belong to exactly s of the sets E1,...,En Show that
n
meas(G) =
I
(
(-1)P-5
p=s
(c):
l
l s la l
11
.
p
Let HS be the set of points which belong to at least
s of the sets E1,-..,En.
Show that
37
CHAPTER 1:
38
n
(_1)p-s(s-11
I
meas(HS) =
P=S
Qp.
l
A0t = VA V = AVA = VAV = AV For every non-empty subset A of {l,...,n} let us set
SOLUTION:
EA = EA - U Ei.
EA = n Ei,
i+A
ieA
The EA are mutually disjoint and it is clear that
E
A
= U EB', BDA
(-1)po
(_1)CardAmeas(EA).
I
=
CardA=p
P
One has:
SOLUTION: (a):
n (-1)pa
p=1
(-1)CardA meas(EA) _
=
p
(-1)CardA
(_1)CardA.
I meas(EB) B
ACB
Note that if p = CardB, then
(_1)CardA
(_1) r(P 1
=
x
r=1
ACB
l
=
- 1.
J
From this it follows that
n (-1)Pa
p=1
meas(EB)
= p
B
meas(E1 U
L
BD A
A
A
UEn.
meas(EB)
MEASURABLE SETS One has:
SOLUTION: (b): n
39
(
l
(-1)p 8Ja p=s
p
l
(-1)CardA(CarsdA lmeas(EA)
=
I
CardA3s
J
(-1)CardA(CardA 1
Card4 s
s
l meas(EB)
E
CardB;s
I
meas(E')
BDA
J
B
(-1)CardA(CardA)
ACB
J
ll
CardA=s On setting CardB = p , s again, one has:
C
ACB
( -1)CardA [ CardA) l s J
( -1)r (r 1 (p )
=
ls)lr
r=S
CardA3s
=
l
s ) I ( -1)r (r - s ) P-8
(-1)s
s)
(-1)rlprs1
0
ifp> s,
(-1)S
if p = a.
From this it results that: n
SOLUTION: (c):
s
meas(E') = (-1)smeas(G ).
= (-1)S
(-l)PI p)Q
p=s
,
CardB=s
s
s
The formula is true for s = 1 (for this is none
other than that obtained in (a)). HSt1 = HS - GS,
Also,
CHAPTER 1:
40
whence, proceeding by induction and taking account of (b),
n
(-i)P-s-1 j(
meas(Hs+l) = I
l
lsJ
p=s n
-1 [P-111ap
(-1)P-s-1(p $ 1 l
p=s+1
l
op.
J
For all c > 0 construct an open set U everywhere
EXERCISE 1.2:
dense in R, and such that meas(U) < e.
ovo = vov - ovo = vov = AVA
SOLUTION:
Let (rn) be the sequence of rational numbers, and for
all n let us denote by In the open interval with centre rn and length e2-n.
The union U of the In is an everywhere dense open
set of 3R (for it contains all the rational numbers) and furthermore,
meas(U) S
e2-n
1
= e.
n=1
EXERCISE 1.3:
Let (En) be a sequence of measurable sets such
that
E meas(E n
<
Show that the set of points which belong to an infinity of En's has measure zero (The Borel-Cantelli Lemma). AVA = VAV = AVA = VAV = ova
FIRST SOLUTION:
The set considered is
MEASURABLE SETS
41
A n [UEJ p mead U E (pan
]
I
-<
p
pan
meas(E ), p
and consequently meas(A) = lim mead U Epn
J
(pan lim
I
n pan
meas(E ) = 0. P
Let cpn be the characteristic function of En.
SECOND SOLUTION:
By virtue of the theorem on the term by term integration of series of positive functions, one has
i cn = I
Jcpn = I meas(En) < -.
J
From this it follows that the set of points where
Pn = W has
measure zero.
But this set if precisely A.
EXERCISE 1.4:
Let (En) be a sequence of measurable sets such
that
G
meas(En) <
n For every integer s, H
is denoted as the set of points which s
belong to at least s of the sets En.
meas(Hs) 4
G meas(En). n s
Show that
CHAPTER 1:
42
First, suppose we had a finite sequence E1,...,
FIRST SOLUTION:
En and for every non-empty subset A of (1,...,n) let us denote by EA the set of points belonging to the Ei for which i e A and not
belonging to those for which i4 A (in Exercise 1).
smeas(Hs) =
Then
smeas(EA)
E
CardA>,s
n
CardAmeas(E') _ A
CardA>,1
meas(E.). i=1
1
In the general case one considers, for n > s, the sets Hs n formed by the points which belong to at least s of the sets E1,.. -,nIt is clear that Hs
n C Hs n+l
and that
Hs = U Hs,n.
n=s Consequently
meas(H ) = Jim meas(H ) n s,n S
n Jim
<<
n
meas(E.).
meas(E.) = s i=1
i=1
s
SECOND SOLUTION: m = I pn.
Let
q)
be the characteristic function of En and
One has
meas(En) =
smeas(H).
Jp > JH S
EXERCISE 1.5:
sets one sets
For every finite sequence E1,...,En of measurable
MEASURABLE SETS
43
n
n
l
D (E1,...,En) =
E1l
Eil -
I F]
,
J
and if the E. are not all negligeable,
meas(D(E1,...,En) a(E1,...,En) =
meas
Show that if none of the E. is negligeable one has
a(E1,...,E1) E
n
1
1
1
i
MMA = VM0 = A VA = VAV = MMA
SOLUTION:
Let us show, by induction on n, that if x e D(E1,...,En)
then x e D(E1,E)'for at least n - 1 pairs i < j. if n = 2.
x e E2 U
This is clear
Let us assume that it be true up to n - 1 and that
One then has x e D(Ei,Ei) for at least
UEn, x e El.
n - 2 pairs such that 2 < i < j 4 n and for at least one pair (l,i) with i Z 2.
From this it follows, by Exercise 4, that
meas(D(E1,...,En)
n
1
L
meas(D(Ei,Ej)).
i<
meas(E1U UE) 3 meas(EiUEj ), from this one deduces that
a(E1,"''
C
1
n
meas(D(E.,E
n - 1 i
1
E
i
UEn
44
CHAPTER 1:
EXERCISE 1.6:
Consider a sequence (En) of measurable sets such
that
mead( U E n) < n and inf meas(E ) = a > 0. n n
Show that the set A of points that belong to an infinity of sets E
n
is measurable and that meas(A) > a.
ovo = vov = ovo = vov = AV
SOLUTION:
We have
A=ni(UEsl n=1 s=n which shows that A is measurable.
Now we also have
m
mead U
ES]
. meas(En) > a.
s=n
As the sets UU ES form a decreasing sequence and have finite sin measure, we have
meas(A) = lim measi U ES1 > a. n s=n
REMARK: The first condition is essential; for example, consider the sequence En = [n,n + 1[.
EXERCISE 1.7:
Let A be the set of real numbers x for which there
MEASURABLE SETS
45
exists an infinite number of pairs (p,q) of integers such that
q;i, 1and P
x
q
Show that A is negligeable.
4V4 = VAV = AVA = VAV =-000
Since
SOLUTION:
x + n -ng =x -q , by setting
B = An [0,1] from the former relation one deduces that
A = U (B + n). n=-w Hence it suffices to prove that B is negligeable. For p and q integers, and q
Ip,q
=
LP-
q
-
1q3 P+ 1J ' q q3
and let us note that x e I
p,q
la let us set
' is equivalent to
(1) qx-2
q
Since in an interval of length 2/q2 there can be only at most one or three integers according as q >. 2 or q = 1, from this it
CHAPTER 1:
46
results that x e B if and only if x belongs to an infinite number of sets
Bq = [0,1[n (U
Ip,q).
p By Exercise 1.3 it therefore suffices to show that
(2)
G
meas(B ) <
q
q
Now by virtue of Equation (1) above, the integers for which Ip,gfl[0,l[ + O are such that
-2. p. q + 21 1
q
q
When q > 2 this is equivalent to
0
q.
Consequently
1) meas(Bq ) < 2(q + 3 q
which proves Equation (2) above.
EXERCISE 1.8:
Let (D ) be a sequence of discs in the plane, of n unit diameter and mutually disjoint. The number of discs that
are contained within the disc with center 0 and radius n will be denoted v(n).
Show that if
liminf v(2) = a > 0 n n
MEASURABLE SETS
47
there exists a half-line issuing from 0 that meets an infinite number of these discs D
n
AVA = V AV = AVL = V AV - AVA
SOLUTION:
Let us consider the cones Cn with vertex 0 generated
by the Dn, and let An be the intersection of Cn with the circle with center 0 and unit radius.
For some integer p and q, 1< p < q,
let A be the union of the An corresponding to the discs Dn conP,q tained in the disc with center 0 and radius q and not contained
in that of radius p.
Furthermore, let us set
U>p A P,q
BP
Everything reduces to proving that the B , which form a deP
creasing sequence, have a non-empty intersection.
The union of
the radii of the disc with center 0 and radius q that meet A P,q covers (v(q) - v(p)) discs of unit diameter and mutually disjoint. From this it follows that if meas(A ) denotes the Lebesgue meaP,q sure (evaluate in radians) of A the unit circle, one has p ,q ,q
2 iq meas(AP,q)
Since A
p,q
C A p,gt1
it
(v(q) - v(p)).
,
one therefore has
) meas(B ) = lim meas(A p,q P q
2 liminf q
v(q)
q2 v(p)
From this it follows that
measl n BP) = lim meas(BP) l p
P
2
2
CHAPTER 1:
48
which proves that the intersection of the Bp is non-empty.
EXERCISE 1.9:
n +
Let A be the set of real numbers of the form
p E at-p,
p=1 where n ea and ap = 0 or 1. (a):
Show that A is negligeable.
(b):
From this deduce that there exist two negligeable sets,
the sum of which is fit. evo = VtV = AVA = VAV = eve SOLUTION: (a):
Let B = Afl[0,1[.
Then
A = fg, (B + n). n=-w Hence it suffices to show that B is negligeable.
Let u be the
function, with unit period, equal to zero on [0,2'[ and to unity on [12,1[.
For every integer p >. 1 the number
ap(x) = u(2p-1 x)
is the p-th term in the binary development of x e [0,1[.
Whatever may be the numbers el,...,ep (ei = 0 or 1), the set of x's such that 0 4 x < 1 and al(x) = ell
has measure 2-P.
ap(x) = cp,
From this it follows that the set Bp of the x's
such that 0 4 X < 1 and
MEASURABLE SETS
49
a1(x) = 0,
a3(x) = 0,
. ,
a 2p-1 (x) = 0,
has as its measure 2-2p+1 x
2p-1
= 2-p.
Since B is the intersection of the Bp one certainly has meas(B) =0.
SOLUTION: (b):
The set A' = 2B is also negligeable.
Now A' is
the set of numbers of the type
Y
a 2-(2p+1)
p
p=0
a p = 0 or 1.
As every real number may be written
n+
ap=0or1,
ap2p,
p
it follows that ]R = A + A'.
EXERCISE 1.10:
Let f be a complex measurable function on ]R such
that f(x + 1) = f(x) for almost all x.
Show that there exists a
function g such that f = g almost everywhere and g(x + 1) = g(x) for all x.
AVA = DA4 = ADA = DAD = ADA
Let E = {x,f(x) + f(x + 1)} and let us set
SOLUTION:
+co
(E t n),
F= n=-co
as well as
CHAPTER 1:
50
- f(x) if xeF. It is clear that g has all the properties desired.
EXERCISE 1.11:
Let A be a bounded set of ]R
Sip = meas{x: I Ix II
and let
r 1},
where lixll denotes the Euclidian norm of x. Let (xi)i>.1 be a sequence of points of A; we set
do = inf{Ilxi - xi II1 1 4 i < ,j 4 n}. Prove that
liminfndP n n <
1 meas(A)
ap
Qp
where (
p
1
cp = 2 -P{ 1 + p10 ( 1 + t l
dt}
For this one will consider a number y such that ndP > y for n n sufficiently large, as well as the balls B. with center xi and radius ri, where
if 1i
Y1/Pn - 1/P r. _ i11/P(2i-1/P
-
n-1/p)
if n < i E 2Pn.
DOA = DAO = ADA = DAV = ADA
MEASURABLE SETS SOLUTION:
51
It will be noted that the ri form a decreasing sequence.
If 1< i < j N< n one has
11xi - xj II y do > y1/Pn - 1/P whenever n is large enough, and
yl/pn-1/P.
ri t rj =
n, one has
When 1 4 i < j E 2pn and j
- 1/p, IIxi - xi II > dj > yl/Pj and we also have
ri + r.
yl/Pn- 1/P
zyl/P(2j-1/p
+
- n-1/P)
=
yl/Pj -1/p.
From this it follows that the balls Bi are mutually disjoint. If one sets A(e) = {x:d(x,A) .< e), one will then have
urn - 1 2
LL
n
Pn (2i-1/P + 2C
- n 1/P)Pl
< meas(A(e )), n
J
i==n
where
e
n
=
2yl/Pn - 1/p
We have 2Pn 1 (2i-'/P -
lnm
i=n
n-1/p)P
Pn [2rnl -1/p - 1]p = 2C = lim n i==n
Il
J
(Contd)
CHAPTER 1:
52
r2P (2,-l /P - 1)Pdx
(Contd) JI
1
=
(1 - t)P
1tt
p1
dt,
(the last integral is obtained by setting x = (1 t t)P).
On the
other hand, the sets A(En) are decreasing and have A as intersection; furthermore, as A is bounded they are of finite measure. From this it follows that
meas(A) 1
c
St
p
P
and consequently that
liminfndp 4
n
n
EXERCISE 1.12: = 1.
1 meas(A)
c
P
SZ
P
Let X be a measurable set of ]RS such that meas(X)
Let u be a bijection of X onto itself such,that for every
subset E of X, E is measurable if and only if u(E) is measurable, and then meas(E) = meas(u(E)).
Furthermore, assume that if N is
a measurable subset of X and if u(x)e N for almost all the points of N, then N or X - N is negligeable. Let E be a measurable subset of X such that meas(E) > 0, and
if x e X if uP(x)$ E for all p > 1, inf(p:p >, 1,uP(x)e E}
(one sets u1 = u, uPt1 = u0up).
Show that
otherwise
MEASURABLE SETS
53
1 f n(x)dx=1. E
A00 = 0A0 = A0A = VAV = A00
SOLUTION:
E
n
For 1 .< n 6 W let us set
= {x:x e E and n(x) = n}.
Let us also set GO = E,
n
u_P(E),
Gn = E - U
n . 1,
p=1
where u -p = (U-1 )P).
It is clear that
14n
En=Gn-1Gn, E
= o GG.
For n > 1 one has n-1 un(Gn)
=
un(E)
- U
up(E).
P=O From this it results that the sets un(G ), n 3 0, are mutually n
disjoint, and that
y = l_J un(G ) n=0
= lJ un(E). n=0
Since u(y) C -y and meas(y) > meas(E) > 0, the second hypothesis
54
CHAPTER 1:
made on u implies that meas(y) = 1; in other words, since u preserves the measure, m
Go
1 =
meas(G.). E E measun(G ) = n=O n=0 n n
Let us note that the Gn are decreasing; consequently meas(E ) = lim meas(G ) = 0.
n
n
Thus one has
N n(x)dx =
E
n=1
nmeas(E ) = lim n
N
X
n=1
N
(N-1
= liml I
N- In=O
(n + 1)meas(G )
n
rN-1
-
I
n=O
l
nmeas(G )J
l
= liml I meas(Gn) - Nmeas(GN)J N n=O 1
It remains to be observed that
meas(Gn) = 1
L
and
meas(Gn) 3 meas(Gn+1)
implies Nmeas(GN) -} 0.
Thus
I n(x)dx = 1. J
meas(G )} n
n{meas(G n-1
E
REMARK: We have used the following classical result:
n
MEASURABLE SETS If un
55
> un+l >
0 and G Un < m, then nun -> 0.
We shall briefly recall the proof: let c > 0 and let p be such that
(n - p)un S up+1 + ... + Un S E
for all n
p; then
liminfnun s E. QED n
EXERCISE 1.13:
One says that a set A of]Rp is ALMOST OPEN if
almost all the points of A are interior points of A. Let f be a real function defined on an open set U of Iltp.
Prove that the following conditions are equivalent: (a): f is continuous at almost all the points of U;
(b) :
For all a e]R the sets (f > a) and (f < a) are almost open sets. AVA = VAV = AVA = V0V = AVA
SOLUTION: (a):
Let E be a negligeable set in U such that f is
continuous at every point of -U - E.
If x e (f > a) - E one' will
have f(y) > a for all the points y of a neighbourhood of x, therefore x is interior to (f > a).
This set is therefore almost open.
One argues similarly for (f < a). (b):
f is continuous at x if for every rational number
r < f(x), x is interior to (f > r), and if, for every rational number s > f(x) x is interior to (f > s).
If r is rational, let
us denote by Ar (reap. Br) the set of points of (f > r) (reap. (f < r)) not interior to this set.
If these sets are neglige-
able then so is their union, which, by the preceding, contains the set of points of discontinuity of f.
56
CHAPTER 1:
EXERCISE 1.14:
One says that a bounded real function f defined
on ]R is ALMOST EVERYWHERE CONTINUOUS if the set of its points of discontinuity is negligeable. (a):
Give an example of a function that is almost everywhere
continuous and such that there exists no continuous function coinciding with it almost everywhere. (b):
Show that in order for a bounded real function f to be
almost everywhere equal to an almost everywhere continuous function, it is necessary and sufficient that there exists a set A of R such that ]R - A is negligeable, and that the restriction of f to A is continuous. (c):
Deduce from (b) that f is measurable and that there ex-
ists a sequence of continuous functions fn which is convergent at every point of R and whose limit is almost everywhere equal to f. (d):
Show that a right-continuous function is continuous
except at the points of a set that is at most denumerable, and therefore is almost everywhere continuous. A0A = V AV = A0A = 0M4 = AVI
SOLUTION: (a):
If f(x) = 0 for x < 0 and f(x) = 1 for x > 0, f
cannot coincide almost everywhere with a continuous function g, for with the complement of a negligeable set being everywhere dense in R, one would be able to find two sequences xi < 0 < yi tending to zero, and such that g(xi) = f(xi) = 0,
g(yi) = f(yi) = 1.
On passing to the limit one would have g(0) = 0 and g(O) = 1 at the same time, which is absurd. SOLUTION: (b):
The condition is evidently necessary.
show that it is sufficient.
To do that let us set:
Let us
MEASURABLE SETS
57
g(x) = Iim{sup(f(y):y e A,Iy - xI < a)}. a-*0 a>0
This definition has.a meaning, for A is everywhere dense. one has f = g.
On A
Furthermore, if x e A and if c > 0 there exists
a > 0 such that f(x) - c .< f(y) . f(x) + c if yeA and I y -xI < a. Then if ix - x '
l y - x' I
l
< a one has
l y - x l
< a' = a - Ix - x' I
< a for all y e ll such that
.
From this it follows that g(x) - e .< g(x') .< g(x) + c,
which proves that g is continuous at each point of A. SOLUTION: (c):
For every x and all c > 0 there exists a > 0 such
that f(y) .< g(x) + c if y e A, ly - xI < a.
As above, from this one deduces that
g(x') . g(x) + e if Ix' - xl < a.
The function g is therefore bounded and upper
semi-continuous.
There then exists (cf., a course on Topology)
a sequence fn of continuous functions that converges everywhere towards g.
SOLUTION: (d):
Let us assume that f is right-continuous.
If A
is a non-empty set of 3t we shall denote by e(A) the diameter of A, that is to say the upper bound of the numbers la - bl for a e A,
b e A.
For all x e]R let us then set w(x) = inf{8(f(V)):V a neighbourhood of x},
58
CHAPTER 1:
(one says that w(x) is the OSCILLATION OF f AT x).
The set of
points of discontinuity of f is then:
{w> 0} = U JW >n} n=l
JJJJ
I t therefore suffices to prove that for all a > 0 the set A = {w > a} is denumerable.
By reason of the right-continuity of f,
for all x e A there exists a
x
> 0 such that
a(f(]x,x + ax[)) < a. But then:
]x,x+ax[(nA=0. Let us choose a rational number rx in ]x,x + ax [ .
If y e A and
x < y, one therefore has
rx < y < r y , which proves that x e A - rx is an injection of A into Q, and con sequently that A is denumerable.
EXERCISE 1.15: < -.
Let A be a measurable set of ]R such that meas(A)
Show that the function x y meas(A(1]-co,X]) is continuous.
AVA = V AV = AVA = V AV - AVl
SOLUTION:
If xn is a decreasing sequence, and tends to x, one
has
A(1]-co,x] = n {An]-.,x n]}, n
and consequently
MEASURABLE SETS
59
meas(Afl]--,x]) = 1im meas(Afl]-m,xn]) n If xn is a strictly increasing sequence that converges towards x then
Ar)]-.,x[ = U {Afl]-W,xn]}, n and consequently, because meas({x}) = 0, meas(A r)]-W,x]) = meas(Af)]-o,x[) = lim meas(A r)]-m,x ]),
n
n
which proves that the function x 1+ meas(Afl]-oo,x]) is continuous.
EXERCISE 1.16:
Let 0 < A < 1.
For any measurable sets A,B C[0,1]
of positive measure, do there exist 0 < x < y < 1 such that
The answer is affirmative if and only if A = 1/n, where
SOLUTION:
n = 2,3,...
First of all let us assume that A = 1/n.
.
Since the
function f(x) = meas(A r) [O,x])
is continuous and increasing on [0,1], there exist points 0 =t t1 <
< to = 1 such that
f(ts) =
meas(A).
n
< 0
60
CHAPTER 1:
Under these conditions one has:
meas(A), meas(A 0 [ts,ts+1 ]) = n
0 .<
s F n - 1.
Furthermore, the n numbers meas(B n [ts,ts+1]) have meas(B) as
Therefore they cannot all be strictly less than or all
sum.
strictly greater than (1/n)meas(B).
Therefore there exists an s
(1 < s 6 n - 1) such that one of the numbers meas(BI)[ts-l' ts]) and meas(Bf) [ts'ts+l]) is less than (1/n)meas(B) and the other is greater.
The idea of the proof is to 'vary x continuously' from ts_1 to is and y from is to ts+l in such a way that f(y) - f(x) = meas(Af)[x,y]) always remains equal to (1/n)meas(A).
meas(B n [x,y] ) will then vary continuously between two values on
opposite sides of (1/n)meas(B), and must therefore take this value at least once.
The rigorous proof reduces to proving that in the rectangle is-1 <
x < ts,
is 6 y < ts+l the set E of points such that
meas(A)
f(y) = f(x) +
n is connected.
Let us note that if ts_1 < x < is one has
f(ts) = f(ts_1) +
meas(A) s f(x) +
meas(A)
n n
E f(x) + f(ts+1) - f(ts)
E f(ts+1). From this it follows that meas(A)}
EX = {y:ts 4 y 4 ts+l'f(y) = f(x) + n
MEASURABLE SETS
61
is a non-empty compact interval, and that E is the union of the Ex's (if EX is identified with {x} x Ex).
Note that E is compact;
let us assume that it is the disjoint union of two non-empty comHaving seen that EX is compact and connected,
pact sets E1,E2.
one has either EX C E1 or EX C E2.
In other words, the projec-
tions of E1 and E2 onto the x-axis are two non-empty compacts sets that are disjoint and have the union [ts_l,ts] which is absurd.
Let us now assume that A is not of the above form; then there exists an integer n such that
Let a,$ > 0 be two numbers such that
(n+1)a+ns= 1, and let us decompose [0,1] into 2n + 1 contiguous intervals, al-
ternatively of length a and B, the two extreme intervals having length a.
Let A be the union of intervals of length a, and B
that of intervals of length S.
If
meas(Afl [x,y]) = Ameas(A), then [x,y] contains at least one of the intervals of length S, as otherwise one would have
meas(Afl[x,y]) < a =
n + 1
meas(A) < Ameas(A).
But then,
meas(Bf)[x,y]) >, B =
meas(B) > Ameas(B). n
EXERCISE 1.17:
Let I be a compact interval of 3R such that meas(I)
> 0 and 0 < S < 1.
We shall say that the operation T(s) is carried
62
CHAPTER 1:
out on I if one subtracts from I the open interval having the same centre as I and of length Smeas(I).
More generally, if I is
a disjoint union of a finite number of compact intervals of nonzero lengths, to apply T(S) to I consists in carrying out this operation on each of the intervals forming I.
Now let (0n) be a sequence of real numbers 0 < Sn < 1; we shall denote by In the compact set obtained by successively carrying out the operations T(S1),T(S2),...,T(Sn)
starting from the interval
[0,1].
Show that
(a):
In+1 C In,
meas(In)
2)...(1 -
1)(1 -
n
and that every interval contained in In has a length less than 2-n
From this deduce that
(b):
K
R In n
is compact, non-empty, nowhere dense, has no isolated point, and that
meas(K) = lim(l n (c):
1
S) n
)(1 - a
Assume that Sn = 1/3 for all n.
Show that meas(K) = 0
and that X e[0,1] belongs to K if and only if it can be written in base three uniquely using only the digits 0 and 2.
From this
deduce that K is not countable.
(d):
If Bn = 1 -
al/n(n+l),
0 < a < 1, show that meas(X) = a
(which proves the existence in [0,1] of nowhere dense compact sets whose measure is arbitrarily close to 1).
MEASURABLE SETS (e):
63
Deduce from part (d) the existence in [0,1] of a se-
quence An of sets of the first category that are mutually disjoint and such that: (i): meas(An) = 2-n n
(ii): Kn = U A
is a nowhere dense compact set;
i=1
(iii): Every interval contiguous with n (that is to say, every connected component of the complement of Kn in [0,1]) contains a set in An+1 of measure greater than zero. From this dedua
A
that
U An n=1
is of first category, has measure one, and that its complement in [0,1] is a set of second category of measure zero.
(f) :
Let
E nU-0 A2nt1'
Show that for every interval I contained in [0,1] and of nonzero length there holds 0 < meas(E()I) < meas(I).
(g):
From part (f) above deduce the existence of Borel sets
E C IR such that for every interval I of non-zero length one has 0 < meas(E(lI) < meas(I). (h):
Can one have meas(E) < m for such sets?
Deduce from the preceding that there exist positive
functions that are Lebesgue integrable, but that are not limits almost everywhere of increasing sequences of positive step functions.
CHAPTER 1:
64
SOLUTION: (a):
It is clear that
In+l C In
and
meas(In) _ (1 - S1)...(1 - On).
Moreover, n is formed by 2n mutually disjoint intervals of equal lengths; when the latter property. SOLUTION: (b):
The set K is compact and non-empty by virtue of a
well known theorem in Topology.
If I is an interval contained in
K one has I C I for all n; by Question (a) one thus has meas(I) < 2-n and consequently meas(I) = 0.
In other words the interior
of K is empty, which is the definition of a nowhere dense compact set.
If x e K and e > 0, for large enough n one of the intervals
forming In will be contained in ]x - e,x + e[; for this it is sufficient that 2-n < e.
Now, the two endpoints of this interval
belong to K, which shows that x is not an isolated point of K. Lastly,
meas(K)= lim meas(In) = lim (1 -
Sn).
n
SOLUTION: (c):
In this case, one has: ( ln
meas(K) = liml3J
n
= 0.
Furthermore, I1 is equal to the set of x's which are written in base three as
x = O.ala2...an...'
with a1 = 0 or 2 (it will be noted that 1/3 = 0.0222 and that
1 = 0.222 ).
Similarly it is seen that x e In if and only if
ai = 0 or 2 for 1 i i < n.
From this one deduces that x e K if for
all i ai = 0 or 2; the expansion of x in this form is then unique. If with every set A C iN one associates xA =
0
MEASURABLE SETS
65
if i eA and ai = 2 if i4 A, a bijection between P(v) and K is realised; K is therefore not denumerable. a1/n(n+l),
SOLUTION: (d):
If Bn = 1 -
0 < a < 1, one has
a
meas(K) = lima n, n
with
1 _
1
n+1
(n
whence meas(K) = a. SOLUTION: (e):
By the preceeding it is seen that in every non-
empty open interval I there exists a nowhere dense compact set whose measure is imeas(I).
The let Al be a nowhere dense compact
set of [0,1] such that meas(A1) = z.
In each interval contiguous
to Al let us choose a nowhere dense compact set the measure of which is half that of this interval, and let us denote by A2 the union of these compact sets; A2 is of first category (for the set of intervals contiguous to a compact set is denumerable) and its measure is 1.
Furthermore, K2 = Al U A2 is closed; in fact,
if x is a limit point of K2 and does not belong to Al it belongs to an interval I contiguous to A
1
and is therefore a limit point
of IflA2, which is compact, whence x e A2.
On the other hand, it
is clear that [0,1] - K2 is dense in [0,1] - K1, which itself is dense in [0,1], which proves that K2 is nowhere dense.
Quite
generally, An+1 will be constructed by choosing in every interval contiguous to the compact set K
n
a nowhere dense compact set of
CHAPTER 1:
66
measure equal to half the length of this interval, and by taking the union of these compact sets. is of first category and that Kn+l
As above, it is seen that An+1 = Kn U An+1 is
Further-
closed.
more,
n
2-i
meas(Kn) _
= 1 - 2-n, i=1
whence:
z[1 - (1 - 2-n)] = 2-(n+1)
meas(An+l)
Finally, Condition (iii) is satisfied by construction.
A
If
U An n=1
this set is of first category, for it is the countable union of sets of first category, and
-
W
meas(A) =
2-n = 1.
1
n=1
Its complement is not of first category, by a theor m of Baire, and its measure is zero. SOLUTION: (f):
E
n
Let
A2n+1'
F
i A2n' l
If I is an interval of length greater than zero-contained in [0,1], its intersection with E or F is of measure greater than zero, since meas(E U F) = 1.
meas(If1E) > 0.
Let us assume, for example, that
Then ICE contains at least two points x < y.
Let n be such that these two points belong to Al U A2 U
U A2n+1'
MEASURABLE SETS
67
They therefore belong to K2n+1, and as this set is compact and nowhere dense there exists an interval J contained in [x,y] which is contiguous to it; one then has meas(If)F) > meas(JflA2n+2) > 0.
Since
meas(I) = meas(IfE) + meas(If)F), it follows from this.that meas(Ef)I)
< meas(I).
If E is the set studied in Question (f) above,
SOLUTION: (g): then
E= U (E + n) nea
answers the question.
By going back to the proof of Question (f)
again one sees that for E one can take the set
EN
N
n
A2n+1'
If
E = U (EINI + N), ne2Z
then for every interval I of positive length one has: 0 < meas(If1E) < meas(I), and furthermore, W
m
m
2-(2n+1) = 10
2-(2n+1) + 2
meas(E) _ n=0
N=1 n=N
9
CHAPTER 1:
68
SOLUTION: (h):
Let f be the characteristic function of the set
E defined above, and let cp be a positive step function such that rp 4 f almost everywhere.
If I is an interval of length greater
than zero on which 9 is equal to a constant, since meas(I - E) > 0 this constant is zero.
Thus cp = 0 almost everywhere.
From this
it follows in particular that f cannot be the limit almost everywhere of an increasing sequence of step functions. REMARK:
To prove the existence of a nowhere dense compact set of
[0,1], the measure of which is arbitrarily close to unity, one may also consider the set E = [0,1] - Q.
There holds
1 = meas(E) = sup{meas(K):K compact, K C E}, and every compact set contained in E is evidently nowhere dense.
EXERCISE 1.18:
Consider a double sequence (fm n) of measurable
complex functions on X = [0,1] such that for all m the sequence converges almost everywhere towards a function gm and
(fm
that the sequence (gm) converges almost everywhere towards a function h.
Show that there exist two sequences of strictly increasing inconverges n s' S Generalise this result to the case where
tegers (ms) and (ns) such that the sequence (fm almost everywhere to h.
X =P. (orEP).
wA=vw=w0=0AV =MA SOLUTION: set E C X.
First assume that every convergence is uniform on a Then there exists ml < m2 <
such that on E there holds
h - g ms Thus
<1s
- fmsons
gmS
and n1 < n2 <
and
MEASURABLE SETS
Ih-
fm
s
,n
69
I
<
on E,
s
s
n -
which proves that fm '
h on E.
s
We are going to show that for all e > 0 there exists such a set E, which, moreover, is measurable and satisfies meas(X - E) < S.
In fact, by Egoroff's Theorem there exist measurable sets
EO,E12... such that
meas(X - E ) m
<
e2-(m+1)
> 0,
and, on the other hand, one has limgm = h
uniformly on E.,
Iimfm,n
uniformly on En if m > 1.
= gm
On setting
E=nm m=O
the desired result is obtained.
Let us now use an argument known as the "diagonal process". By applying the preceding result to e = 1 one first determines a measurable set E1 and two strictly increasing mappings T1,-5 1 of
N* into itself, such that meas(X - E1) < 1 and
f91(n),O1(n) ' h
on E1.
Now applying the same result to e = Z and the double sequence
f
(m) 0 (n) one obtains a measurable set E2 and two strictly in1
creasing mappings 92102 of 1V* into itself such that meas(X - E2) <
I and
CHAPTER 1:
70
f1Plo(P 2(n),91082(n) -
h
on E2.
Proceeding thus repeatedly, one obtains finally a sequence (Er) of measurable sets and two sequences (pr) and (0r) of strictly increasing mappings of]N* into itself, such that meas(X - E ) < r 1/r and
fY10...oq) r(n),O1o...o8r(n) -> h
on Er.
Let us then set
E = U r, r mS = 910...op (a),
ns = 010...o0s(a).
S
The set E is measurable and meas(X - E) = 0. every r 3 1 the sequence (fm
Furthermore, for
)s-r is a subsequence of the se--
s,n
s
From this it follows
quence that
fm in s s
-> h
on E.
It will be noted that in order to use Egoroff's Theorem it suffices to assume that X C 1R
and meas(X) < -.
By writing ]R
as
the union of a sequence of such sets, a new application of the diagonal process allows this result to be generalised to the case
where X = ]R
.
EXERCISE 1.19:
sets of]R set K of ]R
A mapping t -- F(t) ofIR into the set FP of closed
is called a MEASURABLE MAPPING if for every compact the set {t:F(t)f1K 4 01 is measurable.
MEASURABLE SETS
71
(a):
Show that {t:F(t) = O} is measurable.
(b):
Show that t -; F(t) is measurable if and only if
{t:F(t)f1 B * 0} is measurable for every open ball B of ]RP, or
again if t - F(t)1K is measurable for every compact set K of ]R Show that if t -
(c):
.
F1(t) and t - F2(t) are measurable
mappings of]R into the closed sets of)R
and]Rq respectively,
then t - F1(t)x F2(t) is measurable. Show that if t - K(t) is a measurable mapping of ]R into
(d):
the compact sets of ]R , and if f is a continuous mapping of IlRP into 3R q, then t - f(K(t)) is measurable.
Show that if t -> Fn(t) are measurable, then
(e):
t - n Fn(t) n is also measurable. (f):
Show that t - F(t) and t -
of Et into the sets of ]R
K(t) are measurable mappings
which are respectively closed and compact,
then t -; F(t) + K(t) is measurable.
1V = VV =Ova=V V =AVA SOLUTION: (a): n.
Let Bn be the closed ball with center 0 and radius
We have: Co
{t:F(t) = O} =]R - U {t:F(t)f1Bn * 0}. n=1
SOLUTION: (b):
If t -; F(t) is measurable and if V is an open set
of]RP, there exists a sequence of compact sets Kn of which V is the union, and:
CHAPTER 1:
72
{t:F(t) n V 4 0) = U {t:F(t) n n # 0}. n=1 If now {t:F(t)n B 4 0} is measurable for every open ball B, as every open set V is the countable union of such balls, the set {t:F(t)n V 4 0) is measurable.
If K is compact, let us consider
the open sets Vn = {x:d(x,X) < 1/n} where d(x,K) denotes the distance from x to K for a norm on]R .
One has:
{t:F(t)f1K 4 0) = U {t:F(t)f1Vn 4 o}. n=1
In fact it is clear that the first set is contained in the second; moreover, if for all n there exists xn a F(t) n Vn one can find a
subsequence x
n
such that x .
It is clear that x e K (as
-> X. .
K) = 0), and that x e F(t) also, for F(t) is
d(x,K) = linrl(x Z
n
1
closed.
If t -> F(t) is measurable and K is compact, t - F(t)n K is
evidently measurable (this is true, moreover, if K is closed). If t + F(t)n K is now measurable for every compact set K, by considering afresh the balls Bn (cf., Question (a) above), we have
{t:F(t) n K # 0} = U {t:(F(t)f1K)n Bn # 0}, n=1 which proves that t SOLUTION: (c):
F(t) is measurable.
Note that we have not specified the norm on ]R P.
If we choose the norm II(xl,...,x P of ]Rp ]R
)II
= Maxlxil, every open ball
is of the form B1 xB2, where B1 and B2 are open balls of
and itq respectively.
Then:
{t:(F1(t)xF2(t))n(B1xB2) 4 0} = n {t:Fi(t)nB. 4 O}. i=1,2
Consider first the case of two measurable mappings
t - F1(t) and t - F2(t).
Let us denote the 'diagonal' of RpxRq
For every compact set K of Rp,
by A.
{t:F1(t)nF2(t)nK = 0} = {t:(F1(t) x (F2(t)nK))nt + f}. By what has gone before, t + F1(t)x (F2(t)n K) is measurable, and on the other hand 0 is a countable union of compact sets.
that t -} F1(t)n F2(t) is measurable.
the result for a finite intersection.
This shows
By recurrence one obtains In the general case one
writes n
{t: ( n F (t))nK + y} = n {t:(n Fr(t))nK + 0}, n=1
n=1
r=1
an equality that results because the decreasing compact sets
n
n (Fr (t)nK) r=1
have a non-empty intersection-if and only if each of them is nonempty.
SOLUTION: (f):
It is known (see a course on Topology) that
F(t) + K(t) is closed if F(t) is closed and K(t) is compact. Moreover, if u denotes the mapping (x,y) -; x + y of Rp x Rq into
Rp, we have u(F(t)x K(t)) = F(t) + K(t) By the preceding, t -r F(t) x K(t) is measurable, and it is seen,
as in Question (d) above, that u(F(t)x K(t)) is also.
CHAPTER 1:
74
Let Fp be the set of closed sets of R
EXERCISE 1.20:
set of non-empty compact sets of i2P, x ->
l
ix1
1
,
KP the
a norm on ]R
and
,
for every non-empty set F of Fp d(F) = Min{ IjxII:x a F}. For F E Fp write:
(a):
cp0(F) _
{x:x eF, jjxjj = d(F)}
if F # 0,
{0}
if F = 0.
Show that t u cp0(F(t)) is measurable if t - F(t) is measurable
(for the definition of the measurability of a one-parameter family of closed sets see the preceding Exercise).
Let ei(x) =X. if x = (x1,. .. ,xp) e:IR
(b) :
.
For K e K0 set
e.(K) = Min{e.(x):x a K} and cpi(K) = {x:x a K,ei(x) = ei(K)}.
Show that t -
(K(t)) is measurable if t - K(t)E K0 is measurP
able.
From this deduce that there exists a mapping 9:F
(c):
->iRp
P
such that:
(i) : ip(F) e F if F e Fp and F $ 0; (ii): t - q(F(t)) is measurable if t y F(t) is.
Let f:jRp +R be continuous.
(d):
a 2Rq x F
->
P
Show that there exists
PP such that :
(i) : If F e F and x e f(F), then a(x,F) e F and f(a(x,F)) = x; P
(ii): If t H x(t)e]i
,
t i+ F(t)e Fp are measurable, then
t * a(x(t),F(t)) is.
(e):
Assume that the mapping t f* Fi(t)e FP (i = 1,2) and
MEASURABLE SETS
75
t -> x(t) eF1(t) + F2(t) are measurable. Show that there exist t -> xi(t)e Fi(t) that are measurable and
such that x(t) = x1(t) + x2(t) for all t. A0A = V AV = A0A = 0A0 = A0A
SOLUTION: (a):
Let K be a non-empty compact set of ItP.
The set
{t:g0(F(t))nK # 0} is equal to
{t:F(t)f1K # 0,d(F(t)) = d(F(t)f1K)}
(*)
when 0 4K, otherwise it would be necessary to add {t:F(t) = O}. The latter set is measurable (cf., Exercise 1.19(a)).
Since the
set {t:F(t)n K + 0} is measurable by definition, and as
K
is measurable it suffices to prove that t -} d(F(t)) is measurable on its defining set.
Now, for all a 3 0
{t:d(P(t)) < a) = {t:F(t)n Ba # 0},
where B
a
SOLUTION:
denotes the closed ball with centre 0 and radius a.
(b):
Similarly, on the measurable set {t:K(t)n K # 0}
one has:
{t:ei(K(t))n K # 0} = {t:ei(K(t)) = ei(K(t)f1K)},
and one sees, as above, that t - e .(K(t)) is measurable (by note
ing that {t:K(t)n F # y} is measurable for F closed). SOLUTION: (c):
1op0)(F) is reFor every F e F the set (cp P P It is clear that F -> q(F) has all the re-
duced to a point p(F). quired properties.
76
CHAPTER 1:
SOLUTION: (d):
Let us, for x ERq and F e Fp, write:
a(x,F) = p(f 1(x)t F). It is clear that Condition (i) is satisfied.
To prove that (ii)
is also satisfied it suffices to show that t -> f 1(x(t))r)F(t) is
Now, for every compact set K of Rp
measurable.
{t:f 1(x(t))r)K 4 O} = {t:x(t)e f(K)}, and f(K) is compact.
SOLUTION: (e) :
Let f 2R x]R -> Rp be defined by (x,y) -> x + y,
and let a be defined as above. Let us set a(x(t),F1(t)x F2(t)) = (x1(t),x2(t)).
Then x1(t) and x2(t) satisfy the properties required.
EXERCISE 1.21:
In this Exercise it is proposed, by consideration
of the Axiom of Choice, to prove the existence of non-measurable The Axiom of Choice appears in the following form:
sets of R.
Given a non-empty family (B ) of mutually disjoint non-empty sets i
of ]R there exists a set E of 3R which contains one and only one
point of each Bi. (a):
Show that there exists a set E C [0,1] such that for
every x eR there exists an unique y e E such that x - y is rational. (b):
Let S be the union of the sets E + r, where r runs over
the set of rational numbers lying between -1 and 1. Show that [0,1] C S C [-1,2],
and that if r,s are two distinct rational numbers, then E + r and
MEASURABLE SETS
77
E + s are disjoint. (c):
Deduce from this that E is not measurable.
ovo = vov = ovo = vov = ovo
SOLUTION: (a):
Let Q be the set of rational numbers, and if x,y
are real numbers let us express that x - y eQ by writing x ti y. This defines an equivalence relation on]R which all the equivalence classes intersect [0,1].
The existence of E follows from
this and the axiom of choice. SOLUTION: (b):
It is clear that
E + r C [0,1] + [-1,1] = [-1,2], and therefore S C [-1,2].
On the other hand, if 0 _< x s 1 there
exists y e E and re Q such that x = y + r.
1 which proves that [0,1] C S.
One has Iri = Ix - yJ
Lastly, if r # s one has (E + r)
fl (E + s) _ 0, otherwise there would exist z eIIZ and y1 a E, Y2 e E such that
z =y1+r=y2+s, and consequently y1 4 y2, which contradicts z being equivalent to a single element of E. SOLUTION: (c):
If E were measurable S would also be measurable,
and
meas(S) =
I
meas(E + r).
reQfl [-1,1] Now, all the numbers meas(E + r) are equal to meas(E), so that meas(S) = 0 if meas(E) = 0, and meas(S) = - if meas(E) > 0.
This
is absurd, because by virtue of the inclusions proved in (b) one would have to have 1 4 meas(S) 4 3.
CHAPTER 2
v-Algebras and Positive Measures
EXERCISE 2.22:
Let (X,C) be a measurable space, (xi) a sequence
of points of X, and (mi) a sequence of real numbers mi > 0.
For
every set E e C set: V(E) =
I
m..
x.eE i
(a):
Show that u is a measure on C.
(b):
Show that if {xi}e C for all i then one has C,1 = P(X),
and conversely.
AVA = VAV = AVA = VAV = AVA
SOLUTION: (a):
It is clear that u(O) = 0 (as usual, the conven-
m = 0). Let (E ) be asequence of mutuE i n ieo For every inally disjoint sets of C, and let E be their union. tion is adopted that
teger N one evidently has: N
11 (E1U...UEN) =
1 n=1
(E),
ir(En)
79
80
CHAPTER 2: a-ALGEBRAS
whence
u(En)
E
u(E).
n=1 Moreover, for every finite set A of 1N one has
E
m
X eE
= 1
i
m. 4
I
G
n=1 x.eE
n
i
1
1 n=1
u(En
ieA
ieA
whence M
u(E) = sup
X
m. <
G
1
n=1
A x ieE
U(E ) .< u(E). n
ieA
SOLUTION: (b):
A set E C X is measurable if and only if there
exist A,B e C such that A C E C B and xi $ B -A for all i.
If for
an index i one has {xi} $ C then {x.} is not u-measurable, because A can only be the empty set and so B D {xi} implies that xi e B -A On the other hand, if {xi} e C for all i and if E is an ar-
= B.
bitrary set of X, let
A = {xi:xi a E},
C = {xi:xi * E}.
Then A and B = X - C belong to C, A C E C B and {xi} $ B - A for all i.
From this it follows that E is u-measurable.
EXERCISE 2.23:
Let a be the set of positive, nul or negative
integers. (a):
Show that the set of subsets.A of a such that for
every integer n 3 1 one has 2n e A if and only if 2n + 1e A is a a-algebra.
AND POSITIVE MEASURES
81
Show that the mapping f of a into itself defined by
(b):
f(n) = n + 2 is a measurable bijection, but that f_1 is not meas-
urable. AVA = DAD = AVA = VAV = ADA
SOLUTION: (a):
It is clear that 0 possesses the property, and
that the property is preserved under complementation.
If the Ai
have the property for i = 1,2,..., it is impossible that for an n > 1 one of the two integers 2n,2n + 1 belongs to one of the A. and that the other belongs to none of them.
This shows that the
union of the Ails also possesses the property. SOLUTION: (b):
It is clear that f is a bijection.
Furthermore,
if A has the property and if n 3 1 one has 2n e fl(A) if and only if 2(n + 1)e A, hence if and only if 2(n + 1) + 1 e A, that is to say, if and only if 2n + 1 e f 1(A). urable.
This proves that f is meas-
Finally note that A = {0} has the property, but that
f(A) = {2} does not, since 2e f(A) and 3 * f(A), so f_1 is not measurable.
EXERCISE 2.24:
Let C be a family of subsets of a set X.
If
M C X, set: CM = (Mf)E;E a C}. (a):
Show that if C is a a-algebra on X, CM is a a-algebra
on M (CM is called the a-algebra INDUCED on M by C). (b):
If Me C give a simple characterisation of CM.
(c):
If C is generated by a family A of subsets of X, show
that CM is generated by AM. (d):
Deduce from part (c) that if M is a subset of a topo-
CHAPTER 2: a-ALGEBRAS
82
logical space X, the Borel a-algebra associated with the topology induced by X on M is equal to the a-algebra induced on M by the Borel a-algebra of X.
Consider in particular the case where M is
a Borel set.
AVA = VAV = AVA = VtV = tVt
SOLUTION: (a):
0 = MflO, and:
M - (MnE) = Mfl (X - E),
(*)
U (MnEn) = Mfl(U En), n n which shows that CM is a a-algebra on M.
SOLUTION: (b) :
If M e C, then
CM = {E:E e C and E C M}.
SOLUTION: (c):
Since AM C CM, C(AM) C CM.
subsets E C X such that MflEe C(AM).
Let C
be the set of 0
Evidently one has 0 e Coy
and equalities (*) and (**) show that C0 is a a-algebra on X. Since A C CO, one has C C CO, which proves that CM C C(AM).
SOLUTION: (d):
This follows from TM being the topology induced
on M by the topology T of X.
If M is a Borel set, the Borel sets of M can be interpreted either as the Borel sets of X contained in M or as the Borel sets of the topological
EXERCISE 2.25:
ability on C.
subspace M of X.
Let C be a a-algebra on X and let p be a probLet M C X be such that E e C and E D M implies
p(E) = 1. Show that a probability 11
M
is defined on CM by setting
83
AND POSITIVE MEASURES
PM(MnE) = u(E) for all E e C (the induced a-algebra CM was defined in the preceding Exercise).
Let us first prove that if E e C, F e C, and Mr) E = Mf1 F, then u(E) = u(F). Now, in this case one has SOLUTION:
(X - E) U (EnF) > M,
(X - F)U(EfF) > M, whence
1-u(E)+u(EnF)=1, 1 - u(F) + u(E n F) = 1,
and consequently u(E) = u(F). thus defined unambiguously.
Recall that the most natural method is to apply Abel's
Theorem to the series development of log(1 t x).
EXERCISE 3.34:
Show that if a 3 0 then
s i n a x
f 0x(x + 1)
dx =
(1 - e-a). 2
(Consider the function of a defined by the left hand side). AVA = VAV = 40A = VAV = OVA
FUNDAMENTAL THEOREMS SOLUTION:
95
Let cp(a) be this left hand side.
cosax I
Since
1
x2 +1
x +1
the function (p is continuously differentiable on Ft
and
x2 W cosax dx.
(p, (a) _
fo
+1
Furthermore, if a > a0 > 0, then for A > 1:
xsinax
2
Aa0
JAx2+1
(by the second mean-value formula, since x(x2 +
1)-1
is decreas-
This proves the uniform convergence of the im-
ing for x z 1).
proper integral
xsinax 1
0x2+1
on [a0,=[.
dx
Consequently 9 is twice continuously differentiable
for a > 0, and f0mxsinax fW xsinax
Ox2+1
dx
rW sinax dx + 0
x
On [0,oo[ rp is therefore a solution of the system
9(0) = 0,
9'(0) = 2
(x2+1)
dx
96
CHAPTER 3: THE
(Indeed, as 9 coincides on ]0,-[ with a solution of gyp" - cp = -n/2
and as, furthermore, it is continuous on [0,co[, it satisfies this From this it follows that
differential equation on [0,oo[). (1-e-a).
cp(a)=2
EXERCISE 3.35:
For every integer n 3 1 and all real x, let 2e-2nx
fn(x) = e- T'x -
Show that the series with the general term fn(x) is convergent for all x > 0, and calculate its sum f(x).
Next, show that each
fn, as well as f, is integrable on (0,co), and compare
ff(x)dx
and
n1
J:ffl(X)d.X.
A0A = V AV = AVA - VAV = A00
SOLUTION:
L
n=1
For x > 0 one has:
f (x)= n
e
-x
-2
1 - e -x 1
ex-1 1
ex + 1 On the other hand,
_
e
1-
-2x
e-2x
2
e2x-1
FUNDAMENTAL THEOREMS
10fn(x)dx =
97
- 2 2n = 0,
n so that:
0.
n=1 JJ 0 However, x
Jm
dx
Oex +1
ex
[log
=
e
x
+1x=0
= log2.
J
Hence one must have:
00,
n=1
0
which is immediately verified by observing that Jm
1
olfn(x)Idx=nJol1-2uldu=n
EXERCISE 3.36: (a):
Let (fn)nal be the sequence of functions de-
fined on ]R by
if 04x4n
2
n x
fn(x)
=
- n21x -
n 0
if n s< x s n,
ifx>' 2n '
Calculate the four numbers: A = lim infJJJfn ,
n
A' = Ilim inff , n n JJJ
CHAPTER 1: THE
98
B' = Juim supfn.
B = lnm supJfn,
(b):
The same question for the sequence (gn)n>0 defined by:
gn = 810,4] if n is even,
otherwise.
gn = IL11 1]
(c):
If (hn) is a sequence of positive measurable functions,
what can be said of the four numbers A,A',B,B', and more particIs the Fatou Lemma true for sequences of
ularly about B and B'?
real measurable functions with arbitrary sign?
AV = 000 = 00A = VAV = 00A
SOLUTION: (a):
For all n one has
Jfn = 1,
so that A = B = 1.
On the other hand,
lim inffn = lnm supfn = 0,
and A' = B' = 0. SOLUTION: (b):
In particular, A' < A and B' < B. In this case Jgn is alternately equal to I and 4, Furthermore,
so that A = a, B = 4. lim infgn = 0
and
lim supfn
= Il[0,1]'
whence A' = 0, B' = 1. Thus A' < A < B < B'. SOLUTION: (c):
A' < A < B
There always holds: and
A' < B'.
(The inequality A' 4 A is Fatou's Lemma, the two others are clear)
99
FUNDAMENTAL THEOREMS
The two preceding examples prove that in general nothing can be said about the relative values of B and B'. However, one can extend Fatou's Lemma in the following way: let us assume that the fn's are real and measurable, and that there exists a measurable real function g such that: for all n,
9- 4f n
jg_ < . In fact 0 .< fn - g, and
One then has A' s A.
Jfn'
- °° < Jg <
therefore we can suppose that ig < W, and in this case
J (fn - g) = Jfn - jg'
The classical Fatou Lemma shows that
Ilim inf(fn - g) = Ilnm inffn - Jg n
.< lim inf Jfn -
1g,
n whence the result.
Jg
=
(Recall that if Jg- < - one sets
Jg_
-
which is an element of
g < h,
fg-
< -
and that
imply
Similarly one shows that if:
Jh_ < -
and
Jg < Jh).
100
CHAPTER 3: THE
fn < g
for all n,
jg+ < W, then
lnm supJfn
fuim supfn.
(Consider the functions g - fn and recall that
lnm inffn.
lnm sup( - fn) Here one again sets
if
Jg = Jg+ - Jg_
Jg+ < W,
and
g < h,
< M
jh
+
EXERCISE 3.37: (a):
imply
Jg+ < -
and
Jg
'C
Jh).
Show that if f is integrable on
,
and if
K is a compact set of this space, then
lim zI I-
(b):
lf(x)ldx = 0.
J K+z
Show that if f is uniformly continuous on]Rm, and that
if there exists p > 0 such that IfIp is integrable, then lim f(x) = 0. IIxII-)W
ovo = VAV = ovo - vov = ovo
FUNDAMENTAL THEOREMS
101
SOLUTION: (a):
If(x)Idx = 0,
liml
r-* since
0
fr
and
If(x)Idx < J
x+z
IfI
If S = sup{IIyII:y a K}
(Lebesgue's Theorem).
J
Ifrl
,
If(x)Idx, IIxIIIIxII-a
whence the result.
+ 0 as IIx II e > 0 and a sequence xn such that (b) :
SOLUTION:
Assume that
If(xn)I
There there exist
>. e
As the function f is uniformly continuous there exists a closed ball B, with centre 0 and radius greater than zero, such that s
If(y) - f(x)I <
2
if
y e B + x.
In particular, IfI > e/2 on the balls B + xn, and consequently P IfIP > (2) meas(B) > 0,
J
B+x
l
n
which contradicts the first part of the Exercise.
EXERCISE 3.38:
Let G be a continuous function on 3R such that
G(0) = 0 and G(x) > 0 if x # 0. Show that if f is an uniformly continuous bounded real func-
tion on 3t and
CHAPTER 3: THE
102
mG(f(x))dx < -,
j
then
1im f(x) = 0.
IIxlIH A0, = V AV = tot = VMv = LVA
If f(x)-J 0 as x
SOLUTION:
there would exist c > 0 and a
sequence xn such that
If(xn)I z E.
IlxnII - -,
As the function f is uniformly continuous there exists a closed ball B, with centre 0 and radius greater than zero, such that
I f(y) - f(x) I< 2 if y e B + x. In particular, if M is the upper bound of IfI on e, one would
have
24IfISM on the balls B + xn.
By virtue of the assumptions made about G,
one has inffG(u):
juI < M) = u > 0,
2 so that
JG(f(x))dx 3 umeas(B), B+x
n
which would contradict that G(f) is integrable (cf.,the preceding Exercise).
FUNDAMENTAL THEOREMS
103
Let f be an integrable function on ]R
EXERCISE 3.39: (a):
.
Show that
meas(IfI > a) = o11)
(b):
as a --* W.
Show that if f is measurable (and almost everywhere
finite) on F? then it is integrable if and only if
2nmeas(2n-1
< IfI i 2n)
X
100 = VAV = MMA = VAV = MMA
This results from
SOLUTION: (a):
I f(x)Idx = 0, a-J lim lfl >a and the inequality
ameas(IfI > a) E J
If(x)Idx. Ifl
Let us set
SOLUTION: (b):
An =
>a
(2n-1
< Ifl 6 2n),
B = (f = 0),
These sets are mutually disjoint, with union ]R ure zero.
J
R
whence
Then
IfI =
T
J" If
n=-W A
n
C=(IfI ,
and C has meas-
104
CHAPTER 3: THE +m
(
2
2nmeas(An)
n=-m
1
1A s
I
2nmeas(An),
n=-m.
IR
which proves that the condition is necessary and sufficient.
EXERCISE 3.40:
Let f be an integrable positive real function on
Pp. Show that there exist measurable sets An C Rp (n ea) with finite measure such that
f(x) _
n=-
for all x e]R
2')LA (x) n
.
AVA = DAD = AV4 = 4AV = DAD
Observe that if for all t elR+ and n e2z one sets:
SOLUTION:
an(t) = IL[z,11(2-(n+1)t - [2-
where [x] denotes the integral part of x, then
2nan(t).
t
n=(Note that an(t) = 0 whenever 2n > t; the an(t) are simply the digits of the binary expansion of t).
In particular,
2nan(f(x)).
f(x) _
n=-m If one sets A urable, and
= {x:x ep
and a (f(x)) = 1), then A
n is meas-
FUNDAMENTAL THEOREMS
f(x) =
105
2)LA (x).
.1
n
n= Also,
2nmeas(An) <
so A
n
fRpf(x)dx,
has finite measure.
EXERCISE 3.41:
Let X be a measurable set of SRp such that meas(X)
= 1.
Show that if f is integrable on X then
f(x)dx = 0 JX
if and only if
JIi + zf(x)Idx
1
for every complex number z. AVA = DAD = AVA = V AV = AVA
SOLUTION:
The condition is necessary, because
+ zf(x))dx= Ii + zl fdxI = 1.
+zf(x)ldx Jxl
X
Let usnow show that it is sufficient. written in the form:
11 + peiof(x)I - 1 dx 3 0, J
X
p
X
The inequality can be
106
CHAPTER 3: THE An easy calculation shows that:
where p > 0, 8 eat.
11 + pzI p
- 1 _ 2Re(z) + pIzI2
1+pz + 1
if p > 0, z e0, so that
lim
- 1 = Re(eiof(x)).
11 + peI f(x)I p
p-)-0
Moreover,
11+ p egs f(x) I- 1 I< I f(x) I. p
Therefore, by Lebesgue's Theorem,
Re(eloJ f(x)dx)
=
JRe(f(x))dx X
x
limJ
I1 + pe10f(x)I - 1
p->0 X
dx
p
>. 0,
and this holds for all e em.
el$Jxf(x)dx = -
By choosing 0 so that
Jf(x)dxl X
one obtains
f f(x)dx = 0.
JX
EXERCISE 3.42:
Let f be an integrable function on at and let a'>0.
Show that for almost all x eat the series
FUNDAMENTAL THEOREMS
107
+W
n=- f[.+nI
(*)
is absolutely convergent, and that its sum F(x) is periodic with period a and is integrable on (O,a).
tVA = VOV = AVA = VAT - OVA
SOLUTION:
Setting u =
+ n on obtains
a
I J:If l
+
n]
=a
I
dx
n=-w
C'fxI
=a
JI
.f (x) I dx
so that the series in (*) converges absolutely at almost all points of (O,a).
As this series does not change when x is replaced by
x + a, it follows from this that it is absolutely convergent at almost all points of ]R, and that its sum coincides with a function
F of period a (setting, for example, F(x) = 0 at the points x where the series is not absolutely convergent) that is integrable on
(0,a). Since one can integrate term by term, one obtains by proceeding as above: fa
i
F(x)dx =
J(x)dx.
0
EXERCISE 3.43:
Let f be an integrable function on Ir, and let a > 0.
Show that for almost all x e 1R:
limn af(nx) = 0.
n-*-
CHAPTER 3: THE
108
DOA - VAV = X00 = VAV = 400
SOLUTION:
t
r(
(x)Idx,
so
in-CL
n=1
f(nx) I dx <
m
I
From this it follows that for almost all x eat
I
n-ajf(nx)I
<
n=1
and in particular that
n-af(nx) - 0. EXERCISE 3.44:
Let f be a measurable complex function oniR with
period T > 0 and such that:
T
I f(x)l dx < m
A= J
(a):
0
Show that for almost all x,
limn-2f(nx) = 0.
n
(b):
From this deduce that for almost all x,
limlcosnxll/n = 1.
n
FUNDAMENTAL THEOREMS
109
MVA = VAV = AVA = VAV = MVA
SOLUTION: (a):
Set
ppn(x) = 2 f(nx). n Then:
1T
T
1 nT
Jk(x)dx = 2J If(nx)Idx = 3f 0 n 0 n 0
If(x)Idx
A
= n2 Hence
T Ipn(x)Idx <
n
0
The series I Pn (x) therefore converges for almost all x, and in
particular
n(x) = n-2f(nx) -> 0
for almost all x. SOLUTION: (b):
Consider the function:
(log1cosxl)2.
f(x) _ It has period it and is integrable on [0,n], for in the neighbour-
hood of n/2 it is equivalent to (loglx - 271)2.
lim(n 1loglcosnxl)2 = 0, n i.e.,
By the preceeding,
CHAPTER 3: THE
110
limlcosnxll/n = 1 n for almost all x.
EXERCISE 3.45:
Let (xn)nal be a sequence of points of [0,1].
If
0 < a < b 4 1 and N a 1 is an integer, one denotes by v(N;a,b) the number of integers n such that 1 4 n 4 N and a S xn < b.
The se-
quence (xn)n31 is called an EQUIDISTRIBUTED SEQUENCE if, for any
0s as b5 1, lim v(N;a,b) = b - a. N
N-
Show that the following conditions are equivalent:
(a):
(1): The sequence (xn)na1 is equidistributed; (ii): For every continuous function f on [0,1] 1 J
f(x)dx = 1imN 0 N-?-
N
-1
Y
f(xn);
(*)
n=1
(iii): For every integer p a 1, -1
limN N-
N I
2nipxn e
= 0.
n=1
APPLICATIONS: Investigate whether the sequences xn = na - [na]
(a irrational)
and
x
n
= logn - [logn]
are equidistributed (here [a] denotes the integral part of a).
FUNDAMENTAL THEOREMS (b):
113
Show that if (xn)n31 is equidistributed and if f is
Riemann integrable, then (*) still holds. ovo = vov = ovo = vov = ovo
SOLUTION:
We shall prove part (b) first of all, which will also
prove the step (i) => (ii) of part (a).
We may assume that f is
Let us note that if (xn) is equidistributed then (*) holds
real.
for every characteristic function of an interval [a,b], 0 4a 4b4 1.
This formula therefore also holds, by linearity, for every step function.
Now for every e > 0 there exists a step function 9 such
that: 1
J TSe+
f5 cp,
J0
J0
f
From this one deduces that:
-1
N
l im supN
E
n=1
N
f (x n )
.<
l imN
-1
1
N
I
9_(X
n=1
N
)1 S E +
J
n
f,
0
and consequently
1 im supN
N-
t
1
N
n=1
f (Xn ) <
J
f.
0
Replacing f by -f, this yields:
N lim infN 1
N--
1
f(x ) > J f,
n=1
n
0
which shows that (*) holds for f. That (ii) => (iii) is trivial upon setting f(x) =
e2nipx
in
N. Finally, let us show that (iii) => (i).
We first show that
CHAPTER 3: THE
112 (iii) => (ii)t.
Now, if (iii) holds, then (*) is true for every
trigonometric polynomial.
If f is continuous on [0,1], and if
e > 0, there exists such a trigonometric polynomial 9 for which If - 91
