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Contenido Introducci´ on on
1
1 Longitud y Medida
3
1.1
Longitud . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Medida Exterior . . . . . . . . . . . . . . . . . . . . . . . .
12
1.3
Conjuntos Medibles
. . . . . . . . . . . . . . . . . . . . . .
13
Ejercicios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
2 Medi Medida da:: Definic Definici´ i´ on, Uso y Propiedades
23
2.1
σ -´ algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
2.2
Medidas y sus Propiedades . . . . . . . . . . . . . . . . . .
24
2.3
Funciones Medibles . . . . . . . . . . . . . . . . . . . . . . .
28
2.4
Integral de Lebesgue . . . . . . . . . . . . . . . . . . . . . .
35
2.5 Conver Convergenci genciaa Mon´ otona . . . . . . . . . . . . . . . . . . . .
38
Ejercicios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
41
3 Funciones Integrables
43
3.1
Teoremas de Convergencia . . . . . . . . . . . . . . . . . . .
45
3.2
Otros Modo odos de Convergencia . . . . . . . . . . . . . . . . .
46
Ejercicios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
1
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4 Descom Descomposic posici´ i´ on de Medidas y el Teorema de Rad´ onNiko dym 55 4.1
Medidas con Signo . . . . . . . . . . . . . . . . . . . . . . .
56
4.2
El Teorema de Radon-Nikody odym . . . . . . . . . . . . . . . .
58
Ejercicios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
62
5 Me Medidas Pro ducto y el Teorema de Fubini
63
5.1
Espacios Producto . . . . . . . . . . . . . . . . . . . . . . .
63
5.2
Medidas Producto . . . . . . . . . . . . . . . . . . . . . . .
66
5.3
El Teorema de Fubini . . . . . . . . . . . . . . . . . . . . .
69
Ejercicios . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
A De Abiertos, Abiertos, Cerrados Cerrados y Aproximaci´ Aproximaci´ on de Conjuntos Medon ib l e s 73 A.1 Aproximac Aproximaci´ i´ on por Abiertos . . . . . . . . . . . . . . . . . .
75
A.2 Aproximac Aproximaci´ i´ on por Cerrados . . . . . . . . . . . . . . . . . .
78
A.3 Aproximac Aproximaci´ i´ on por Compactos . . . . . . . . . . . . . . . . .
79
A.4 Aproximac Aproximaci´ i´ on por por Intervalos . . . . . . . . . . . . . . . . . .
80
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Introducci´ on on El presente librito se prepar´o para servir como notas del curso Teor´ eor´ıa de la Medida que se dict´o en el IMCA como parte de la Escuela EMALCA realizada en Lima entre el 18 y 29 de Febrero del 2008. En el primer cap´ cap´ıtulo de este libro se desarrolla la medida de Lebesgue en la recta. recta. La noci´ noci´on on de medida en general est´a relacionada con la comparaci´ on con algo que se tiene a bien llamar unidad. Desde hace mucho ya on sabemos medir intervalos por su longitud, que es equivalente a comparar un intervalo dado con el intervalo unitario [0, [0 , 1]. La medida de Lebesgue es una generalizaci´on on de este concepto, en el sentido que nos ense˜na na la manera mas conveniente de “aplicar” esta manera de medir a subconjuntos de la recta recta que no sean intervalos. intervalos. Lamentable Lamentablemen mente te no todos los subconjunt subconjuntos os de la recta est´an an en esta esta clase. clase. Los conjun conjuntos tos medible medibless seg´ un un Lebesgue ser´ an la mayor clase de subconjuntos que se puedan medir conservando las an propiedade propiedadess de la medida de interv intervalos. alos. trata trata el concepto concepto que se tiene tiene de medida A partir del segundo cap´ cap´ıtulo desarrollamos la noci´on on de medida como un concepto m´as as general que puede ser aplicado en una diversidad de situaciones. Se ver´a en el Cap´ Cap´ıtulo 2 que la propiedad de ser contablemente aditiva es el ingrediente principal en la definici´on on de medida y en las propiedades de la integraci´on on a la que estamos acostumbrados desde el C´alcu a lculo lo.. En el Cap´ Cap´ıtulo 3 veremos c´omo omo la noci´on on de medida e integraci´on on introducen nuevos modos de decir como una sucesi´on on de funciones se acerca a otra. Los teoremas desarrollados en este cap´ cap´ıtulo son resultados que se usan y aplican en probabilidad y en an´ alisis alisis funcional. En el Cap´ıtulo ıtulo 4 veremos algunas relaciones que pueden existir entre medidas definidas en el mismo
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2
´n introduccion o
espacio. espacio. Aqu´ Aqu´ı se desarrolla desarrolla el importante importante Teorem Teoremaa de Rad´ on-Nikodym on-Nikodym que es fundamental en Teor´ eor´ıa Erg´ odica y para otros resultados como el odica Teorema de Representaci´on on de Riesz. En el Cap´ Cap´ıtulo 5 damos un modo de construir medidas producto en espacios producto. pro ducto. Aqu´ Aqu´ı el modelo es R R y el resultado principal es el que se refiere a integrales iteradas en espacios productos. Este resultado es conocido como el Teorema de Fubini.
×
Asumimos que el lector ya est´a familiarizado con las nociones y terminolog´ nolog´ıa de conjuntos. Es decir, las operaciones op eraciones de uni´on on (“ ”), intersecci´on on (“ ”) y diferencia (“ ”) de conjuntos y con la definici´on on de complemento c de un conjunto (“ ”). Eventualmente usaremos u saremos el s´ımbolo ımbolo para enfatizar que la uni´ on on realizada se est´a haciendo entre conjuntos dos a dos disjuntos.
∩
\
∪
En el desarrollo de este trabajo usamos las notaciones m´as comunes posibles. En m´ as as de una ocasi´on on ser´a necesario considerar los reales exten¯ didos R = R , en los cuales se define las operaciones de suma y multiplicaci´ on de la siguiente forma: on
∪{−∞ ∞}
a
todo a ∈ R , ±∞±∞ = ±∞ ± ∞ = ±∞ para todo a ±∞ · 0 = 0 , a · ∞ = ∞ si a > 0. 0.
Y el resto resto de las operaciones operaciones se hace de forma usual, como lo har´ har´ıamos si fuesen elementos de R. No est´an an definidas definidas las operaciones + .
±∞ ∓∞
Un hecho hecho que simplifica simplifica las cosas cosas en los reales reales extendido extendidoss es que ah´ ah´ı, cualquier sucesi´on on mon´ otona otona es convergen convergente, te, pues puede conver convergir gir a un n´ umero umero a R, a + o´ a .
∈
∞ −∞
Finalizo agradeciendo la confianza recibida por parte del Prof. Marcelo Viana y el Comit´e Cient´ Cient´ıfico de UMALCA para la realizaci´ realizac i´on on de este curso.
Roger Metzger Alv´an an Lima, Febrero 2008
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Cap´ıtulo 1
Longitud y Medida 1.1 1.1
Longi ongitu tud d
La medida de un intervalo se denomina usualmente longitud. Est´a claro lo que ser´ ser´ıa la longitud de un intervalo intervalo acotado cualquiera, comenzamos pues longitud tud de un interv intervalo alo con extremos escribiendo esta definici´on. o n. La longi a, b es el n´ umero umero ((I ) = b a.
−
Con esto esto ya sabemos sabemos medir medir inter interv valo alos, s, y entre entre ellos a los inter interv valo aloss abiertos. Ahora, ¿cu´al al es la manera m´as as razonable de definir la longitud de la uni´ on de dos intervalos abiertos disjuntos y acotados?. Es decir ¿cu´al es on la longitud de G de G = = (a, b) (c, d) donde b donde b < c?. Naturalmen Naturalmente te se debe tener que que ((G) = b a + d c.
−
−
∪
La definici´ on anterior sirve para el caso de uni´on on on de dos intervalos abiertos disjuntos, pero ¿c´omo omo deber´ıa ıa ser la definici´ definici ´on on cuando se trate de abiertos en general? Se puede mostrar que un conjunto abierto y acotado de R se descompone de manera unica u ´ nica en una uni´on on disjunta, a lo m´as as numerable, de intervalos abiertos, vea el Lema A.0.1. Por lo tanto definimos para cualquier conjunto abierto G abierto G R, la longitud de G como
⊂
∞
(G) =
(J )
(1.1)
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4
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
donde G = n∞=1 J n es la descomposici´on on (´ unica) unica) de G de G como uni´ on on de una colecci´ on on disjunta, a lo m´as as numerable de intervalos abiertos.
∪
Una observaci´on on importante es que si G es G es un conjunto abierto acotado entonces (G) < y por lo tanto la suma en (1.1) es una serie absolutamente convergente.
∞
En la siguiente proposici´on on tenemos la propiedad principal de esta manera de definir longitud de abiertos. Proposici´ on on 1.1.1 Dados dos conjuntos abiertos G abiertos G 1 y G G 2 acotados y dis juntos, se tiene que (G1
(G1 ) + (G2 ) . G2) = (
Demostraci´ on. Como on. Como los conjuntos G conjuntos G1 = n∞=1 I n y G y G2 = n∞=1 J n son acotados y disjuntos, entonces si definimos K 2n = I n y K 2n−1 = J n tendremos que K n n∞=1 es una uni´on on numerable de intervalos acotados disjuntos dos ∞ a dos y adem´as G as G 1 G2 = n=1 K n de modo que
∪
{ }
(G1
∪
∞
G2) =
(K n ) ,
n=1
y como la suma su ma es absolutamente absol utamente convergente, p odemos odem os reordenar reorden ar t´erminos erminos y obtenemos (G1
G2 ) =
∞
∞
n=1
que es lo que quer´ıamos ıamos demostrar. demostra r.
(I n ) +
(J n ) ,
n=1
Observaci´ on on 1.1.1 Otras propiedades importantes de la longitud de un abierto acotado son las siguientes: 1. El vac´ vac´ıo es considerado como el intervalo abierto ( a, a). Por lo tanto la longitud l ongitud del conjunto vac´ vac´ıo es cero.
⊂ G2, se tiene
2. Si G Si G 1 y G 2 son dos conjunto abierto y acotados, y G 1 (G1 ) (G2 ).
≤
∈ R, se tiene que G que G ⊕ x0
3. Si G es un conjunto abierto y acotado y x 0
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5
´ n 1.1. longitud seccion o longitud
Para la longitud de un conjunto abierto cualquiera G (no necesariamente necesariamente acotado), escoja una sucesi´on on de intervalos abiertos y acotados I n n∞=1 tal ∞ que I n I n+1 y n=1 I n = R. Esta ultima u ´ltima condici´on on se escribe como R o en palabras, que la sucesi´on I I n on I n crece hasta R.
{ }
⊂
Definimos entonces la longitud de G de G como (G) = lim lim (G n→∞
∩ I ) . n
El siguiente lema nos dice que la definici´on on anterior no depende de la sucesi´ on on de intervalos intervalos escogida. Y tambi´ tambi´en en que la longitud de un intervalo abierto existe como n´ umero umero real extendido.
{ }∞=1 y {J }∞=1 dos ⊂ J +1, I ⊂ I +1
Lema 1.1.1 Sean G un G un subconjunto abierto de R y I n sucesiones de intervalos abiertos acotados, tales que J n para todo n N, y
∈
∞
∞
J n = R =
n=1
En este caso se tiene
lim (G
n→∞
n n
n
n
n
n
I n .
n=1
∩ J ) = n
⊂
lim lim (G
n→∞
∩ I ) . n
Demostraci´ on. on. Como J n J n+1 son conjuntos abiertos, se tiene que (G J n ) (G J n+1 ) para todo n todo n N de modo que limn→∞ (G J n ) existe como n´umero umero real extendido y lo mismo vale para limn→∞ (G I n ).
∩
≤
∩
∈
∈ ∩ ⊂ ∩
∩ ∩
∩
Como para cada n N el conjunto G J n es abierto y acotado, existe K N tal que G que G J n G I k para todo k todo k > K pues pues los conjuntos I conjuntos I k son intervalos cada vez m´as as grandes (crecen hasta R).
∈ ∈
∩ J ) ≤ (G ∩ I ) para todo k todo k > K , por lo tanto (G ∩ J ) ≤ lim (G ∩ I ) →∞ y como esta ultima u ´ ltima desigualdad vale para todo n todo n ∈ N tenemos que lim (G ∩ J ) ≤ lim (G ∩ I ) . →∞ =1 Tenemos entonces entonces ((G
n
k
n
n
k
k
n
n
n
De igual manera se obtiene la desigualdad contraria y el lema est´a
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6
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
Observaci´ on o n 1.1.2 1.1.2 Algunas Algunas propiedade propiedadess que vienen vienen directame directamente nte del Lema 1.1.1 y de la definici´on on son: 1. Si G Si G es un conjunto abierto acotado, las dos definiciones de longitud para abiertos que hemos visto coinciden. coinciden.
⊂ G2 son dos conjuntos abiertos entonces entonces ((G1 ) ≤ (G2 ).
2. Si G1
∞
3. Un intervalo intervalo no acotado G puede tener medida infinita ( ( (G) = ), como en el caso en que G = (0, (0, ); o puede puede tener medida medida finita ((G) < ), como en el caso en que G que G = = n∞=1 (n 2n1 2 , n + 2n1 2 ) que tiene longitud longitud ((G) = n∞=1 n12 < .
∞ ∪ − ∞ 4. Si G Si G es es un conjunto abierto de R y x 0 ∈ R entonces entonces ((G ⊕ x0 ) = ( (G). ∞
Lema 1.1.2 Si el conjunto G acotado, tenemos que: (G
⊂ R es abierto e I es I es un intervalo abierto y
∪ I ) ≤ (G) + (I ) . ∈
R y a < b. Supo Demostraci´ on. on. Denotemos I = (a, b), con a, b Suponngamos primero que G es abiert abiertoo y acot acotad ado. o. En ese caso tenem tenemos os que G = n∞=1 (an , bn ), con a con a n , bn R y a n < bn para todo n todo n N.
∩
∅
∪ I ) =
∈
Ahora bien, si G si G I = , entonces G entonces G que por la definici´on on de longitud se tiene
∪ I =
∞
∈
n=1 (an , bn )
∞
(G
(bn
−a
n)
+b
− a = = ((G) + (I ),
I , de modo
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7
´ n 1.1. longitud seccion o longitud
≤ b.
(ii) a < a n0 < bn0 (iii) a < a n0
n0
Demostraremos solamente el primer caso, pues los dem´as as son similares y se deja como ejercicio al lector.
{ ∈ ≤ ∅ Luego, si B = ∅, entonces
∩ ∅ }
Considere B = n N : (an , bn ) I = , n = n0 . Enton Entonces ces para para cada n B se tiene que bn0 an en cualquier otro caso se tendr´ tendr´ıa que (an , bn ) (an0 , bn0 ) = .
∈ ∩
∞
G
∪ I =
(an , bn )
n = 1
(a
n0 , b),
= n 0 n
que es una uni´on on a lo sumo numerable de conjuntos dos a dos disjuntos. Luego, por definici´ on on se tiene que: ∞
(G
∪ I )
=
(bn
−a
+b
(bn
−a
+ (b (b
n = 1
n)
− a ≤ n0
n = n 0
∞
≤
n = 1 = n 0 n
n)
− a) + (b(b − a n0
n0 )
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8
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
que tambi´ en en es una uni´ on o n disjunta y a lo m´as as numerable numerable de interv intervalos alos abiertos, y por lo tanto: ∞
(G
∪ I ) =
(bn
n = 1
−a
n)
+c
−a
n0 ,
= n 0 n
y similarmente al proceso anterior es f´acil ver que (G
∪ I ) ≤ (G) + (I ) . { }∞=1
Finalmente para el caso en que G que G sea sea un abierto no acotado sea J n una sucesi´on on de intervalos abiertos y acotados que crece a R. Luego
n
∪ I ) ∩ J ) = (( ((G G ∩ J ) ∪ (I ∩ J )) ≤ ((G ((G ∩ J )) + ((I ((I ∩ J )) pues G∩J es un conjunto abierto y acotado y cae en el caso ya desarrollado, puesto que I que I ∩ J tambi´en en es un u n intervalo abierto abie rto y acotado. aco tado. El lema queda (G
n
n
n
n
n
n
n
demostrado demostrad o tomando toman do l´ımites ımites y usando usand o la definici´ definici on o´n de longitud de conjuntos abiertos en este caso. Corolario 1.1.1 Sean n N y I k abiertos y acotados. Entonces
∈
n
n k
{ { } =1 una familia finita de intervalos n
≤ I k
k=1
k =1
(I k ) .
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9
´ n 1.1. longitud seccion o longitud
Entonces Entonces tenemos tenemos que G es abierto y acotado y por lo tanto G = J n n∞=1 es una sucesi´on on de intervalos disjuntos abiertos y acotados. Similarmente para cada n cada n,, Gn = n∞=1 I rn , con los I los I rn intervalos abiertos disjuntos y acotados.
∞ n=1 J n donde
{ }
Tome ε > 0. 0. Como Como (G) < tanto, existe un N N tal que
∈ ∈
∞ sigue que
∞
∞
n=1
(J n ) <
(J n ) < ε/2 ε/2 .
∞ y por lo
(1.2)
n=N +1
¯n sea comPara cada J n tome K n un intervalo cerrado (esto hace que K ε pacto) tal que (J n ) < (K n ) on o n a (1.2) 2N y si juntamos esta relaci´ tenemos N N ε (G) < (J n ) + < (K n ) + ε . (1.3) 2 n=1 n=1
−
¯ Como los I rn son abiertos y cubren N n=1 K n que es compacto, existe un conjunto finito I rn11 , . . . , I rn ss que lo sigue cubriendo y por lo tanto tamb ta mbi´ i´en en a N n=1 K n , es decir
{ }
{
}
N
N
⊂
K n
n=1
Tenemos entonces que
n=1
¯n K
n1 r1
N
n=1
ns rs .
⊂ I ∪ · · · ∪ I
(1.4)
s
(K n )
≤
(I rn11
∪···∪
I rnss )
≤ j =1
(I rnjj ), ) ,
(1.5)
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10
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
Demostraci´ on. on. Como primer paso haremos la demostraci´on on del teorema en el caso en que G1 es un intervalo acotado y G2 es acotado y uni´on on de un n´ umero finito de intervalos (abiertos acotados y disjuntos). umero Estamos afirmando entonces que si G si G 1 = (a, b) y G2 = ni=1 I i entonces (1.7) vale para cualquier colecci´on on finita de intervalos I i ni=1 disjuntos dos a dos.
{ }
Demostraremos esta afirmaci´on on por inducci´on o n en n. Para ara n = 1, es decir si G1 = (a, b) y G2 = (c, d), el result resultado ado es obvio obvio.. Suponga Suponga que la afirmaci´ on on vale para n para n,, mostraremos ahora que vale para n + 1. Se tiene entonces que G 2 = Si I Si I
n+1 i=1 I i y
hacemos I hacemos I = (a, ( a, b) = G1 .
on de longitud, se tiene ∩ G2 = ∅ entonces, de la definici´on n+1
∪
(I G2 ) = ( (I ) +
I i = = ((I ) + (G2 )
i=1
∩ ∅
∪
Si I Si I G2 = , escribimos primero G primero G 2 = I 1 G2 , donde G donde G 2 = modo que (I ) + (G2 ) + (I 1 ) = ( (I ) + (G2 ) .
n+1 i=2 ,
de
(1.8)
Sin p´ erdida erdida de generalidad se puede sup oner que I I intersecta a I 1 . (De no ser as´ as´ı se re-enumeran los I i de modo que I 1 sea un intervalo tal que I I j = .)
∩ ∅ Si I Si I ∩ I 1 = ∅ se tiene que I que I ∪ I 1 es un intervalo abierto y acotado, y por
la hip´ otesis otesis de inducci´on on
(I ) + (G2 ) + (I 1 ) = (I I 1 ) + (I I 1 ) + (G2 )
∪
(I I
∩
((I I ) G ) + ((I
G ) + (I I )
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11
´ n 1.1. longitud seccion o longitud
∈ ∈ N tal que
Para cada ε cada ε > 0 existe N existe N ∞
j =N +1
˜1 = Defina G N ˆ 2 = ∞ I i , G i=1
N j =1
ε (J j ) < , y 2
ˆ1 = J j , G
i=N +1 I i .
∞
∞
Mostremos ahora que el teorema vale para G1 = ∞ i=1 I i , conjuntos abiertos arbitrarios pero acotados. ∞
(I i ) <
i=N +1
j =N +1 J j ,
j =1
J j , y G2 =
ε . 2
˜2 = y de la misma manera G
Entonces para i = 1, 2 se tiene
˜ i ) + (G ˆ i ), y (G ˆ i ) < ε/2 (Gi ) = ( (G ε/2.
De la primera parte tenemos que ˜ 1 ) + (G ˜ 2 ) = ( ˜ 1 (G (G
∪ G˜2) + (G ˜ 1 ∩ G˜ 2) ˜1 ∪ G ˜ 2 ⊂ G 1 ∪ G2 y G ˜1 ∩ G ˜ 2 ⊂ G 1 ∩ G2 la igualdad anterior se y como G
convierte en
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12
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
y como estas desigualdades valen para todo ε > 0 tenemos (G1 ) + (G2 )
≥ (G1 ∪ G2) + (G1 ∪ G2)
y que junto con (1.10) nos da la igualdad (1.7) para G1 y G2 abiertos acotados. Para el caso cuando G cuando G1 y G y G2 son abiertos no acotados se usa la definici´on on de longitud en este caso.
1.2 1.2
Medi Medida da Exte Exteri rior or
A continuaci´on on definimos la funci´on on m∗ : (R) [0, [0 , exterior y estudiaremos algunas de sus propiedades.
P → ∞], llamada medida
Sea E P ( P (R). La medida exterior ) , se define por el exterior de E , m∗ (E ), n´ umero umero real extendido:
⊂
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13
´ n 1.3. conjunto seccion o conjuntos s medibles medibles
Dado ε > 0, por la definici´on on de medida exterior, exterior, existen existen conjunto conjuntoss abiertos G abiertos G 1 y G 2 con E con E 1 G1 y E 2 G2 y tal que
⊂
m∗ (E i ) +
⊂
ε > (Gi ) 2
para i = 1, 2.
Sumando estas desigualdades tenemos m∗ (E 1 ) + m∗ (E 2 ) + ε > (G1 ) + (G2 ) = = ((G1
∩ G2) + (G1 ∪ G2).
Pero E 1 E 2 G1 G2 y E 1 E 2 G1 G2 , de modo que por las propiedades (2) y (3) de la medida exterior tenemos
∪
⊂ ∪
∩
⊂ ∩
m∗ (E 1 ) + m∗ (E 2 ) + ε > m∗ (G1
∩ G2) + m∗(G1 ∪ G2)
que vale para todo ε > 0 y por lo tanto implica la ecuaci´on on 1.11.
{ { }∞=1 una sucesi´ on de subconjuntos de R . En-
Proposici´ on on 1.2.2 Sea E n tonces:
n
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14
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
de R que satisfagan lo que deseamos, m´as as a´ un un la funci´ on on as´ı obtenid obte nidaa ser´ s er´a contablemente aditiva. Se dice que una funci´on on definida sobre subconjuntos contablemente aditiva si para toda sucesi´ de R es contablemente on on E n n∞=1 de con∞ ∞ juntos dos a dos disjuntos se tiene que m que m ∗ ( n=1 E n ) = n=1 m∗ (E n ), que es precisamente la propiedad que hace funcionar la teor´ teor´ıa de la integraci´on, on, como se podr´a percibir posteriormente.
{
}
Sea m∗ la medida exterior definida en los subconjuntos de R. Un conon on de Caratheodo Caratheodory ry si se verifica la junto E R satisface la condici´ igualdad siguiente
⊂
m∗ (A) = m∗ (A para todo A todo A
∩ E ) + m∗(A ∩ E ) c
(1.12)
⊂ R.
Los conjuntos que satisfacen la condici´on on de Caratheodory Caratheo dory tambi´en en son llamados medibles, y a la clase conjuntos Lebesgue medibles la denotaremos por = E R : E es es Lebesgue medible
L { ⊂ ⊂
}
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15
´ n 1.3. conjunto seccion o conjuntos s medibles medibles
Demostraci´ on. on. La parte (i) es f´acil acil puesto que se cumple la siguiente igualdad para todo A todo A R
⊂ m∗ (A) = m∗ (A ∩ ∅) + m∗ (A ∩ ∅ ) = m∗ (A ∩ R) + m∗ (A ∩ X ) . c
c
Para la parte (ii) basta observar que la condici´on (1.12) es sim´etrica etrica con c respecto respecto a E y E . Para Para la parte parte (iii) (iii) lo haremo haremoss por partes partes como anter anterior iormen mente. te. Primer Primeroo mostraremos que si A si A y B est´an an en entonces E entonces E F F est´an an en .
L
Sean, entonces dos conjuntos medibles E y F . F . m∗ (A) = m ∗ (A
∩
L
∩ E ) + m+(A ∩ E ) para todo A todo A ∈ L c
y de ah´ı m∗ (A
∩ E ) = m∗(A ∩ E ∩ F ) F ) + m∗ (A ∩ E ∩ F ) para todo A ∈ L usando A usando A ∩ E en en el lugar de A de A en la ecuaci´on on anterior. Es decir c
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16
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
y por inducci´on on que m∗ (A E 1 son conjuntos dos a dos disjuntos.
∩ ∩ · · · ∩ E ) = n
n i=1
m∗ (A
∩ E ) si los A i
i
Ahora bien, dada una familia numerable de conjuntos E n medibles, se puede formar una nueva sucesi´on on de conjuntos dos a dos disjuntos, de la siguiente manera F 1 = E 1
F i = E = E i E i
\
1 para
todo i todo i > 1
{ }
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17
´ n 1.3. conjunto seccion o conjuntos s medibles medibles
Teorema 1.3.2 La restricci´ on de m∗ a
L satisface
(i) m∗ ( ) = 0
∅
≤ m∗(E ) ≤ ∞ para todo E ∈ ∈ L. {E }∞=1 es una familia numerable de subconjuntos de L L dos a dos (iii) Si { (ii) 0
n n
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18
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
de modo que m∗ (A)
≤ m∗((A ((A ∩ I ) (A ∩ I )) ≤ m∗(A ∩ I ) + m∗(A ∩ I ) Tamb Tambi´ i´en A ∩ I = (A ( A ∩ I ) (A ∩ (I ∩ I ). Y de ah´ı, ı, n
n
n
c n
c
c
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19
´ n 1.3. conjunto seccion o conjuntos s medibles medibles
⊂ ⊂
Teorema 1.3.5 Si E, F son F son medibles y si E F F entonces m(E ) y si adem´ as m(E ) < se tiene m(F E ) = m( m (F ) F ) m(E ).
∞
\
≤ m(F ) F )
− Demostraci´ on. Como on. Como m m es aditivo en L y F = E (F \ E ) vale m vale m((F ) F ) = m(E ) + m(F \ E ), ), de donde m donde m((E ) ≤ m(F ) F ) y si m si m((E ) < ∞ se puede restar m(E ) en ambos miembros y queda m( m (F ) F ) − m(E ) = m( m(F \ E ). ).
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20 pero como
cap´ ıtulo ıtulo 1. Longitud Longitud y Medida
∞
k=1
\ ∩∞=1F ) se tiene tambi´en en que
E k = F 1 ( m(
k
k
∞
∞
E k ) = m( m(F 1 )
k=1
que junto a la ecuaci´on on anterior nos da
− m(
k =1
F k )
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56
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
Como ejemplo de esta situaci´on on podemos dar la medida de Lebesgue m Lebesgue m en el intervalo I = [0, [0, 1] y la medida δ δ concentrada λ en el punto p = 12 . Recordemos que λ(E ) = 0 si p A y λ(E ) = 1 si p E . Tenemos enemos que δ m pues podemos tomar A = I p y B = p y tenemos δ (A) = m(B ) = 0.
∈
⊥
∈ { }
\{ }
Nuestro Nuestro objetivo objetivo sera probar versione versioness simplificad simplificadas as del Teorema eorema de Radon-Nikodym y del Teorema de descomposici´on on de Lebesg Lebesgue. ue. Una de las simplificaciones consiste en que nos restringiremos a medidas finitas, pero los teoremas valen para espacios de medida σ -finitos. -finitos. Un espacio espacio de medida (Ω, (Ω, , µ) es σ -finito si Ω es uni´on on numerable de conjuntos con medida finita.
A
4.1 4.1
Medi Medida dass con con Sign Signo o
A
Dado un espacio medible (Ω, (Ω , ), decimos que la funci´on µ on µ : :
A → R es una
ta mbi´ i´en en medid medida real (o tamb medida a con signo signo), si µ( ) = 0 y es contable-
∅
mente aditiva, esto es ∞
µ
∞
E k
=
k =1
µ(E k )
(4.1)
k=1
para cualquier colecci´on on disjunta de conjuntos medibles E k en Ω.
∞
∞
Observe que de la definici´on on se tiene que µ que µ(Ω) (Ω) < < , m´as as a´ un, µ un, µ((E ) < para cualquier conjunto medible E . Esto Esto implica implica que la serie serie en el lado derecho de la ecuaci´on on 4.1 es absolutamente convergente. Un subconjunto P de P de Ω es denominado conjunto positivo de la medida real µ si para todo conjunto medible E se se tiene µ( µ (E P ) P ) 0. Similarmente un conjunto N conjunto N de de Ω se denomina negativo si µ( µ (E N ) N ) 0 para todo E todo E medible. medible.
∩ ≥ ∩ ≤
Teorema 4.1.1 (Descomposici´ on on de Hahn) Si λ es una medida real, entonces existen conjuntos P y N en Ω tal que Ω = P N , N , P N = , y tal que P es P es positivo y N es N es negativo respecto de λ.
∪ ∪
P
∩ ∩
∅
Demostraci´ on. Considere on. Considere la clase de de todos los conjuntos positivos, esta clase es no vac´ vac´ıa pues el conjunto est´ a en ella. Tome α Tome α = = sup λ(A) : A
∅
{
∈
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57
´ n 4.1. medidas seccion o medidas con signo
P}
{ }
P
y una sucesi´on on An de conjuntos en que que aproxime este supremo, i.e. lim λ(An ) = α. Podemos Podemos definir ahora ahora P = n∞=1 An y como la uni´on on de dos conjuntos positivos es positivo, podemos tomar la sucesi´on on An como una sucesi´on on creciente de modo que
∩
λ(E P ) P ) = λ( λ (E
∞
∞
∩
An ) = λ( λ (
n=1
n=1
∩
∩
E An ) = lim λ(E An )
≥0
lo que demuestra que P que P es es un conjunto positivo y adem´as as con la propiedad que lim λ(An ) = µ( µ (P ) P ) = α < .
∞
\
S´ olo queda por demostrar que el conjunto N = Ω P olo P es un conjunto negativo negativo.. Si no lo fuera, existe un subconjunto subconjunto E de N con λ(E ) > 0. El conjunto E conjunto E no no puede ser positivo pues si hacemos P hacemos P E obtend obt endrr´ıam ıamos os un conjunto positivo con medida estrictamente mayor que α, α, lo que contradice la definici´on on de α de α.. Por lo tanto E tanto E contiene contiene un conjunto E conjunto E 1 medible tal que λ(E 1 ) 1/n1 .
∪ ∪
≤−
Como λ Como λ((E ) = λ( λ (E 1
λ (E 1 ) + λ(E \ E 1 ) tenemos que ∪ E \ E 1) = λ( λ(E \ E 1 ) = λ( λ (E ) − λ(E 1 ) > λ(E ) > 0 > 0 .
\
Otra vez, E E 1 no puede ser positivo por un argumento an´alogo alogo al mencionado para E . Por lo lo que que E E 1 contiene conjuntos con λ-medida negativa. Tome n Tome n 2 el menor numero natural tal que E que E E 1 contiene un con junto E junto E 2 con medida λ medida λ((E 2 ) 1/n2 . Como antes E antes E (E 1 E 2 ) no puede ser positivo y se tomamos n tomamos n 3 el menor entero tal que E que E (E 1 E 2 ) contiene un conjunto E conjunto E 3 con λ con λ((E 3 ) 1/n3 . Repitiendo este argumento obtenemos un sucesi´ sucesi´ on on (E k ) de conjuntos medibles tales que λ que λ((E k ) 1/nk .
\
\ \
≤− ≤−
Ahora hacemos F hacemos F =
∞ k=1
λ(F ) F ) =
E k de modo que
∞
k=1
∞
λ(E k )
≤−
k=1
1 nk
∪ \ ∪ ≤−
≤0
→
lo que nos da que necesariamente 1/n 1 /nk 0. Si G es un subconjunt subconjuntoo medible de E de E F con λ con λ((G) < 0, < 0, entonces existe un n k de modo que λ que λ((G) < 1/(nk 1) para un k suficientemente suficientemente grande. Pero como G E (E 1 E k ) tenemos una contradicci´on on pues nk deber deb er´´ıa ser el menor natural tal que E (E 1 E k ) contiene un conjunto con medida menor que 1/nk .
\
− − ···∪ \ ∪ · · · ∪ −
⊂ \
∪
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58
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
\
≥ −
De este modo, todo subconjunto G subconjunto G de de E E F debe debe satisfacer λ satisfacer λ((G) 0, lo que indica ind icarr´ıa que E que E F es F es positivo para λ para λ.. Como λ Como λ((E F ) F ) = λ( λ (E ) λ(F ), F ), tendr´ ten dr´ıamos ıa mos que qu e P (E F ) F ) es un conjunto positivo y con medida mayor que α que α,, lo cual es una contradicci´on. on.
\ \
\
Sigue que el conjunto N = Ω P P es negativo para λ y ya tenemos la descomposici´ on on requerida.
\
o n de Hahn Hahn de Los conjuntos P y N se N se denominan una descomposici´on Ω con respecto a la medida real λ. λ . En general dicha descomposici´on on no es u unica, ´ nica, pero esto no causa mayor inconveniente pues se puede demostrar que los diferentes descomposiciones difieren por conjuntos con medida nula.
4.2 4.2
El Teo Teore rema ma de de Rado Radonn-Ni Nik kodym odym
Diremos que la funci´on on integrable f integrable f representa a la medida λ medida λ si λ(E ) =
∈ A,
f dµ,
E
E
(4.2)
Teorema 4.2.1 (Radon-Nikodym) Sean λ λ y µ y µ medidas finitas en el espacio medible (Ω, (Ω, ) y suponga que λ que λ << µ. µ. Entonc Entonces existe una funci´ on integrable positiva definida en Ω tal Ω tal que f representa f representa a λ. λ . M´ as a´ un Si existe otra funci´ on g que representa a λ se tiene que f = g en casi todo punto.
A
Demostraci´ on. Dado on. Dado c c > 0 denotemos por P por P ((c) y N y N ((c) la descomposici´on on de Hahn de la medida real λ cµ. cµ. Si k N, consideremos los conjuntos medibles
−
∈
k
A1 = N = N ((c),
Ak+1 = N (( N ((k k + 1)c 1)c)
\
Aj .
j =1
k j =1
\
Observe que los A los A k son disjuntos y que
N ( N ( jc) jc ) =
k j =1
Aj .
k −1 −1 Se tiene entonces que Ak = N ( N (kc) kc ) N ( jc) jc ) = N ( N (kc) kc ) jk=1 P ( P ( jc). jc ). j =1 N ( y por consiguiente, si E si E es es un conjunto medible de A de A k , entonces E entonces E N ( N (kc) kc ) y E P (( P ((k k 1)c 1)c) con lo que
⊂ ⊂
−
λ(E )
kcµ(E ) ≤ 0 − kcµ(
y λ(E )
1)cµ((E ) ≥ 0 . − (k − 1)cµ
⊂ ⊂
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59
´ n 4.2. teorem ´ n-nikodym seccion o teorema a de radon-nikodym o
Por lo tanto
− \ (k
≤
1)cµ 1)cµ((E )
∞ Defina ahora B ahora B = = Ω j =1 Aj = para todo k todo k.. Esto implica
0
λ(E )
i j =1
≤ kcµ( kcµ(E ).
nftyP ( jc), jc ), de modo que B que B
kcµ(B ) ≤ λ(B ) ≤ λ(Ω) < (Ω) < ∞, ≤ kcµ(
para todo k todo k
(4.3)
⊂ P ( P (kc) kc)
∈ N.
Sien Si endo do as´ı µ(B ) tiene que tener medida nula, y como λ << µ ta µ tamb mbi´ i´en en se tiene λ tiene λ((B ) = 0. Observe tambi´en en que B que B es disjunto de todos los A los A k y que ∞ B k=1 Ak = Ω.
∪
Ahora estamos en condiciones de comenzar a buscar la funci´on integrable f grable f solicitada en el teorema. Para cada c cada c > 0 de la construcci´on on anterior anterior podemos definir definir f c (x) =
(k 0
− 1)c 1)c
si x si x
∈A ∈ B. k
Ahora, si E si E es es un conjunto medible cualquiera puede ser escrito como uni´on disjunta de conjuntos del tipo E tipo E B y E Ak , k N, de modo que usando la ecuaci´on on 4.3 tenemos
∩
f c dµ
E
≤ λ(E )
≤
∩
∈
(f c + c)dµ
E
≤
f c dµ + cµ(Ω) cµ(Ω)..
E
Ahora formamos una sucesi´on on de funciones tomando c = 2−n para n y las denotamos f denotamos f n . Siendo as´ as´ı, las l as funciones f unciones f n satisfacen
f n dµ
≤ λ(E )
E
para todo n todo n
≤
f n dµ + 2−n µ(Ω)
∈ N, (4.4)
E
∈ N. Dicho de otro modo, se cumple que
E
f n dµ
≤ λ(E )
y
λ(E )
f m dµ + 2−m µ(Ω)
E
para cualquier cualquier medible medible E E y todo m, n suponiendo suponiendo que m que m n, es f´acil acil ver que
≥
≤
(f n
E
− f
∈
≤
m )dµ
N.
Con Con esta esta inform informac aci´ i´ on, o n, y
2−n µ(Ω)
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60
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
+ para cualquier medible E . Escogi Escogiend endo o en cada caso caso E n,m = x Ω : − (f n (x) f m (x)) 0 y E n,m = x Ω : (f n (x) f m (x)) < )) < 0 0 , obtenemos
−
≥ }
+
(f n
E n,m
{ ∈
− f
−
(f n
E n,m
De modo que
m )dµ
Donde, como antes, (f (f n Similarmente
=
− f
(f n
m)
Ω
− f
m)
+
− f
=
(f n
Ω
|
f n
+
dµ
}
≤ 2−
n
{ ∈
µ(Ω)
denota la funci´on on parte positiva de f de f n
m )dµ
Ω
−
−
− f
m)
− f |dµ ≤ 2 · 2−
n
m
dµ
≤ 2−
n
− f
m.
µ(Ω). (Ω).
µ(Ω)
Esto significa que la sucesi´on on f n es de Cauchy en media, y por lo tanto converge en media a una funci´on on positiva f positiva f .. Tambi´ Tamb i´en, en ,
{ }
f n dµ
E
para todo n todo n
− ≤ | f dµ
f n
E
E
− f |dµ
≤ |
f n
− f |dµ,
on 4.4 quiere decir que ∈ N, que junto con la ecuaci´on λ(E ) = lim
f n dµ = dµ =
E
fdµ,
E
para todo E todo E medible, medible, y por lo tanto f tanto f es es la funci´on on buscada. Supongamos ahora que existen dos funciones f funciones f y g que representan a λ a λ respecto de µ de µ,, esto es λ(E ) =
f dµ = dµ =
E
gdµ,
E
para todo E medible. E medible. Podemos Podemos escoger entonces entonces E 1 = x Ω : f ( f (x) > g (x) y E 2 = x Ω : g (x) > f (x) y aplicando la Proposici´on on 2.4.3 obtenemos f = g c.t.p. c.t.p. .
}
{ ∈
}
{ ∈
Teorema 4.2.2 (Descomposici´ on on de Lebesgue) Sean λ y µ medidas finitas en un σ -´ algebra . Entonc Entonces existe una medida medida λ1 que es singular respecto de µ y una medida λ2 que es absolutamente continua respecto de µ y tal que λ = λ = λ 1 + λ2 . Mas a´ un, las medidas λ1 y λ2 son unicas. ´
A A
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61
´ n 4.2. teorem ´ n-nikodym seccion o teorema a de radon-nikodym o
Demostraci´ on. Defina on. Defina ν ν = λ + µ de modo que ν tambi´en en es una u na medida med ida finita. De esta definici´on on tambi´ ta mbi´en en se s e tiene tie ne que q ue tanto ta nto λ λ como como µ µ son son absolutamente continuas respecto de ν de ν de de modo que el Teorema de Rad´on-Nikodym on-Nikodym implica que existen funciones positivas f y g tales que representan a λ y µ respecto de ν de ν ,, esto es: λ(E ) =
f dν,
µ(E ) =
E
gdν
E
para todo medible E . Sean Sean A = x Ω : g (x) = 0 , y B = x g (x) > 0 > 0 de tal forma que A B = , y Ω = A = A B .
{ ∈ ∩ ∅
}
∪
{ ∈ Ω
}
:
Definamos Definamos ahora dos medidas medidas λ 1 y λ 2 de la siguiente forma
∩
∩
λ1 (E ) = λ( λ (E A)
λ2 (E ) = λ( λ (E B ) ,
para todo E todo E medible. medible. Queda claro entonces que λ = λ = λ 1 + λ2 .
⊥
Veamos que λ que λ 1 µ. En efecto, tenemos que µ( µ (A) = A gdν = gdν = 0 pues g pues g es id´enticament entic amentee nula nu la en A en A y λ 1 (B ) = λ( λ (B A) = 0 pues A pues A B = .
∩
∩
∅
Veamos ahora que λ2 << µ. Si µ(E ) = 0 se tiene que E gdν = 0, de modo que g (x) = 0 para ν c.t.p. x E . Entonces Entonces,, ν (E B ) = 0 lo que implica λ(E B ) = 0 pues λ pues λ << ν . Esto Esto ultimo u´ltimo es lo mismo que λ2 (E ) = λ( λ (E B ) = 0, implicando que λ que λ 2 << µ.
− −
∩ ∩
∩
∈
∩ ∩
Ahora s´olo olo falta la unicidad de esta descomposici´on. on. Vamo amoss a usar usar el hecho de que si α es una medida satisfaciendo α << µ y α µ entonces α = 0. La prueba de esta afirmaci´on on la dejamos como ejercicio.
⊥
Ahora bien, suponga que existan dos descomposiciones para λ = λ 1 + λ2 = γ 1 + γ + γ 2 , donde λi y γ i , i = 1, 2, son medidas satisfaciendo λ1 µ, λ2 << µ, γ 1 µ y γ 2 << µ.
⊥
⊥
Tenemos que λ que λ 1 γ 1 = γ = γ 2 λ2 es una medida con signo y consideramos P y N N la descomposici´on on de Hahn de esa medida medida.. Defina Defina h+ (E ) = (λ1 γ 1 )(E )(E P ) P ) = (γ ( γ 2 λ2 )(E )(E P ), P ), de modo que h que h + es una medida positiva (vea + + el Ejercicio 4.2) y tal que h µ y h y h << µ, de modo que h que h + (E ) = 0 para todo E medible. medible. Similarmen Similarmente, te, si definimos definimos h− (E ) = (λ1 γ 1 )(E )(E N ) = − (γ 2 λ2 )(E )(E N ), N ), obtenemos que h que h (E ) = 0 para todo E todo E medible. medible.
∩
−
−
−
−
∩
⊥
∩
−
−
Juntando Juntando estas relaciones relaciones tenemos que para todo E medible E medible 0 = h+ (E ) + h− (E ) = (λ1 (λ )(E )(E )
)(E ∩ P ) P ) + (λ ( λ1 − γ 1 )(E )(E ∩ N ) − γ 1)(E
∩
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62
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
= (γ 2 = (γ 2
− λ2)(E )(E ) − λ2)(E )(E ∩ P ) P ) + (γ ( γ 2 − λ2 )(E )(E ∩ N ) N ) = h + (E ) + h− (E ) = 0 ,
lo que indica que λ que λ 1 = γ = γ 1 y γ 2 = λ 2 .
Ejercicios 4.1 Si P 1 and P 2 son conjuntos positivos para una medida con signo λ entonces P 1 P 2 tambi´ en en es un conjunto positivo p ositivo para la misma medida.
∪
4.2 Si P es P es un conjunto positivo de la medida real ν entonces ν + (E ) = ν (E P ) P )
∩
define ν + como una medida (positiva) sobre la misma σ -´ algebra. algebra. 4.3 Demuest Demuestre re que si λ si λ 1 << µ y λ 2 << µ entonces λ entonces λ 1 + λ2 << µ. 4.4 Demuest Demuestre re que si λ si λ 1 << µ y λ 2
entonces λ 1 ⊥ λ2 . ⊥ µ entonces λ 4.5 Demuest Demuestre re que si λ si λ << µ y µ y λ ⊥ µ entonces λ = 0. 4.6 Sean λ Sean λ y µ medidas σ medidas σ -finitas en en el espacio medible (Ω, (Ω , ), y tales que λ < < µ. µ. Si f f es la funci´on o n dada por el teorema de Rad´ononNikodym, entonces pra toda funci´on on no negativa definida en Ω, se tiene gdλ = gdλ = gfdµ.
A
4.7 Demostrar Demostrar que si tenemos dos medidas, µ medidas, µ y λ, λ , satisfaciendo µ satisfaciendo µ << λ y λ << µ entonces existe una funci´on on g tal que si E = x Ω : g(x) = 0 se tiene
{ ∈
}
µ(A) =
A∩E
para todo A todo A medible.
gdλ
y
λ(A) =
1 dµ, A∩E g
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Cap´ıtulo 5
Medidas Producto y el Teorema de Fubini El ob jetivo principal de este cap´ cap´ıtulo es demostrar el Teorema de Fubini, el cual es de suma importancia para la evaluaci´on on de integrales de funciones definidas en espacios medibles generados por el producto de espacios medibles. El modelo que uno debe tener presente es una construcci´on del espacio medible R R. Es por eso necesario dar las definiciones y algunos resultados de los espacios producto.
×
5.1 5.1
Espa Espaci cios os Produ Product cto o
Si tenemos dos espacios medibles (Ω 1 , 1 ) y (Ω2 , 2 ) al producto cartesiano angulo angulo medible medible de A1 A2 con A con A i i , i = 1, 2, le llamaremos un rect´ del espacio producto Ω = Ω1 Ω 2 . La Lass unio unione ness finit finitas as de rect rect´ angulos ´angulos medibles ser´a denotada por 0 y es usualmente llamada la clase de los conjuntos elementales.
×
A
∈ A
A
× A
Definimos como 1 algebra algebra en Ω1 2 como la menor σ -´ contiene todos los rect´angulos angulos medibles.
A × A
clase mon´ otona otona Una clase
Si E Si E
M E
× Ω 2 que
M es una colecci´on on de conjuntos tales que
E
para todo i todo i
entonces A N entonces A
∞
E
M
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64
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
• Si E ∈ M, E ⊃ E +1 para todo i todo i ∈ N entonces A entonces A = = ∞ =1 E ∈ M. ⊂ Ω 1 × Ω2 y puntos x ∈ Ω1 e y ∈ Ω2, definimos Dado un conjunto E ⊂ la secci´ on on x como E = { y ∈ Ω2 : (x, y ) ∈ E } la secci´ on on y como E = {x ∈ Ω 1 : (x, y ) ∈ E }. Es evidente evidente de la definici´ on on que E ∈ Ω 2 y E ⊂ Ω1 . Proposici´ on on 5.1.1 Si E ∈ ∈ A1 × A2 entonces E ∈ A2 y E ∈ A1, para todo x ∈ Ω1 e y ∈ Ω2 . i
i
i
i
i
x
y
x
y
x
y
Demostraci´ on. Consider on. Consider la clase de conjuntos siguiente
B = {E ∈ ∈ A1 × A2 : E ∈ A2 para todo x ∈ Ω1}. Si E = A1 × A2 , es f´acil acil ver que E = B cuando x ∈ A 1 y E = ∅ si x Esto es, es, B es una clase de conjuntos que ∈ A1, y por lo tanto E ∈ ∈ B . Esto x
x
x
contiene a los rect´angulos angulos elementa elementales. les. Adem´ as, as, no es dif´ dif´ıcil mostrar que se satisfacen las siguientes propiedades:
× Ω2 ∈ B. (b) Si E ∈ (E ) = (E ) ∈ A2 para todo x ∈ Ω1 y por lo ∈ B entonces (E tanto E ∈ B . ∞ (c) Si E ∈ B, y E = ∞ =1 entonces E = =1 (E ) ∈ A2 para todo x ∈ B. y por lo tanto E tanto E ∈ Con estas tres propiedades la clase B es una σ -´ algebra algebra y contiene a los rect´ angulos medibles, por lo tanto B = A 1 × A2 . Es deci angulos decirr para para todo E (a) Ω1
c
x
x
c
c
i
x
i
i
i x
medible se tiene que E x es medible. medible. Un procedimiento procedimiento similar demuestra demuestra la propiedad para la secci´on E on E y .
A A × A
Teorema 5.1.1 La σ -´ algebra 1 onica que 2 es la menor clase monot´ contiene a todos los conjuntos elementales.
B
Demostraci´ on. Llamemos on. Llamemos a la menor clase monot´onica onica que contiene a onica onica tal que 0 , los conjuntos elementales, esto es la menor clase monot´ . Para Pa ra probar prob ar el teorema teore ma debemos debem os probar prob ar que = 1 A 2 . 0 Para esto, observe que como tambi´ clase monot´ onica onica
A A ⊂B
A
A
B
A ×
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65
B ⊂ A ×
que contiene los elementales entonces A2 . Para Para demostrar demostrar que 1 bastar´a mostrar que es un σ un σ-´ -´ algebra pues ya contiene a los algebra 1 2 rect´ angulos angulos medibles.
A × A ⊂ B
B
Dejamos al lector el ejercicio de demostrar que la intersecci´on on de dos rect´ angulos angulos es un rect´angulo angulo y que la diferencia de rect´angulos angulos es la uni´on on 1 disjunta de una cantidad finita de rect´angulos angulos . Por consiguiente si P y Q son conjuntos elementales, entonces tambi´en en son elementales los conjuntos P Q y P Q y como P como P Q = (P Q) Q tambi´en en tenemos tene mos que P Q es elemental.
∩ ∩
\
∪ ∪
\ ∪
∪ ∪
Defina para cualquier conjunto P
P ) como ∈ ∈ Ω1 × Ω2 el conjunto H(P ) H(P ) ∈ B , P ∪ ∪ Q ∈ B} P ) = {Q ⊂ Ω1 × Ω2 : P \ Q ∈ B , Q \ P ∈ Nuestro primer afirmaci´on on es que si Q si Q ∈ B entonces B ⊂ H(Q). Para esto describamos primero dos propiedades simples de H(P ). P ). (a) Q ∈ H(P ) P ) si y solamente si P si P ∈ H(Q). (b) Como B es una clase monot´onica, onica, H(P ) P ) tambi´en en lo es. Ahora, fije P ∈ A0 un conjunto conjunto elemental. elemental. De las propiedades propiedades de los conjuntos elementales es f´acil acil ver que Q ∈ H (P ) P ) para todo Q elemental. Esto es A0 ⊂ H(P ) P ) si P si P es es elemental y como H(P ) P ) es una clase monot´onica onica se tiene que B ⊂ H(P ) P ) para todo P ∈ A0 . Pero Pero si fijamos fijamos Q ∈ B lo que probamos es que Q que Q ∈ H(P ) P ) para todo P todo P elemental, elemental, lo que por la propiedad (a) dice que P ∈ H(Q) para todo P P elemental. elemental. Nuevamen Nuevamente, te, entonces entonces,, A0 ⊂ H(Q) y as´ı B ⊂ H(Q) para todo Q todo Q ∈ B , demostrando la afirmaci´on. on. Usando Usando la afirmaci´ afirmaci´ on precedente tenemos que si P y Q est´an on a n en B entonces P \ Q y P ∪ an en B. ∪ Q est´an Ahora Aho ra s´ı veamos veamo s que B es una σ una σ -´ algebra. algebra. • Se tiene que A1 × A2 ∈ A0 y por lo tanto A tanto A 1 × A2 ∈ B . • Si Q Si Q ∈ B entonces Q entonces Q = A 1 × A2 \ Q ∈ B . • Si P ∈ B y P = ∞=1, entonces Q = =1 P est´a en B y de esta manera P manera P = Q ∈ B de la definici´on on de clase mon´otona. otona. c
i
i
n
1
n
n i
i
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66
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
B A × A ⊂ B A × A B
Concluyendo, como es una σ-´algebra algebra que contiene a los elementales entonces 1 y como tambi mbi´´en en es 2 1 2 , pues 1 2 ta una clase mon´otona otona que contiene a los elementales, tenemos finalmente, que 1 = . 2
B ⊂ A × A
× →
A × A
∈
→
R y cada x R Para cada funci´on f on f : A 1 A2 cada x A1 definimos f definimos f x : A2 tal que f x (y ) = f ( f (x, y). De modo simila similarr f y es la funci´ on on definida en A2 tal que f que f y (x) = f ( f (x, y).
Teorema 5.1.2 Sea f f una funci´ on
Ω 1 × Ω2 entonces: A A1 × A2-medible en Ω
∈ Ω1, f es A A2-medible. A1-medible. (ii) Para todo y ∈ Ω2 , f es A (i) Para todo x
x
y
Demostraci´ on. on. De la definici´on o n de que f f es medible se tiene que para todo abierto abierto V R el conjunto Q = (x, y ) : f ( f (x, y ) V = f −1 (V ) V ) es medible. Usando la Proposici´on on 5.1.1 se tiene que
⊂
Qx
= =
{
{y {y
: (x, y ) : f x (y )
∈ }
f (x, y ) ∈ V } = ∈ Q} = {y : f ( −1 V ) ∈ V } = (f ) (V ) es medible y por consiguiente f es una funci´on on A 2 -medible. -medible. Usando Usando un argumento similar se demuestra que f es una funci´on on A1 -medible. x
x
y
5.2 5.2
Medi Medida dass Produ Product cto o
El teorema siguiente es la base para la definici´on on de medida producto, como se podr´a apreci apreciar ar posterior posteriormen mente. te. Lo que dice dice es que dada dada una funci´ funci´ on on
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67
A
∈ A × A
Demostraci´ on. on. Sea clase de conjuntos Q 1 2 para los cuales la conclusi´on on del teorema teorema se cumple. cumple. Queremos Queremos mostrar mostrar que la conclusi´ conclusi´on on del teorema vale para todos lo medibles, es decir queremos mostrar que = 1 Para esto primero veamos veamos que se cumplen cumplen las siguientes siguientes 2 . Para propiedades.
A A × A
(a) Todo rect´ angulo angulo medible est´a en
A.
⊂
⊂ ··· , y si cada Q est´a en A entonces el conjunto ∪ A. (c) Si (Q (Q ) es una colecci´on on disjunta numerable de conjuntos de A entonces el conjunto Q conjunto Q = ∪Q est´a en A. (d) Si µ(A) < ∞, λ(B ) < ∞, y A × B ⊃ Q 1 ⊃ Q2 ⊃ ·· ·, con todos los Q en A entonces el conjunto Q = ∩Q est´a en A. = A × B es un rect´angulo angulo medible, entonces parte (a) Si Q = A (b) Si Q1 Q2 Q = Qi est´a en
i
i
i
i
i
λ(Qx ) = λ( λ (B )ξ A (x) y µ(Qy ) = µ( µ(A)ξ B (x) y as´ı las integrales en (5.2) (5.2 ) son ambas a mbas iguales igua les a µ a µ((A)λ(B ), lo que demuestra la parte (a). parte (b)
Sean ϕi y ψ i las funciones asociadas (como en el enunciado del teorema) a Q a Q i . La propiedad de las medidas µ medidas µ y λ muestra que lim ϕi (x) = ϕ( ϕ (x),
lim ψi (y ) = ψ( ψ (y )
siendo la convergencia mon´otona otona creciente en cada punto. Del Teorema de la Convergencia Mon´onota, onota, tenemos que
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68
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
=
dλ
Ω2
χQ1 Q2 ···Qn dµ( dµ(x)
Ω1
˜ N = N Qi . Siendo Por lo tanto se cumple para la sucesi´on on creciente Q Siendo i=1 as´ as´ı, podemos usar la parte parte (b) para ver que tambi´ tambi´ en en se cumple cumple para el ∞ ∞ ˜ conjunto Q = i=1 Qi = N =1 QN . parte (d)
Es similar a la parte (b) s´olo olo que esta vez debemos usar el Teorema de la Convergencia Dominada, lo cual es posible debido a que µ(A) < y λ( λ (B ) < .
∞
∞
Siguiendo con la demostraci´on on del teorema. teorema. Como los espacios espacios de me∞ dida son σ son σ -finitos tenemos que Ω1 = i=1 E i , donde los conjuntos E i forman una colecci´on on eneumerable disjunta de conjuntos medibles con medida ∞ finita y Ω2 = j =1 F j , donde los conjuntos F j , son conjuntos disjuntos, medibles, con medida finita.
∩
× ∈ A
∈
M
Defina Q Defina Q m,n = Q (E m F n ) para cada m, cada m, n N y la clase de los Q que Q m,n para todo m, todo m, n. Tenemos lo siguiente: 1 2 tales que Q
∈ A ×A • Las afirmac afirmacion iones es (b) y (d) ante anterio riores res,, mon´ otona. otona.
dicen dicen que
M es una clase
• las afirmaciones (a) y (c) dicen que los conjuntos elementales est´an en M. De la definici´on on tenemos que M ⊂ A1 × A2 y como los hechos anteriores junto con el Teorema 5.1.1 implican que M ⊃ A1 × A2 obtenemos que M = A1 × A2. Hasta aqu´ aqu´ı, hemos demostrado que para todo Q todo Q ∈ A× A2 los conjuntos Q est´an a n en A . A partir partir de aqu´ aqu´ı, y consideran considerando do que Q es uni´on o n de m,n
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69
5.3 5.3
El Teore eorema ma de Fubin ubinii
Teorema 5.3.1 Sean (Ω1 , finitos, y f f una funci´ on 1
espacios A1, µ) y (Ω2, A2, λ) espacios A A × A2-medible en Ω1 × Ω2.
(a) Si 0
de medida medida σ-
≤ f ≤ ≤ ∞, y si
ϕ(x) =
f x dλ,
ψ(y ) =
Ω2
f y dµ,
(x
Ω1
A A1-medible, ψ es A A2-medible y ϕdµ = ϕdµ = f d(µ × λ) =
∈ Ω1, y ∈ Ω2). (5.3)
entonces ϕ es
Ω1
Ω1 ×Ω2
ψdλ
(5.4)
Ω2
× Ω2-integrable -integrable entonces entonces es λ-integrable para µ-casi todo x ∈ Ω1 , es µ-integrable para λ-casi todo y ∈ Ω2 ,
(b) Si f es f es una funci´ on real Ω1
• f • f • ϕ, que est´ a definida µ- c.t.p. , es µ-integrable, • ψ, que est´ a definida λ- c.t.p. , es λ-integrable. x
y
M´ as a´ un, se cumple la ecuaci´ on (5.4). (5.4). Demostraci´ on. El on. El teorema anterior implica que la parte (a) vale cuando f = χ Q es una funci´on on caracter´ caracter´ıstica medible. Por linealidad de la integral tambi´en en vale para par a funciones funcion es simples. simple s. Ahora, si f f es un funci´ on on no negativ negativa, a, entonces entonces existe una sucesi´ sucesi´on on mon´ otona de funciones simples que tales que otona f entonces f entonces por el Teo-
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70
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
Para probar la parte (b) descomponga f descomponga f en en su parte positiva y su parte + − negativa, esto es f = f f , de modo que f = f + + f − . Asociem Asociemos os + ϕ1 a f del mismo modo en que ϕ est´a asociados a f . f . De modo modo similar similar definimos ϕ definimos ϕ 2 .
−
| |
Usando la parte (a)
ϕ1 dµ = dµ =
Ω1
+
f d(µ
× λ)
f − d(µ
× λ)
Ω1 ×Ω2
ϕ2 dµ = dµ =
Ω1
Ω1 ×Ω2
≤ || ≤ ||
f <
∞
f <
∞
Como la funci´on on ϕ1 es no negativas, se tiene que ϕ1 es µ-integrables. M´ as as a´ un, un,
ϕ1 (x)dµ( dµ(x) <
Ω1
∞
lo que que impl implic icaa que que ϕ1 x) < para µ- c.t.p. c.t.p. x Ω1 . Pero ϕ1 (x) = + + ( y )dλ( dλ(y) es decir f x es λ-integrable en los mismos x donde ϕ(x) Ω2 f x (y es finita. Un razonamien razonamiento to an´ alogo alogo demuestra que (f (f y )+ es µ-integrable en los y los y donde ϕ donde ϕ 2 (y) es finito, esto es λ- c.t.p. c.t.p. y Ω2 .
∞
∈
∈
Como f Como f x = (f + )x (f − )x y f x = (f ( f + )x + (f − )x tendremos que f que f x es λ-integrable para los x donde tanto ϕ1 (x) como ϕ2 (x) son finitos. Lo que sucede µ- c.t.p. c.t.p. x Ω1 , pues ϕ1 y ϕ2 son integrables. integrable s. Observe tambi´ ta mbi´en en que ϕ que ϕ((x) = ϕ 1 (x) ϕ(x) lo que implica que ϕ es µ-integrable. µ -integrable.
−
| |
∈ −
Una vez comprobados estos hechos para ϕ, ϕ , la mitad de la ecuaci´on on 5.4 ya est´a probada
ϕdµ = ϕdµ =
f d(µ
× λ)
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71
Ejercicios 5.1 Sean (Ω1 , 1 ) y (Ω2 , 2 ) dos espacios espacios medibles. medibles. SI Aj para j = 1, 1 , . . . , n, n, entonces el conjunto 2 para j
A
A
A
∈ A1 y B ∈ j
n
(Aj , Bj )
j =1
×
pued puedee ser ser escr escrit itoo como como uni´ uni´ on o n disj disjun unta ta de una una cant cantid idaa finit finitaa de rect´ angulos angulos en Ω1 Ω2 .
× 5.2 Sea A Sea A ⊂ Ω1 y B ⊂ Ω2 , j = 1, 2. entonces (A1 × B1 ) \ (A2 × B2 ) = [(A [(A1 ∩ A2 ) × (B1 \ B2 )] ∪ [(A [(A1 \ A2 ) × B1 ] j
j
y tamb ta mbi´ i´en en (A1
× B1) ∩ (B1 × B2) = (A1 ∩ A2) × (B1 ∩ B2).
5.3 Dada una funci´
f
R
R medibl medibl
cto a la medida
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72
´ n de Medidas cap´ ıtulo ıtulo 4. Descomposi Descomposici cion o Medidas
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Ap´ endice A
De Abiertos, Cerrados y Aproximaci´ on on de Conjuntos Medibles
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74
´ ´ n de conjunto apendice endice A. aproximaci aproximacion o conjuntos s medibles medibles
C (y, 2n1 z ), con la mitad mitad del radio radio que el anter anterior ior.. Como Como Q es denso en 1 R existen infinitos puntos racionales en G C (y, 2n ). Sea Sea ω el primero z (considerando el orden de la enumeraci´on on de Q). Luego
∩
∈ G ∩ C (y, 21
ω
nz
) lo que implica que C ( C (ω,
1 2nz )
⊂ C (y,
1 nz
) y adem´as as
∈ C (ω, n1 ) ⊂ C (ω, n1 ) ⊂ G0
y
z
lo que implica que G que G
ω
⊂ G0 y el lema esta probado.
Teorema A.0.2 Todo abierto y todo cerrado de R es Lebesgue medible. Demostraci´ on. Todo on. Todo abierto es medible pues es uni´on on numerable de intervalos abiertos, que son medibles. Todo cerrado es complemento de abiertos y por lo tanto medible.
⊂ ⊂ R con m∗(E ) = 0 se llama conjunto con medida
Un subconjunto E
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75
´ n A.1. aproxima ´ n por abiertos seccion o aproximaci cion o abiertos
y tamb ta mbi´ i´en en x
c
c
⊕ B = (x ⊕ B) . Si tenemos x tenemos x,, z con x con x = −z , entonces de la invariancia de m de m ∗ resulta m∗ ((z ((z ⊕ A) ∩ B ) = m∗ (A ∩ (x ⊕ B )) . Sea E Sea E un un conjunto medible, usando los resultados anteriores con B = B = E E c y B = E obtenemos m∗ (A) = m∗ (z A) = m∗ ((z ((z A) E ) + m∗ ((z ((z c ∗ ∗ = m (A (x E )) )) + m (A (x B )) = m∗ (A (x E )) )) + m∗ (A (x E )c ).
⊕ ∩ ⊕ ∩ ⊕
⊕ ∩ ∩ ⊕ ∩ ⊕
c
⊕ A) ∩ E )
Las ecuaciones anteriores valen para todo A medible, A medible, por lo tanto x tanto x E es ∗ ∗ medible, y como m como m (x E ) = m (E ), ), tenemos que m que m((x E ) = m( m (E ). ).
⊕
A.1 A.1
Apro Aproxi xima maci ci´ on ´ on por Abiertos
⊕
⊕
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76
´ ´ n de conjunto apendice endice A. aproximaci aproximacion o conjuntos s medibles medibles
que junto con la definici´on on de ´ınfimo ınfimo prueba esta parte del lema.
∈
Prueba Prueba de (ii). (ii). Para Para cada n N llamamos G llamamos G n al abierto obtenido usando la parte (i) del lema con ε = ε = 1/n, /n, es decir m(Gn ) Definimos el conjunto H conjunto H como
≤ m∗(A) + n1 .
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77
´ n A.1. aproxima ´ n por abiertos seccion o aproximaci cion o abiertos
∞ se tiene m(G \ E ) = m( m(G) − m(E ) < ε. Ahora, si m si m((E ) = ∞ definimos E definimos E 1 = E ∩ {x : ||x|| ≤ 1}, E = E ∩ {x : n − 1 < ||x|| ≤ n}, y E = E = E . De modo que para todo n todo n ∈ N podemos tomar G tomar G abierto con E con E ⊂ G ∞ y tal que m(G \ E ) < 2 . Si definimos G definimos G = = G , tendremos que G que G =1 ∞ ⊂ G y G \ E ⊂ ⊂ =1(G \ E ) y por lo tanto es abierto, E abierto, E ⊂ y como m como m((E ) <
n
N
n
n
n
ε
n
n
n
n
n
n
n
n
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78
A.2 A.2
´ ´ n de conjunto apendice endice A. aproximaci aproximacion o conjuntos s medibles medibles
Apro Aproxi xima maci ci´ on ´ on por Cerrados
⊂ ⊂
Teorema A.2.1 Un conjunto E R es Lebesgue medible si, y solamente si, para cada ε > 0, 0, existe un conjunto F F cerrado, con F E y
⊂ ⊂
m∗ (E F ) F ) < ε .
\
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79
´ n A.3. aproxima ´ n por compactos seccion o aproximaci cion o compactos
A.3 A.3
Apro Aproxi xima maci ci´ on ´ on por Compactos
Recordamos el hecho b´asico asico de que un compacto tiene medida finita pues esta contenido en una intervalo suficientemente grande. Teorema A.3.1 Un conjunto E conjunto E R con m m ∗ (E ) < es Lebesgue medible si, y solamente si, para todo 0 existe un conjunto C , compacto tal que
⊂ ⊂
∞
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80
A.4 A.4
´ ´ n de conjunto apendice endice A. aproximaci aproximacion o conjuntos s medibles medibles
Apro Aproxi xima maci ci´ on ´ on por Intervalos
Teorema A.4.1 Si E E es Lebesgue medible con medida finita y ε > 0, 0 , enn tonces existen intervalos abiertos (limitados) I i i=1 tal que si K K = ni=1 I i entonces m(E K ) < ε .
{ }
∪
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