MATH 361 Homework 9
Royden 3.3.9 First, we show that for any subset E subset E of of the real numbers, E c + y = (E + E + y )c (translating the complement is equivalent to the complement of the translated set). Without loss of generality, assume E can E can be written as c an open interval (e (e1 , e2 ), so that E that E + y is represen represented ted by the set { x|x ∈ (−∞, e1 + y ) ∪ (e2 + y, +∞)}. This is equal to the set { x|x ∈ / (e1 + y, e2 + y)}, which is equivalent to the set (E (E + + y)c . Second, Let B = A − y. From Homework Homework 8, we know that outer measure is invarian invariantt under translation translations. s. Using this along with the fact that E that E is is measurable: m∗ (A) = m ∗ (B ) = m ∗ (B ∩ E ) + m∗ (B ∩ E c ) = m ∗ ((B ((B ∩ E ) + y ) + m∗ ((B ((B ∩ E c ) + y ) = m ∗ (((A (((A − y ) ∩ E ) + y) + m∗ (((A (((A − y ) ∩ E c ) + y) = m ∗ (A ∩ (E + + y)) + m∗ (A ∩ (E c + y )) = m ∗ (A ∩ (E + + y)) + m∗ (A ∩ (E + + y )c ) The last line follows from E from E c + y = (E + + y)c .
Royden 3.3.10 First, since E since E 1 , E 2 ∈ M and M is a σ -algebra, E -algebra, E 1 ∪ E 2 , E 1 ∩ E 2 ∈ M. By the measurability of E E 1 and E and E 2 : m∗ (E 1 ) = m ∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∩ E 2c ) m∗ (E 2 ) = m ∗ (E 2 ∩ E 1 ) + m∗ (E 2 ∩ E 1c ) m∗ (E 1 ) + m∗ (E 2 ) = 2m∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∩ E 2c ) + m∗ (E 1c ∩ E 2 ) = m ∗ (E 1 ∩ E 2 ) + [m [ m∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∩ E 2c ) + m∗ (E 1c ∩ E 2 )] Second, E 1 ∩ E 2 , E 1 ∩ E 2c , and E 1c ∩ E 2 are disjoint sets whose union is equal to E 1 ∪ E 2 . As above, above, since since ∗ ∗ c c c c E 1 , E 2 ∈ M, E 1 , E 2 ∈ M and hence E 1 ∩ E 2 , E 1 ∩ E 2 ∈ M. From clas class, s, m (∪n E n ) = n m (E n ) for measurable sets E sets E n . Therefore:
m∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∩ E 2c ) + m∗ (E 1c ∩ E 2 ) = m ∗ (E 1 ∪ E 2 ) Combining the two arguments above: m∗ (E 1 ) + m∗ (E 2 ) = m ∗ (E 1 ∩ E 2 ) + [m [ m∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∩ E 2c ) + m∗ (E 1c ∩ E 2 )] = m ∗ (E 1 ∩ E 2 ) + m∗ (E 1 ∪ E 2 )
Royden 3.3.11 Define E Define E n = (n, +∞). i. Empty intersection: ∞
E = n
∅
i=1
For any x any x ∈ R, we can choose a natural number n > x such that x ∈ / E n . Therefore, there is no x ∈ R such that x that x ∈ E n for all n all n.. This implies that the intersection intersection stated above above is empty empty. ii. By definition, definition, m m ∗ (E n ) = +∞, as each interval is an open interval containing + ∞. 1
Royden 3.3.12 i. From lecture, we established the following for a countable sequence of E i : n
n
m A E = m (A ∩ E ) ∗
∗
i
i
i=1
i=1
For the infinite case, we use the monotonicity property: n
∞
A E ⊃ A E m A E E ≥m A m (A ∩ E ) ≥ i
i
i=1
i=1
n
∞
∗
∗
i
i
i=1
i=1
n
∗
i
i=1
Since this is true for all n ∈ N, letting n → ∞: ∞
∞
A E ≥ m (A ∩ E ) ∗
i
i
i=1
i=1
ii. The reverse inequality is true by countable subadditivity: ∞
∞
A E ≤ m (A ∩ E ) ∗
i
i
i=1
i=1
From the two inequalities in parts (i) and (ii), we can conclude: ∞
∞
A E = m (A ∩ E ) ∗
i
i
i=1
i=1
Royden 3.3.13 a. Showing (i)⇒(ii)⇔(vi). (i)⇒(ii): By proposition 5 in Royden, for all sets E , there exists an open set O such that E ⊂ O and m∗ (O) ≤ m∗ (E ) + . Since E is measurable, for such a set O: m∗ (O) = m ∗ (O ∩ E ) + m∗ (O ∩ E c ) m∗ (O ∩ E ) + m∗ (O ∩ E c ) ≤ m∗ (E ) + m∗ (E ) + m∗ (O \ E ) ≤ m∗ (E ) + m∗ (O \ E ) ≤ To make the inequality above strict, we can take = 2 for any given > 0 and use the same reasoning above. (ii)⇒(iv): Since O is open, it can be written as a countable disjoint union of open intervals. We pick an open O such that m ∗ (O\E ) < /2. We consider two cases: Case 1: O is an infinite union of open intervals: ∞
I O = m (I ) m (O) ≤ n
n=1 ∞
∗
∗
n
n=1
2
From (ii), it is given that m ∗ (E ) < ∞. So: m∗ (O) ≤ m∗ (E ) + m∗ (O \ E ) <∞ Since the outer measure is finite, the infinite sum above must converge. Therefore, there exists some N such that for all n > N , ∞ N I n < 2 . Define:
N
I U =
n
n=1
Case 2: O is a finite union of K intervals. Then define: K
I U =
n
n=1
The symmetric measure can be decomposed into a union of disjoint sets: U ∆E = (U \ E ) ∪ (E \ U ) m (U ∆E ) ≤ m∗ (U \ E ) + m∗ (E \ U ) ≤ m∗ (O \ E ) + m∗ (O \ U ) < m∗ (O \ E ) + /2 < ∗
(vi)⇒(ii) By Proposition 5, there exists some open set Q such that (E \ U ) ⊂ Q and m∗ (Q) ≤ m∗ (E \ U ) + . Define O = U ∪ Q. The set O covers E , since (U ∩ E ) ⊂ U and (E \ U ) ⊂ Q. Then: m∗ (O \ E ) = m ∗ ((U ∪ Q) \ E ) = m ∗ ((U \ E ) ∪ (Q \ E )) ≤ m∗ (U \ E ) + m∗ (Q \ E ) ≤ m∗ (U \ E ) + m∗ (Q) ≤ m∗ (U ∆E ) + m∗ (E \U ) + ≤ 2m∗ (U ∆E ) + < 3 b. Showing (i)⇒(ii)⇒(iv)⇒(i) (i)⇒(ii): Shown in part (a) above. (ii)⇒(iv): Define G as follows: G =
O
n
n
such that m∗ (O \ E ) < n1 for all n ∈ N. The existence of such On is guaranteed by the condition givne in (ii). Since E ⊂ On for each n, E ⊂ G, and by the monotonicity property: m∗ (G \ E ) ≤ m∗ (On \ E ) 1 < n Since the inequality above holds for all n ∈ N, we must have m ∗ (G \ E ) = 0. (iv)⇒(i): Since G is a countable intersection of open sets, it is measurable. All sets with measure zero are measurable, so given that m∗ (G \ E ) = 0, the set G \ E = G ∩ E c is also measurable. Therefore, (G ∩ E c )c is measurable, and G ∩ (G ∩ E c )c = E is also measurable. 3
c. Showing (i)⇒(iii)⇒(v)⇒(i) (i)⇒(iii): Since E is measurable, E c is also measurable. By (ii), there exists some O such that m ∗ (O \ E c ) < , or equivalently, m ∗ (O ∩ E ) < . Define F = O c . Then: > m∗ (O ∩ E ) > m∗ (E \ Oc ) > m∗ (E \ F ) Since O is open, F is closed. Since E c ⊂ O, we know O c ⊂ E , or equivalently, F ⊂ E . (iii)⇒(v): From (iii), there exists a closed set F n with F n ⊂ E and m∗ (E \ F n ) < Define the following: F =
1
n.
for all n ∈
N.
F
n
n
Since F n ⊂ E for all n ∈ N, F ⊂ E . By monotonicity: m∗ (E \ F ) ≤ m∗ (E \ F n ) 1 < n Since the inequality holds for all n ∈ N, we can conclude m ∗ (E \ F ) = 0. (v)⇒(i): Since F is measurable and m∗ (E \ F ) = 0, E \ F is also measurable. Since E = F ∪ (E \ F ), the union of disjoint, measurable sets, E is also measurable.
Royden 3.3.14 n
a. Define E 0 = [0, 1], E 1 = [0, 31 ] ∪ [ 23 , 1], E n = [0, 31 ] ∪ · · · ∪ [ 3 3−1 , 1]. The Cantor set is equal to the intersection of E n for all n ∈ N. In particular, E n is a descending sequence of measurable sets, as E n+1 ⊂ E n , and m(E 1 ) is finite. By a proposition proven in lecture: n
n
∞
m
E n
n=1
= lim m(E n ) n→∞
Each E n is a union of disjoint closed intervals I n . Since closed intervals are measurable, and m(I ) = (I ), we know that m(E n ) = m(∪2n I n ) = 2n=1 m(I n ) Therefore, it is sufficient to show that the sum of the intervals which make up E n as n → ∞ is equal to zero in order to show that the Cantor ternary set has measure zero. For any n, E n is a union of 2 n closed invervals each with length 31 . The sum of the lengths of each interval is ( 23 )n and therefore m(E n ) = ( 23 )n . n
n
n
For any > 0, take n to be the first natural number such that n > log 2/3 . This forces m(E n ) < , which then implies limn→∞ m(E n ) = 0. b. F is equal to a countable union of closed intervals, and is therefore a closed set. To show F c is dense, define F n to be the remaining closed intervals at each stage n after the middle interval of length 3α is removed. Then F is a countable union of disjoint intervals, each with length strictly less than 21 . Therefore, given any x ∈ [0, 1] and > 0, choose N > log 1/2 . This will ensure that the interval (x − 21 , x + 21 ) contains some point y that was removed in the n th step. n
n
Using the same reasoning as in part (a), the measure of F is equal to the sum of the disjoint intervals whose union is equal to F . At any stage n, there are 2 n−1 intervals before any deletions are made. 4
Therefore, 2n−1 intervals of length
α 3n
are removed. Therefore: ∞
α m(F ) = 1 − 2 3 α 2 =1 − n−1
n
n=1
n
∞
2
3
n=1
=1 −α
Question 3 i Since A 1 is measurable: m∗ (A2 ) = m ∗ (A2 ∩ A1 ) + m∗ (A2 ∩ Ac1 ) m∗ (A1 ) = m ∗ (A1 ) + m∗ (A2 ∩ Ac1 ) m∗ (A2 ∩ Ac1 ) = 0 ii Given that m∗ (B) = 0, where B is a subset of R, B must be measurable, since for any subset C , m∗ (C ∩ B) ≤ m∗ (B), so m ∗ (C ∩ B) = 0. m∗ (C ∩ B) + m∗ (C ∩ B c ) = m ∗ (C ∩ B c ) ≤ m∗ (C ) iii From the two parts above, we can conclude that the set A2 ∩ Ac1 is a measurable set. Hence, we can write A 2 as the union of two measurable sets, A 1 and A 2 ∩ Ac1 . From the lecture notes, the collection of measurable sets, M is closed under taking unions. Therefore, A 2 = A 1 ∪ (A2 ∩ Ac1 ) is also measurable.
Question 4 By countable additivity: ∞
∞
m B m (B ) ≤ ∗
∗
n
n
n=1
n=1
To show the reverse inequality, define B = ∪∞ n=1 Bn and we note that N
∞
⊃ B B A A n
n
n=1
n=1
By the countability of A n : ∞
m
∗
Bn
n=1
∞
= m B A A ≥m B m (B ∩ A ) ≥ m (B ∩ A ) ≥ m (B ) ≥ ∗
n
n=1 N
∗
n
n=1
∗
n
n
∗
n
n
∗
n
n
5
n
