MATH 361 Homework 9

Royden 3.3.9 First, we show that for any subset E of the real numbers, Ec + y = (E + y)c (translating the complement is equivalent to the complement of the translated set). Without loss of generality, assume E can be written as c an open interval (e1, e2), so that E + y is represented by the set {x|x ∈ (−∞, e1 + y) ∪ (e2 + y, +∞)}. This c is equal to the set {x|x∈ / (e1 + y, e2 + y)}, which is equivalent to the set (E + y) . Second, Let B = A − y. From Homework 8, we know that outer measure is invariant under translations. Using this along with the fact that E is measurable:

m∗(A) = m∗(B) = m∗(B ∩ E) + m∗(B ∩ Ec) = m∗((B ∩ E) + y) + m∗((B ∩ Ec) + y) = m∗(((A − y) ∩ E) + y) + m∗(((A − y) ∩ Ec) + y) = m∗(A ∩ (E + y)) + m∗(A ∩ (Ec + y)) = m∗(A ∩ (E + y)) + m∗(A ∩ (E + y)c)

The last line follows from Ec + y = (E + y)c. Royden 3.3.10

First, since E1,E2 ∈ M and M is a σ-algebra, E1 ∪ E2,E1 ∩ E2 ∈ M. By the measurability of E1 and E2:

∗ ∗ ∗ c m (E1) = m (E1 ∩ E2) + m (E1 ∩ E2) ∗ ∗ ∗ c m (E2) = m (E2 ∩ E1) + m (E2 ∩ E1) ∗ ∗ ∗ ∗ c ∗ c m (E1) + m (E2) = 2m (E1 ∩ E2) + m (E1 ∩ E2) + m (E1 ∩ E2) ∗ ∗ ∗ c ∗ c = m (E1 ∩ E2) + [m (E1 ∩ E2) + m (E1 ∩ E2) + m (E1 ∩ E2)]

c c Second, E1 ∩ E2, E1 ∩ E2, and E1 ∩ E2 are disjoint sets whose union is equal to E1 ∪ E2. As above, since c c c c ∗ P ∗ E1,E2 ∈ M, E1,E2 ∈ M and hence E1 ∩ E2,E1 ∩ E2 ∈ M. From class, m (∪nEn) = n m (En) for measurable sets En. Therefore:

∗ ∗ c ∗ c ∗ m (E1 ∩ E2) + m (E1 ∩ E2) + m (E1 ∩ E2) = m (E1 ∪ E2)

Combining the two arguments above:

∗ ∗ ∗ ∗ ∗ c ∗ c m (E1) + m (E2) = m (E1 ∩ E2) + [m (E1 ∩ E2) + m (E1 ∩ E2) + m (E1 ∩ E2)] ∗ ∗ = m (E1 ∩ E2) + m (E1 ∪ E2)

Royden 3.3.11

Define En = (n, +∞).

i. Empty intersection:

∞ \ En = ∅ i=1

For any x ∈ R, we can choose a natural number n > x such that x∈ / En. Therefore, there is no x ∈ R such that x ∈ En for all n. This implies that the intersection stated above is empty. ∗ ii. By definition, m (En) = +∞, 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 Ei:   n  n ∗ \ [ X ∗ m A Ei = m (A ∩ Ei) i=1 i=1 For the infinite case, we use the monotonicity property: ∞ n \  [  \  [  A Ei ⊃ A Ei i=1 i=1   ∞    n  ∗ \ [ ∗ \ [ m A Ei ≥ m A Ei i=1 i=1 n X ∗ ≥ m (A ∩ Ei) i=1 Since this is true for all n ∈ N, letting n → ∞:  ∞  ∞ \ [ X ∗ A Ei ≥ m (A ∩ Ei) i=1 i=1 ii. The reverse inequality is true by countable subadditivity:  ∞  ∞ \ [ X ∗ A Ei ≤ m (A ∩ Ei) i=1 i=1

From the two inequalities in parts (i) and (ii), we can conclude:  ∞  ∞ \ [ X ∗ A Ei = m (A ∩ Ei) 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 ∩ Ec) m∗(O ∩ E) + m∗(O ∩ Ec) ≤ m∗(E) +  m∗(E) + m∗(O \ E) ≤ m∗(E) +  m∗(O \ E) ≤  0  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: ∞ [ O = In n=1 ∞ ∗ X ∗ m (O) ≤ m (In) 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 P∞  such that for all n > N, N In < 2 . Define: N [ U = In n=1 Case 2: O is a finite union of K intervals. Then define:

K [ U = In 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 = On n ∗ 1 such that m (O \ E) < n 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 ∩ Ec is also measurable. Therefore, (G ∩ Ec)c is measurable, and G ∩ (G ∩ Ec)c = E is also measurable.

3 c. Showing (i)⇒(iii)⇒(v)⇒(i)

(i)⇒(iii): Since E is measurable, Ec is also measurable. By (ii), there exists some O such that m∗(O \ Ec) < , or equivalently, m∗(O ∩ E) < . Define F = Oc. Then:

 > m∗(O ∩ E) > m∗(E \ Oc) > m∗(E \ F )

Since O is open, F is closed. Since Ec ⊂ O, we know Oc ⊂ E, or equivalently, F ⊂ E.

∗ 1 (iii)⇒(v): From (iii), there exists a closed set Fn with Fn ⊂ E and m (E \ Fn) < n . for all n ∈ N. Define the following: [ F = Fn n

Since Fn ⊂ E for all n ∈ N, F ⊂ E. By monotonicity:

∗ ∗ m (E \ F ) ≤ m (E \ Fn) 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

1 2 1 3n−1 a. Define E0 = [0, 1], E1 = [0, 3 ] ∪ [ 3 , 1], En = [0, 3n ] ∪ · · · ∪ [ 3n , 1]. The Cantor set is equal to the intersection of En for all n ∈ N. In particular, En is a descending sequence of measurable sets, as En+1 ⊂ En, and m(E1) is finite. By a proposition proven in lecture:

∞  \  m En = lim m(En) n→∞ n=1

Each En is a union of disjoint closed intervals In. Since closed intervals are measurable, and m(I) = `(I), n 2n P2 we know that m(En) = m(∪n In) = n=1 m(In) Therefore, it is sufficient to show that the sum of the intervals which make up En as n → ∞ is equal to n zero in order to show that the Cantor ternary set has measure zero. For any n, En is a union of 2 closed 1 2 n 2 n invervals each with length 3n . The sum of the lengths of each interval is ( 3 ) and therefore m(En) = ( 3 ) .

For any  > 0, take n to be the first natural number such that n > log2/3 . This forces m(En) < , which then implies limn→∞ m(En) = 0. b. F is equal to a countable union of closed intervals, and is therefore a closed set.

c To show F is dense, define Fn to be the remaining closed intervals at each stage n after the middle α interval of length 3n is removed. Then F is a countable union of disjoint intervals, each with length 1 strictly less than 2n . Therefore, given any x ∈ [0, 1] and  > 0, choose N > log1/2 . This will ensure that 1 1 th the interval (x − 2 , x + 2 ) contains some point y that was removed in the n step.

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 2n−1 intervals before any deletions are made.

4 n−1 α Therefore, 2 intervals of length 3n are removed. Therefore:

∞ X  α  m(F ) = 1 − 2n−1 3n n=1 ∞ n α X 2 = 1 − 2 3 n=1 = 1 − α Question 3

i Since A1 is measurable:

∗ ∗ ∗ c m (A2) = m (A2 ∩ A1) + m (A2 ∩ A1) ∗ ∗ ∗ c m (A1) = m (A1) + m (A2 ∩ A1) ∗ c m (A2 ∩ A1) = 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 ∩ Bc) = m∗(C ∩ Bc) ≤ m∗(C)

c iii From the two parts above, we can conclude that the set A2 ∩ A1 is a measurable set. Hence, we can c write A2 as the union of two measurable sets, A1 and A2 ∩ A1. From the lecture notes, the collection of c measurable sets, M is closed under taking unions. Therefore, A2 = A1 ∪ (A2 ∩ A1) is also measurable. Question 4 By countable additivity:  ∞  ∞ ∗ [ X ∗ m Bn ≤ m (Bn) n=1 n=1

∞ To show the reverse inequality, define B = ∪n=1Bn and we note that

∞ N  \  [   \  [  B An ⊃ B An n=1 n=1

By the countability of An:  ∞    ∞  ∗ [ ∗ \ [ m Bn = m B An n=1 n=1   N  ∗ \ [ ≥ m B An n=1 X ∗ ≥ m (B ∩ An) n X ∗ ≥ m (Bn ∩ An) n X ∗ ≥ m (Bn) n

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