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 fxjx 2 (−∞; e1 + y) [ (e2 + y; +1)g. This c is equal to the set fxjx2 = (e1 + y; e2 + y)g, 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 2 M and M is a σ-algebra, E1 [ E2;E1 \ E2 2 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 2 M, E1;E2 2 M and hence E1 \ E2;E1 \ E2 2 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; +1). i. Empty intersection: 1 \ En = ? i=1 For any x 2 R, we can choose a natural number n > x such that x2 = En. Therefore, there is no x 2 R such that x 2 En for all n. This implies that the intersection stated above is empty. ∗ ii. By definition, m (En) = +1, as each interval is an open interval containing +1. 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: 1 n \ [ \ [ A Ei ⊃ A Ei i=1 i=1 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 2 N, letting n ! 1: 1 1 \ [ X ∗ A Ei ≥ m (A \ Ei) i=1 i=1 ii. The reverse inequality is true by countable subadditivity: 1 1 \ [ X ∗ A Ei ≤ m (A \ Ei) i=1 i=1 From the two inequalities in parts (i) and (ii), we can conclude: 1 1 \ [ 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 n E) ≤ m∗(E) + m∗(O n 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∗(OnE) < /2. We consider two cases: Case 1: O is an infinite union of open intervals: 1 [ O = In n=1 1 ∗ X ∗ m (O) ≤ m (In) n=1 2 From (ii), it is given that m∗(E) < 1. So: m∗(O) ≤ m∗(E) + m∗(O n E) < 1 Since the outer measure is finite, the infinite sum above must converge. Therefore, there exists some N P1 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 n E) [ (E n U) m∗(U∆E) ≤ m∗(U n E) + m∗(E n U) ≤ m∗(O n E) + m∗(O n U) < m∗(O n E) + /2 < (vi))(ii) By Proposition 5, there exists some open set Q such that (E n U) ⊂ Q and m∗(Q) ≤ m∗(E n U) + . Define O = U [ Q. The set O covers E, since (U \ E) ⊂ U and (E n U) ⊂ Q. Then: m∗(O n E) = m∗((U [ Q) n E) = m∗((U n E) [ (Q n E)) ≤ m∗(U n E) + m∗(Q n E) ≤ m∗(U n E) + m∗(Q) ≤ m∗(U∆E) + m∗(EnU) + ≤ 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 n E) < n for all n 2 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 n E) ≤ m (On n E) 1 < n Since the inequality above holds for all n 2 N, we must have m∗(G n 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 n E) = 0, the set G n 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 n Ec) < , or equivalently, m∗(O \ E) < . Define F = Oc. Then: > m∗(O \ E) > m∗(E n Oc) > m∗(E n 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 n Fn) < n . for all n 2 N. Define the following: [ F = Fn n Since Fn ⊂ E for all n 2 N, F ⊂ E. By monotonicity: ∗ ∗ m (E n F ) ≤ m (E n Fn) 1 < n Since the inequality holds for all n 2 N, we can conclude m∗(E n F ) = 0. (v))(i): Since F is measurable and m∗(E n F ) = 0, E n F is also measurable. Since E = F [ (E n 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 2 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: 1 \ m En = lim m(En) n!1 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 ! 1 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!1 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 2 [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: 1 X α m(F ) = 1 − 2n−1 3n n=1 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.