Real Analysis Chapter 1 Solutions Jonathan Conder
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1 3. (a) Let M be an infinite σ-algebra of subsets of some set X. There exists a countably infinite subcollection C M, and we may choose C to be closed under taking complements (adding in missing complements if necessary). For each x X, define D x := {C C x C}, so that D x M. Let x, y X and suppose y D x. Then y C for all C C with x C. Moreover, if y C for some C C, then y / C c C, so x / C c and hence x C. This implies that, if C C, then x C iff y C. In particular D x = D y. If x, y X and there exists z D x D y, then D x = D z = D y, so the collection D := {D x x X} is pairwise disjoint. If C C, then C = {D x x C}, since x D x C for all x C. Therefore, the image of the map : 2 D 2 X contains C, so there exists a surjection from some subset of 2 D onto C. If D were finite, then 2 D would be as well, which is a contradiction because C is infinite. Therefore D is an infinite, pairwise disjoint subcollection of M. (b) Let M be an infinite σ-algebra of subsets of some set X. By the previous exercise, there exists an infinite, pairwise disjoint subcollection D of M. Since Q is countable, there exists an injection q : Q D. Define r : 2 Q M by r(a) := a A q(a). Then r is well-defined because each A Q is countable, and injective because D is pairwise disjoint. There exists an injection from R 2 Q, for example the map x (, x] Q (which is injective by the least upper bound property of R). Composing these injections gives an injective map from R M, which shows that card(m) c. 4. Let A be an algebra which is closed under countable increasing unions. To show that A is a σ-algebra, it suffices to show that it is closed under arbitrary countable unions. To this end, let {E n } be a countable collection of members of A. For each n N define F n := n E k. Since A is an algebra it is clear that (F n ) is an increasing sequence of members of A, so E n = F n A as required. Conversely, it is plain that every σ-algebra is closed under countable increasing unions. 5. Let E be a collection of subsets of some set X, and let M be the σ-algebra generated by E. Define N := {A A is the σ-algebra generated by some countable subcollection of E}. It is clear that, X N, E N and N is closed under taking complements. Let {E n } be a countable collection of members of N. For each n N there exists a countable subcollection E n of E such that E n is in the σ-algebra generated by E n. Since union of countably many countable sets is countable, F := E n is a countable subcollection of E. Let A be the σ-algebra generated by F, so that A N. For each n N the σ-algebra generated by E n is a subset of A, because A is a σ-algebra containing E n. This implies that E n A for all n N, and hence E n A N. Therefore N is closed under countable unions, so it is a σ-algebra containing E. This implies that M N. Conversely, let E N. Then E belongs to the σ-algebra A generated by some countable subcollection of E. Since M is a σ-algebra containing this subcollection, A M and hence E M. This shows that N M and hence N = M. 6. Note that µ( ) = µ( ) = µ( ) = 0. Given a sequence (A n ) of disjoint sets in M, there exists a sequence (E n ) of sets in M and a sequence (F n ) of subsets of measure zero sets from M such that A n = E n F n for all n N. Note that (E n ) is pairwise disjoint, and F n is a subset of a measure zero set. Therefore µ( A n ) = µ(( E n ) ( F n )) = µ( E n ) = µ(e n ) = µ(a n ). This shows that µ is a measure. Let A X, and suppose there exists B M such that A B and µ(b) = 0. Then B = E F for some E M and F a subset of a measure zero set N M. It follows that A E N, where E N M and µ(e N) µ(e) + µ(n) = µ(e) + 0 µ(b) = 0. Therefore A = A M, which shows that µ is complete. 1
2 Let λ : M [0, ] be another measure which extends µ, and let A M. Then A = E F for some E M and F a subset of a measure zero set N M. It follows that µ(e) = λ(e) λ(a) λ(e) + λ(f ) λ(e) + λ(n) = µ(e) + µ(n) = µ(e). Therefore λ(a) = µ(e) = µ(a), so λ = µ. This shows that µ is the unique measure on M which extends µ. 7. Set µ := n j=1 a jµ j, and note that µ( ) = 0. If {E i } i N is a pairwise disjoint collection of members of M, then µ( i N E i ) = a j µ j ( i N E i ) = j=1 a j j=1 µ j (E i ) = a j µ j (E i ) = i N i N j=1 i N (note that we are allowed to change the order of summation because all the terms are non-negative; alternatively you µ(e i ) can consider the convergent/divergent cases separately). This shows that µ is a measure on (X, M). 8. Let (X, M, µ) be a measure space and {E n } a countable collection of measurable sets. Then ( k=n E k) is an increasing sequence of measurable sets with µ( k=n E k) µ(e n ) for all n N, so µ(lim inf E n) = µ( ( k=n E k)) = lim n µ( k=n E k) = lim inf n µ( k=n E k) lim inf n µ(e n). Moreover ( k=n E k) is a decreasing sequence of measurable sets with µ( k=n E k) µ(e n ), so µ(lim sup E n ) = µ( ( k=n E k)) = lim n µ( k=n E k) = lim sup n provided that the first term of the sequence has finite measure, i.e. µ( E k) <. µ( k=n E k) lim sup µ(e n ) n 9. Let (X, M, µ) be a measure space and E, F M. Then E = (E F ) (E \ F ) and (E \ F ) F = E F, so µ(e) + µ(f ) = µ(e F ) + µ(e \ F ) + µ(f ) = µ(e F ) + µ(e F ). 10. Clearly µ E ( ) = µ( E) = µ( ) = 0. If {E n } is a pairwise disjoint collection of members of M, then µ E ( E n ) = µ(( E n ) E) = µ( (E n E)) = µ(e n E) = µ E (E n ). Therefore µ E is a measure. 11. Suppose that µ is continuous from below. Let {E n } be a pairwise disjoint collection of measurable sets, and for each n N define F n := n i=1 E n. Since µ is continuous from below and finitely additive, µ( i N E i ) = µ( F n ) = lim n µ(f n) = lim n µ(e i ) = i=1 µ(e i ). i=1 This shows that µ is a measure. Conversely, if µ is a measure it is continuous from below by Theorem 1.8. Now suppose that µ(x) < and µ is continuous from above. If (E n ) is an increasing sequence of measurable sets, then (E c n) is a decreasing sequence of measurable sets, which implies that µ( E n ) = µ(x) µ(( E n ) c ) = µ(x) µ( E c n) = µ(x) lim n µ(ec n) = lim n µ(e n). This shows that µ is continuous from below, so it is a measure by the previous paragraph. Conversely, if µ is a (finite) measure it is continuous from above by Theorem
3 14. Suppose for a contradiction that there exists C (0, ) such that every measurable subset F E satisfies µ(f ) C or µ(f ) =. Set M := sup{µ(f ) F E is measurable and µ(f ) < }, and note that 0 M C. For each n N there exists a measurable subset E n E such that M n 1 µ(e n ) <. Set F n := n i=1 E n for each n N and define F := F n. Note that M n 1 µ(e n ) µ(f ) and also µ(f n ) n i=1 µ(e n) < for all n N, so M µ(f ) = lim n µ(f n ) M. This shows that µ(f ) = M, so µ(e \ F ) =. Since µ is semifinite, there exists a measurable subset A E \ F such that 0 < µ(a) <. This contradicts the definition of M, because A F E but µ(f ) < µ(a) + µ(f ) = µ(a F ) <. Therefore, for any C (0, ) there exists a measurable subset F E such that C < µ(f ) <. 17. Let A, B X be disjoint µ -measurable sets, and let E X. Then (A B) A c = B and hence µ (E (A B)) = µ (E (A B) A) + µ (E (A B) A c ) = µ (E A) + µ (E B). It follows by induction that n µ (E A k ) = µ (E ( n A k)) µ (E ( A k)) for all n N. Therefore µ (E A k ) µ (E ( A k)), which implies that µ (E ( A k)) = µ (E A k ) because µ is subadditive. 18. (a) Let E X and ε (0, ). If µ (E) = then X A A σ, E X and µ (X) µ (E) + ε. Otherwise { } µ (E) = inf µ 0 (A n ) A n A for all n N and E A n, so there exists a sequence (A n ) in A such that E A n and µ 0(A n ) µ (E) + ε. If A := A n, then A A σ, E A and µ (A) µ 0(A n ) µ (E) + ε. (b) Let E X such that µ (E) <. Suppose that E is µ -measurable. For each n N there exists B n A σ such that E B n and µ (B n ) µ (E) + n 1. Define B := B n, so that B A σδ and E B. Since µ (E) <, it follows that µ (B \ E) µ (B n E c ) = µ (B n ) µ (B n E) = µ (B n ) µ (E) n 1 for all n N. This implies that µ (B \ E) = 0. Conversely, suppose there exists B A σδ such that E B and µ (B \ E) = 0. If F X, then µ (F E) + µ (F E c ) µ (F B) + µ (F B c ) + µ (F (B \ E)) µ (F B) + µ (F B c ) = µ (F ) because B A σδ is µ -measurable. Clearly µ (F ) µ (F E) + µ (F E c ), so E is µ -measurable. 19. If E X is µ -measurable, then µ (X) = µ (X E)+µ (X E c ) = µ (E)+µ (E c ) by definition, so µ (E) = µ (E). Conversely, let E X and suppose that µ (E) = µ (E). By the previous exercise, for each n N there exist A n, B n A σ such that E A n, E c B n, µ (A n ) µ (E) + n 1 and µ (B n ) µ (E c ) + n 1. If A := A n, then µ (A \ E) µ (A n B n ) = µ (A n ) µ (A n \ B n ) = µ (A n ) µ (B c n) = µ (A n ) (µ (X) µ (B n )) µ (E) + n 1 µ (X) + µ (E c ) + n 1 3
4 = µ (E) µ (E) + 2n 1 = 2n 1 for all n N, and hence µ (A\E) = 0. Moreover E A and A A σδ, so by the previous exercise E is µ -measurable. 20. (a) If (A n ) is a sequence in M such that E A n, then µ (E) µ ( A n ) µ (A n ) = µ(a n ). Therefore µ (E) µ + (E), by definition. If there exists A M such that E A and µ (A) = µ (E), then µ + (E) µ(a) + µ( ) = µ (A) = µ (E), n=2 so µ (E) = µ + (E). Conversely, suppose that µ (E) = µ + (E). By exercise 18 part (a), for each n N there exists A n M σ = M such that E A n and µ + (A n ) µ + (E) + n 1. Define A := A n, so that A M and E A. Moreover µ + (A) µ + (A n ) µ + (E) + n 1 for all n N, so µ + (A) µ + (E). It follows that µ (A) = µ + (A) = µ + (E) = µ (E), because A M and E A. (b) Let µ 0 be a premeasure on an algebra A M such that µ is induced by µ 0, and let E X. By exercise 18 part (a), for each n N there exists A n A σ M such that E A n and µ (A n ) µ (E) + n 1. Define A := A n, so that A M and E A. Moreover µ (A) µ (A n ) µ (E) + n 1 for all n N, so µ (A) µ (E). It follows that µ (A) = µ (E), because E A. By the previous exercise, this implies that µ (E) = µ + (E). Therefore µ = µ +. (c) Define µ : 2 X [0, ] by µ ( ) = 0, µ ({0}) = 2, µ ({1}) = 2 and µ (X) = 3. Clearly µ (A) µ (B) for all A B X. Moreover µ is subadditive, because µ (X) < µ ({0}) + µ ({1}). Therefore µ is an outer measure. Note that {0} is not µ -measurable, because µ (X) = 3 < = µ (X {0}) + µ (X {0} c ). This implies that µ + ({0}) = inf{ µ (A n ) A n X for all n N and A n = X for some n N} = µ (X) = 3 2 = µ ({0}). 21. Let E X be locally µ -measurable, and let A X. If µ (A) =, then clearly µ (A E) + µ (A E c ) µ (A). Otherwise, by exercise 18 part (a), for each n N there exists a µ -measurable set A n X such that A A n and µ (A n ) µ (A) + n 1. In particular, if n N then µ (A n ) <, and hence A n E and A n E c = A n (A n E) c are µ -measurable (because E is locally µ -measurable). It follows that µ (A E) + µ (A E c ) µ (A n E) + µ (A n E c ) = µ(a n E) + µ(a n E c ) = µ(a n ) = µ (A n ) µ (A) + n 1 for all n N, so µ (A E) + µ (A E c ) µ (A). Clearly µ (A) µ (A E) + µ (A E c ) in either case, so E is µ -measurable. This shows that µ is saturated. 22. (a) Clearly µ M = µ, so by exercise 6 from section 1.3 it suffices to show that M = M. To this end, let A M. Then A = E F for some E M and F X such that F B for some B M with µ(b) = 0. Clearly A E B M = M σδ and µ ((E B) \ A) µ (B) = 0. By exercise 18 part (c), it follows that A M. Conversely, let A M. By exercise 18 part (c) there exists B M σδ = M such that A B and µ (B \ A) = 0. Since B \ A M, exercise 18 part (b) implies that B \ A E and µ (E \ (B \ A)) = 0 for some E M. Note that B \ E M and A E M. Moreover µ (A E) µ (E \ (B \ A)) = 0, so there exists F M such that A E F and µ (F \ (A E)) = 0, by exercise 18 part (b). It follows that µ(f ) = µ (F ) µ (F \ (A E)) + µ (A E) = 0. 4
5 Since A = (B \ E) (A E), it follows that A M. Therefore M = M. (b) Let Ẽ M, and µ be the completion of µ. Given A M with µ(a) <, the converse of the previous exercise implies that A M (using exercise 18 part (b) instead of part (c)). Hence A = E F for some E M and F X such that F B for some B M with µ(b) = 0. In particular µ(a) = µ(e) = µ (E) µ (A) <. Therefore Ẽ A M, as Ẽ is locally µ-measurable. Since µ (Ẽ A) < the previous exercise implies that Ẽ A M (again using exercise 18 part (b) instead of part (c)). This shows that Ẽ is locally µ -measurable, so Ẽ M by exercise 21. Conversely, let Ẽ M and A M be such that µ(a) <. Then A = E F for some E M and F X such that F B for some B M with µ(b) = 0. It follows that µ (Ẽ A) µ (A) µ (E) + µ (F ) = µ(e) + 0 = µ(a) <. This implies that A M (by the previous exercise using part (b) instead of part (c)), so Ẽ A M and hence Ẽ A M (again by the previous exercise). This shows that Ẽ is locally µ-measurable, so Ẽ M. Therefore M = M. By exercise 6 from section 1.3, µ and µ agree on M. If E M \ M, then µ(e) = by definition. Moreover, if E M = M and µ(e) <, then E M by the previous exercise (using part (b) instead of part (c)). This implies that µ and µ also agree on M \ M. 23. (a) Let E := {(a, b] Q a, b R}. Clearly = (0, 0] Q E. If (a 1, b 1 ] Q E and (a 2, b 2 ] Q E then their intersection is (max{a 1, a 2 }, min{b 1, b 2 }] Q E. Moreover, the complement of (a, b] Q E is ((, a] Q) ((b, ] Q), which is a disjoint union of elements of E provided that a b. If a > b then the complement of (a, b] Q is just (, ] Q E. This shows that E is an elementary family of subsets of Q, so the collection of finite disjoint unions of members of E is an algebra. If (a 1, b 1 ] Q E and (a 2, b 2 ] Q E are not disjoint, then their union is (min{a 1, a 2 }, max{b 1, b 2 }] Q E. Therefore A is the collection of finite disjoint unions of members of E, so A is an algebra. (b) Let M be the σ-algebra generated by A. Since Q is countable and M is closed under countable unions, it suffices to show that {x} M for all x Q. Given x Q and n N, it is clear that (x 1 n, x] Q M. Therefore {x} = ((x 1 n, x] Q) M, as required. (c) By definition µ 0 ( ) = 0. Let (E n ) be a sequence of disjoint members of A whose union lies in A. If E n = for all n N, then E n = and hence µ 0 ( E n ) = 0 = µ 0(E n ). Otherwise E n, and E m for some m N, so µ 0 ( ) = and µ 0(E n ) µ 0 (E m ) =. Therefore µ 0 is a premeasure on A. Define µ 1, µ 2 : 2 Q [0, ] by, E µ 1 (E) = 0, E =, E contains 2 n m for some m Z and n N and µ 1 (E) = 0, otherwise. Clearly µ 1 A = µ 0 = µ 2 A, because every non-empty interval contains a rational of the form 2 n m for some m Z and n N. For the same reason that µ 0 is a premeasure, µ 1 is a measure. A very similar argument implies that µ 2 is a measure. But µ 1 µ 2 because µ 1 ({3 1 }) = whereas µ 2 ({3 1 }) = (a) Since µ(a) + µ(a c B) = µ(a B) = µ(b) + µ(b c A), by symmetry it suffices to show that µ(a c B) = 0. To this end, note that E A B c = (A c B) c (if a member of E is not in B c, it is in B E = A E A). So µ(x) = µ (X) = µ (E) µ ((A c B) c ) = µ((a c B) c ) = µ(x) µ(a c B), and hence µ(a c B) = 0 as required. 5
6 (b) It is clear that M E is a σ-algebra on E and that ν( ) = 0. Let {E n } be a pairwise disjoint subset of M E. For each n N there exists A n M such that E n = A n E. Set A := i N j=i+1 (A i A j ) and define B n := A n \ A for each n N. It is easily checked that {B n } is a pairwise disjoint subset of M and that E n = B n E for all n N. Therefore ν( E n ) = ν( (B n E)) = ν(( B n ) E) = µ( B n ) = µ(b n ) = ν(e n ), which shows that ν is a measure on M E. 25. Let (C n ) be a sequence of compact intervals covering R, and fix n N. There exists a G δ set V n M µ and a null set N n M µ such that E C n = V n \ N n. Let (V nk ) k N be a sequence of open sets such that V n = k N V nk. For each k N define an open set U nk := V nk C c n, and set U := n,k N U nk. Then U is a G δ set and U \ E N n. Indeed, if x U \ E there exists n N such that x C n, which implies that x V nk for all k N and hence x V n but x / E C n = V n \ N n. Moreover E U, because E U nk for all n, k N. It follows that E = U \ (U \ E), where U \ E is a null set. If n N, then E C n = H n N n for some F σ set H n M µ and some null set N n M µ. Clearly H n and N n are respectively F σ and null sets. Moreover E = (E C n ) = ( H n ) ( N n ). 26. Let E M µ and suppose that µ(e) <. Given ε (0, ), there exists an open set U M µ such that E U and µ(u) < µ(e) + ε 2. Let (U n) be a sequence of disjoint open intervals such that U n = U. Then µ(u n ) = µ(u) <, so there exists N N such that n=n+1 µ(u n) < ε 2. Define A := N n=1 U n. It follows that µ(e A) µ(e \ A) + µ(a \ E) µ(e \ U) + µ(u \ A) + µ(u \ E) = Let a, b R. Since µ F is continuous from above, n=n+1 µ(u n ) + µ(u) µ(e) < ε. µ F ([a, b]) = µ F ( (a n 1, b]) = lim n µ F ((a n 1, b]) = lim n (F (b) F (a n 1 )) = F (b) F (a ). It follows that µ F ({a}) = µ F ([a, a]) = F (a) F (a ), in which case µ F ([a, b)) = µ F ([a, b]) µ F ({b}) = F (b ) F (a ) and µ F ((a, b)) = µ F ([a, b)) µ F ({a}) = F (b ) F (a). 29. (a) Suppose that E N but m(e) > 0. Define R := Q [0, 1), and for each r R set E r := E + r. Clearly each E r is measurable with m(e r ) = m(e), and r R E r [0, 2). Let r, s R and suppose that E r intersects E s. Then there exists t E r E s, so that t r, t s E N. Since t r = (t s) + (s r) and s r Q, the definition of N implies that t s = t r. Therefore r = s, which shows that {E r } r R is pairwise disjoint. Hence = r R m(e) = r R m(e r ) = m( r R E r ) m([0, 2)) = 2, which is a contradiction. Therefore m(e) = 0. 6
7 (b) Suppose that m(e) > 0, but every subset of E is measurable. Since E = n Z (E [n, n + 1)), there exists n Z such that m(e [n, n + 1)) > 0. Define F := (E [n, n + 1)) n, so that F [0, 1), m(f ) > 0 and every subset of F is measurable. Also define R := Q [ 1, 1], and for each r R set N r := N + r. It is clear that [0, 1) r R N r, and hence F = r R (F N r ). If r R then F N r F and (F N r ) r N, which implies that both subsets are measurable (by containment in F and translational invariance) and have measure zero (by the previous exercise and translational invariance). Therefore m(f ) r R m(f N r) = r R 0 = 0, which is a contradiction so not every subset of E is measurable. 30. Let E L with m(e) > 0, and suppose there exists α (0, 1) such that m(e I) αm(i) for all open intervals I. Without loss of generality m(e) < (since m is semifinite, we may replace E by a subset of finite positive measure). Define ε := m(e)(1 α), so that ε > 0. Since E L and m is outer regular, there exists an open set U R such that E U and m(u) < m(e) + ε. As U is open, there exists a pairwise disjoint collection {I i } i N of open intervals such that U = i N I i, and these intervals are bounded because m(u) <. If if i N then m(i i ) = m(i i \ E) + m(e I i ) m(i i \ E) + αm(i i ) and hence (1 α)m(i i ) m(i i \ E). Since i N (I i \ E) = U \ E has measure m(u) m(e), it follows that (1 α)m(u) = (1 α) i N m(i i ) = i N (1 α)m(i i ) i N m(i i \ E) = m( i N (I i \ E)) < ε = (1 α)m(e) and hence m(u) < m(e) m(u), which is impossible. Thus, for each α (0, 1) there exists an open interval I such that m(e I) > αm(i). The same clearly holds for α (, 0]. 31. Let E L with m(e) > 0, and set α := 3 4. By the previous exercise there exists a open interval I R with endpoints a, b R such that m(e I) > αm(i). Suppose there exists x ( 1 2 m(i), 1 2m(I)) such that x / E E. Then x / E E, and clearly 0 E E, so we may assume x > 0. Moreover E (E c + x) (E c x). Note that (a, a + 2x] I, because 2x < m(i) = b a. Define n := max{k N a + 2kx < b}, so that m(e (a, a + 2nx]) = (m(e (a + 2(k 1)x, a + (2k 1)x]) + m(e (a + (2k 1)x, a + 2kx])) (m((e c x) (a + 2(k 1)x, a + (2k 1)x]) + m((e c + x) (a + (2k 1)x, a + 2kx])) (m(e c (a + (2k 1)x, a + 2kx] x) + m(e c (a + 2(k 1)x, a + (2k 1)x] + x)) (m(e c (a + (2k 1)x, a + 2kx]) + m(e c (a + 2(k 1)x, a + (2k 1)x])) m(e c (a, a + 2nx]). This implies that m(e (a, a + 2nx]) 1 2 (m(e (a, a + 2nx]) m(ec (a, a + 2nx])) = 1 2m((a, a + 2nx]) = nx. Note that 4nx m(i), since otherwise k := 2n satisfies a + 2kx < b (but k > n). It follows that m(e I) m(e (a, a + 2nx]) + m((a + 2nx, b)) m(i) nx m(i) m(i) 4 = 3m(I) 4 = αm(i). This is a contradiction, so there does not exist x ( 1 2 m(i), 1 2m(I)) such that x / E E. 7
8 33. Choose a surjection q : N Q [0, 1], and for each n N define I n := (q(n) 3 n, q(n) + 3 n ). For each n N, define D n = I n \ ( k=n+1 I k). If m, n N and m < n, then D m D n = because D n I n k=m+1 I k. Moreover 2 3 n = m(i n ) m(d n ) + m( k=n+1 I k) m(d n ) + m(i k ) = m(d n ) k = m(d n ) + 3 n, k=n+1 k=n+1 and hence m(d n ) 3 n, for all n N. For each n N there exists a Borel set A n D n such that 0 < m(a n ) < 3 n, which can be found by intersecting D n with an interval of the form [3 n 1 k, 3 n 1 (k + 1)] for some k Z (not all of the intersections can have zero measure). Define A := [0, 1] ( A n ), and let I be a subinterval of [0, 1] with midpoint c. There exists n N such that 4 3 n < m(i), and there are infinitely many rationals in (c 3 n, c + 3 n ), so there exists k N with k n and q(k) (c 3 n, c + 3 n ). It follows that I k (c 2 3 n, c n ) I. Therefore A k A I (since A k I k [0, 1]), so m(a I) m(a k ) > 0. Moreover m(a I) = m(( A n ) I) = m(a k ) + m(( \{k} A n ) I) < m(d k ) + m(i \ D k ) = m(i), because m(a k ) < m(d k ) and \{k} A n \{k} D n Dk c. 8
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