Real Analysis Chapter 3 Solutions Jonathan Conder. ν(f n ) = lim

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1 . Suppose ( n ) n is an increasing sequence in M. For each n N define F n : n \ n (with 0 : ). Clearly ν( n n ) ν( nf n ) ν(f n ) lim n If ( n ) n is a decreasing sequence in M and ν( ) <, then N N n ν(f n ) lim N ν( N). ν( n n ) ν( \ ( \ n n )) ν( ) ν( n( \ n )) ν( ) lim n ν( \ n ) lim n ν( n). 2. Let be measurable, and suppose that is ν-null but ν () 0. Then ν + () ν () ν() 0 and ν + () + ν () ν () > 0, so ν + () ν () > 0. Since ν + ν, there exist disjoint measurable sets A and B covering X such that A is ν + -null and B is ν -null. In particular ν + ( B) ν + ( B) + ν + ( A) ν + () > 0 but ν ( B) ν (B) 0. This implies that ν( B) > 0, which is a contradiction because B and is ν-null. Conversely, suppose that ν () 0. If F is measurable, then ν (F ) ν () 0 so ν + (F ) + ν (F ) 0. In particular ν + (F ) 0 ν (F ), so ν(f ) 0 and hence is ν-null. Suppose that ν µ, so that X A B for some disjoint measurable sets A and B such that A is ν-null and B is µ-null. From above, ν (A) 0, so A is ν -null (because ν is positive) and hence ν µ. Now suppose that ν µ, so that X A B for some disjoint measurable sets A and B such that A is ν -null and B is µ-null. Since ν + ν and ν ν pointwise, A is ν + -null and ν -null. Therefore ν + µ and ν µ. Finally, suppose that ν + µ and ν µ. Then X A B F for two pairs A, B and, F of disjoint measurable sets such that A is ν + -null, is ν -null and B and F are µ-null. Note that B F is µ-null, because every subset of B F B (F \ B) is the disjoint union of two µ-null sets. Moreover X \ (B F ) A, which is both ν + -null and ν -null, and hence ν-null. This shows that ν µ. 3. (a) Let φ L + be a simple function, and write φ n i a iχ i. Then φ d ν a i ν ( i ) a i (ν + ( i ) + ν ( i )) a i ν + ( i ) + a i ν ( i ) i i i i φ dν + + φ dν. () Hence, if f L (ν) then { f d ν sup φ dν + + } φ dν φ L + simple with φ f f dν + + f dν < because L (ν) L (ν + ) L (ν ). Therefore L (ν) L ( ν ). Conversely, if f L ( ν ) it is clear from () that φ dν + φ d ν f d ν for all simple φ L + with φ f, so f dν + f d ν <. This shows that L ( ν ) L (ν + ), and a similar argument shows that L ( ν ) L (ν ). Therefore L ( ν ) L (ν). (b) Let A and B be disjoint measurable sets covering X such that A is ν + -null and B is ν -null. Then gχ A 0 ν + -a.e. and gχ B 0 ν -a.e. for every measurable function g. In particular, from () φ d ν φ dν + for all simple functions φ L + with φ f + χ B or φ f χ B. This implies that fχ B d ν f + χ B d ν f χ B d ν f + χ B dν + f χ B dν + fχ B dν +, and similarly fχ A d ν fχ A dν. Moreover χ A χ B χ X because A B X and A B. Therefore f dν f(χ A + χ B ) dν + f(χ A + χ B ) dν

2 fχ B dν + fχ A dν fχ B d ν fχ A d ν f(χ B χ A ) d ν f(χ B χ A ) d ν f d ν. (c) Define g : χ B χ A. Then g and hence ν () g dν sup{ f dν f } because g dν (χ B χ A )χ dν χ B χ dν + χ A χ dν χ (B ) dν + + χ (A ) dν ν + (B ) + ν (A ) ν + (A ) + ν + (B ) + ν (A ) + ν (B ) ν + () + ν () ν () 0. This completes the proof if ν (). Assume that ν () < and let f be a measurable function with f. Then fχ L ( ν ) L (ν) because f d ν χ d ν ν () <. Hence, by the previous exercise f dν fχ d ν χ d ν ν (). Therefore sup{ f dν f } ν (). 4. Let A and B be disjoint measurable sets covering X such that A is ν + -null and B is ν -null. If is measurable then λ() λ( B) ν( B) + µ( B) ν( B) ν + ( B) ν + ( B) + ν + ( A) ν + (). and similarly µ() µ( A) µ( A) + ν + ( A) λ( A) + ν ( A) ν ( A) ν ( A) + ν ( B) ν (). 5. Clearly v + v 2 is a signed measure, and v + v 2 v + v + v+ 2 v 2 (v+ + v+ 2 ) (v + v 2 ). Since v+ + v+ 2 and v + v 2 are positive measures, the previous exercise implies that v+ + v+ 2 (v + v 2 ) + and v + v 2 (v + v 2 ). Therefore v + v 2 (v + v 2 ) + + (v + v 2 ) v + + v+ 2 + v + v 2 v + v (a) Let A and B be disjoint measurable sets covering X such that A is ν + -null and B is ν -null. Then ν + () ν + ( B) + ν + ( A) ν + ( B) ν( B) sup{ν(f ) F M, F }. Moreover, if F M and F then ν(f ) ν + (F ) ν (F ) ν + (F ) ν + (). Therefore ν + () sup{ν(f ) F M, F }, so ν + () sup{ν(f ) F M, F }. A similar argument shows that ν () ν( A) and hence ν () inf{ν(f ) F M, F }. 2

3 (b) Clearly ν () sup{ n i ν( i) n N and ( i ) n i is a pairwise disjoint sequence in M covering }, as ν( A) + ν( B) ν ( A) + ν + ( B) ν ( A) + ν + ( B) ν + ( A) + ν ( A) + ν + ( B) + ν ( B) ν ( A) + ν ( B) ν (). Now let n N and ( i ) n i is a pairwise disjoint sequence in M covering. Then ν( i ) i ( ν + ( i ) + ν ( i ) ) i ν + ( i ) + ν ( i ) ν + () + ν () ν (). i i This shows that ν () sup{ n i ν( i) n N and ( i ) n i is a pairwise disjoint sequence in M covering }, and hence ν () sup{ n i ν( i) n N and ( i ) n i is a pairwise disjoint sequence in M covering }. 8. Suppose ν µ. If M and µ() 0 then ν(f ) µ(f ) 0 for all F M with F, so is ν-null. By exercise 2, this implies that ν () 0. Therefore ν µ. Now suppose that ν µ. If M and µ() 0, then ν + () + ν () 0 and hence ν + () ν () 0. This shows that ν + µ and ν µ. Finally, suppose that ν + µ and ν µ. If M and µ() 0, then ν() ν + () ν () 0 0 0, and hence ν µ. 9. Suppose that ν n µ for all n N. For each n N there exists a measurable set n such that n is ν n -null and n c is µ-null. Define : n n and note that c n c n; in particular c is µ-null. It is also clear that is a null set with respect to n ν n. Therefore n ν n µ. The second part is trivial.. (a) If f L (µ) then {f} is uniformly integrable, because the finite signed measure f dµ is absolutely continuous with respect to µ. Hence, if {f i } i I is a finite subcollection of L (µ) and ε (0, ), for each i I there exists δ i (0, ) such that f i dµ < ε for all M with µ() < δ i. Set δ : min{δ i } i I, so that f i dµ < ε for all i I and M with µ() < δ. This shows that {f i } i I is uniformly integrable. (b) Let ε (0, ), and choose N N such that f n f dµ < ε 2 for all n N with n N. Define f 0 : f and I : {n} N n0. Since {f n} n I is uniformly integrable, there exists δ (0, ) such that f n dµ < ε 2 for all n I and M with µ() < δ. If n N \ I and M then f n dµ (f n f) dµ + f dµ (f n f) dµ + f dµ f n f dµ + f 0 dµ Since f n f dµ f n f dµ, this implies that f n dµ < ε for all n N and M with µ() < δ. Therefore {f n } n is uniformly integrable. 3. (a) Let M and suppose that µ() 0. Then so m() 0. Therefore m µ. Suppose there exists an extended µ-integrable function f such that dm f dµ. If x X then f(x) {x} f dµ m({x}) 0, so f 0. Therefore m(x) X f dµ 0, which is a contradiction because m(x). 3

4 (b) Suppose that µ has a Lebesgue decomposition λ+ρ with respect to m, with λ m and ρ m. Then ρ({x}) 0 and hence λ({x}) µ({x}) for all x X. There exist disjoint measurable sets A and B covering X such that A is λ-null and m(b) 0. If x A then λ({x}) 0 so A. Therefore m(b) m(x), which contradicts m(b) For each n N let n : {x X f(x) < n }, so that n λ( n ) n dλ n f dλ ν( n ) 0 n and hence µ( n ) λ( n ) 0. It follows that µ( n n) 0, so f 0 µ-a.e. because n n {x X f(x) < 0}. Now set F : {x X f(x) }. Since ν is σ-finite, there is a sequence (F n ) n of subsets of F which cover F such that ν(f n ) < for all n N. Since ν(f n ) f dλ F n dλ λ(f n ) µ(f n ) + ν(f n ) F n it is clear that µ(f n ) 0 for all n N, in which case µ(f ) 0 and hence f < µ-a.e. Without loss of generality f, f L +, so for each M ( f) dλ + ν() dλ λ() µ() + ν(). In particular ( f) dλ µ() for all M with ν() <. This result extends to all M because ν is σ-finite, using the monotone convergence theorem and the additivity of µ. Therefore dµ dλ ( f). Since µ λ (in fact µ λ) and λ µ (as ν µ), it follows that dλ dµ f. This implies that dν dµ dν dλ dλ dµ f f. 7. Define a measure λ on M by λ() : f dµ. Then λ is finite because f L (µ). Define ρ : λ N. Clearly ρ ν, so by the Radon-Nikodym theorem there is an extended ν-integrable function g such that ρ() g dν for all N. In fact g L (ν) since ρ(x) <. Moreover g is unique up to equality ν-a.e. by the Radon-Nikodym theorem. 9. There exist positive measures ρ and σ and complex-valued functions f L (ρ) and g L (σ) such that dν f dρ, d ν f dρ, dµ g dσ and d µ g dσ. Suppose that ν µ. For each pair a, b {r, i} there exist disjoint sets A ab, B ab M such that A ab B ab X, A ab is ν a -null and B ab is µ b -null. It follows that A : (A rr A ri ) (A ir A ii ) is both ν r -null and ν i -null. In particular, if M and A then 0 ν r () Re( f dρ) Re(f) dρ. By applying this to the sets {x A Re(f(x)) > n } and {x A Re(f(x)) < n } for all n N, we can show that Re(f)χ A 0 ρ-a.e., and similarly Im(f)χ A 0 ρ-a.e., giving f χ A 0 ρ-a.e. and hence A is ν -null. Moreover B : A c (A rr A ri ) c (A ir A ii ) c (B rr B ri ) (B ir B ii ) is both µ r -null and µ i -null, and a similar argument shows that B is µ -null. Therefore ν µ. Conversely, suppose that ν µ. There exist disjoint sets A, B M such that A B X, A is ν -null and B is µ -null. It follows that f χ A 0 ρ-a.e., so that Re(f)χ A 0 Im(f)χ A ρ-a.e. and hence A is both ν r -null and ν i -null. Similarly B is both µ r -null and µ i -null, which implies that ν a µ b for all a, b {r, i}. This shows that ν µ. 4

5 Now suppose that ν λ, and let A M satisfy λ(a) 0. If M and A then λ() 0 and hence ν a () 0 for each a {r, i}. It follows that Re(f) dρ Im(f) dρ 0 for all M with A, so f χ A 0 ρ-a.e. which implies that ν (A) 0. This shows that ν λ. Conversely, if ν λ then ν(a) ν (A) 0, and hence ν a (A) 0, for each a {r, i} and all A M with λ(a) 0. This implies that ν r λ and ν i λ, so ν λ. 2. If n N and, 2,..., n M are disjoint with n k, then, 2,... are disjoint, k and ν( k) n ν( k) where n+k : for all k N. This implies that µ () µ 2 (). Moreover, if ( k ) is a sequence of disjoint measurable sets such that k, then f : ν( k ) ν( k ) χ k is well-defined and satisfies f (by disjointness). If a {r, i} and b {+, } then L (ν b a) so f dνa b lim n ν( k ) ν( k ) χ k dνa b lim n ν( k ) ν( k ) νb a( k ) ν( k ) ν( k ) νb a( k ) by the dominated convergence theorem. It follows that f dν f dν r + i f dν i f dν r + f dνr + i f dν + i i f dν i ν( k ) ν( k ) (ν+ r ( k ) ν r ( k ) + iν + i ( k) iν i ( k)) ν( k ) ν( k ) (ν r( k ) + iν i ( k )) ν( k ) ν( k ) ν( k) ν( k ) 2 ν( k ) ν( k ). This implies that f dν ν( k), so µ 2 () µ 3 (). If f is a complex-valued measurable function on X such that f then f dν f d ν d ν ν (), so µ 3() ν (). Conversely, if f : dν d ν (which exists because ν ν ) then f ν -almost everywhere and hence f everywhere (without loss of generality). By generalising the chain rule to complex measures, it follows that µ 3 () f dν f dν d ν d ν ff d ν f 2 d ν d ν ν () ν (). This shows that µ 3 () ν (), so it remains to show that µ 3 () µ (). If f is a complex-valued measurable function on X such that f, there exists a sequence (φ k ) of simple functions which converges pointwise to f 5

6 such that ( φ k ) is increasing to f. For each k N let n k φ k c kj χ kj j be the standard representation of φ k. By the dominated convergence theorem, if a {r, i} and b {+, } then n k f dνa b lim φ k dνa b lim c kj νa( b kj ) and hence f dν lim lim f dν r + i f dν r + n k n k j n k lim j n k lim j j f dν i + i f dν i + i f dνi j f dνr n k n k c kj ν r + ( kj ) c kj νr ( kj ) + i c kj ν i + ( n k kj ) i c kj νi ( kj ) j c kj (ν r + ( kj ) νr ( kj ) + iν i + ( kj ) iνi ( kj )) c kj (ν r ( kj ) + iν i ( kj )) c kj ν( kj ). Since φ k and k, k2,..., knk are disjoint for each k N, it follows that f dν lim n k c kj ν( kj ) lim n k n k c kj ν( kj ) lim ν( kj ) µ (). j Therefore µ 3 () µ (), as required. j j j 22. Assuming f > 0, there exists R (0, ) such that B R (0) f dm > 0 (otherwise the monotone convergence theorem implies that f dm lim N B N (0) f dm 0). If x Rn \ B R [0], then B R (0) B 2 x (x) so Hf(x) A 2 x f (x) m(b 2 x (x)) B 2 x (x) f dm m(b 2 x (0)) B R (0) f dm j x n m(b 2 (0)) B R (0) f dm. Note that C : m(b 2 (0)) B R (0) f dm is positive and independent of x. Now, if α (0, C 2R ) and x R n such that n R < x < ( C α )/n then Hf(x) C x n > C α C α and hence B ( C (0) \ B α )/n R [0] {x R n Hf(x) > α}. Thus ( ) ( ) C m({x R n Hf(x) > α}) m B ( C (0) m(b α )/n R [0]) α Rn m(b (0)) > Cm(B (0)). 2α 23. Let x R n and r (0, ), so that B r (x) is a ball with x B r (x). Then m(b r(x)) B f dm r(x) H f(x) and hence Hf(x) H f(x). Conversely, let y R n and r (0, ) such that x B r (y). Clearly B r (y) B 2r (x), so 2 n f dm f dm f dm 2 n Hf(x) m(b r (y)) m(b r (y)) m(b 2r (x)) B r(y) and hence H f(x) 2 n Hf(x). B 2r (x) 6 B 2r (x)

7 24. Let ε (0, ) and take r (0, ) such that f(y) f(x) < ε for all y B r (x). Then f(y) f(x) dy m(b r (x)) B r(x) f(y) f(x) dy < ε dy εm(b r(x)) m(b r (x)) B r(x) m(b r (x)) B r(x) m(b r (x)) ε. This shows that lim r 0 m(b r(x)) B f(y) f(x) dy 0. Therefore x L r(x) f. 25. (a) Define µ : B R n [0, ] by µ(a) m( A). Clearly µ is a measure with Lebesgue-Radon-Nikodym representation dµ χ dm. Let A B R n and suppose m(a) <. Given ε (0, ), there exists an open set U R n containing A such that m(u) < m(a) + ε and hence m(u \ A) < ε. It follows that µ(u) m( U) m( A) + m(u \ A) < µ(a) + ε. Given A B R n and ε (0, ), choose a sequence (A k ) of Borel sets such that m(a k) < for all k N and A k A. For each k N there exists an open set U k R n containing A k such that µ(u k ) < µ(a k ) + 2 k ε. Clearly A U k and µ( U k \ A) < ε. This implies that µ(a) inf{µ(u) U R n is open and A U}. Therefore µ is regular, so for almost every x R n m( B r (x)) µ(b r (x)) D (x) lim lim r 0 m(b r (x)) r 0 m(b r (x)) χ (x). In particular D (x) for almost all x and D (x) 0 for almost all x c. (b) Given α [0, 4 ) define α : {(x, y) (0, ) 2 y x tan(2πα)}. Fix r (0, ) and note that α B r (0) is a sector of the disk B r (0) between the angles 0 and 2πα. Hence m( α B r (0)) αm(b r (0)) and D α (0) α. For α [ 4, ) just include some of the other quadrants (0, ) (, 0), (, 0) (0, ) or (, 0)2 in α. Now define : n [2 n, 2 n + 2 n ] and fix N N. Then B 2 N (0) nn+ [2 n, 2 n + 2 n ] so m( B 2 N (0)) nn+ while m(b 2 N (0)) 2 2 N 2 N. This implies that m([2 n, 2 n + 2 n ]) nn+ 2 n 2 N, m( B 2 N (0)) m(b 2 N (0)) 2 N 2 N On the other hand B 2 N +2 N (0) ( nn+ [2 n, 2 n + 2 n ]) [2 N, 2 N + 2 N ), which implies that m( B 2 N +2 N (0)) m([2 N, 2 N + 2 N )) + nn+ m([2 n, 2 n + 2 n ]) but m(b 2 N +2 N (0)) 2 (2 N + 2 N ) 2 N + 2 N 3 2 N and hence m( B 2 N +2 N (0)) m(b 2 N +2 N (0)) 2 N 3 2 N 3. Since 2 N and 2 N + 2 N converge to 0 as N, it follows that D (0) does not exist. 2 n 2 N, nn 7

8 28. (a) If x R then { } T F (x) sup F (x k ) F (x k ) n N and x 0, x,..., x n R with x 0 < x < < x n x { } sup µ F ((, x k ]) µ F ((, x k ]) n N and x 0, x,..., x n R with x 0 < x < < x n x { } sup µ F ((x k, x k ]) n N and x 0, x,..., x n R with x 0 < x < < x n x { } sup µ F ( k ) n N and, 2,..., n B R are disjoint with n k (, x] { } sup µ F ( k ) n N and, 2,..., n B R are disjoint with n k (, x] µ F ((, x]) G(x), where the penultimate equality follows from exercise 2. (b) If a, b R and a < b then { } T F (b) T F (a) sup F (x k ) F (x k ) n N and x 0, x,..., x n R with a x 0 < x < < x n b and hence µ F ((a, b]) µ F ((, b]) µ F ((, a]) F (b) F (a) T F (b) T F (a) µ TF ((a, b]). The set A of all finite disjoint unions of left-open, right-closed intervals is an algebra of subsets of R. If x R µ F ((, x]) µ F ((x k, x + k]) µ F ((x k, x+ k]) µ TF ((x k, x+ k]) µ TF ((, x]) and similarly µ F ((x, )) µ TF ((x, )). By the same argument, it follows that µ F () µ TF () for all A (because µ F ( ) µ TF ( )). Define C : { B R µ F () µ TF ()}. If ( k ) is an increasing sequence in C, then µ F ( k) lim µ F ( k ) lim µ F ( k ) lim µ TF ( k ) µ TF ( k) and hence k C. Similarly C is closed under countable decreasing intersections (because µ TF and the real and imaginary parts of µ F are finite). Therefore C contains the monotone class generated by A, which is B R by the monotone class lemma and the fact that B R is generated by A. Thus µ F () µ TF () for all B R. (c) If B R then, by exercise 2, { } µ F () sup µ F ( k ) ( k) is a sequence of disjoint Borel sets such that k { } sup µ TF ( k ) ( k) is a sequence of disjoint Borel sets such that k sup{µ TF ()} 8

9 µ TF (). In particular, if x R then G(x) µ F ((, x]) µ TF ((, x]) T F (x). Therefore G T F by part (a). Since µ F and µ TF are finite, it is straightforward to show that M : { B R µ F () µ TF ()} is a σ-algebra which contains (, x] for all x R. Therefore B R M and hence µ F () µ TF () for all B R. 29. The total variation of µ F as a complex measure is the same as the total variation of µ F as a signed measure, so T F (x) + F (x) µ TF ((, x]) + µ F ((, x]) µ F ((, x]) + µ F ((, x]) 2µ + F ((, x]) and hence µ P ((, x]) P (x) 2 (T F (x) + F (x)) µ + F ((, x]) for all x R. Therefore µ P { B R µ P () µ + F ()} is a σ-algebra containing a set which generates B R. Similarly µ N µ F. µ + F, because 30. Let q : N Q be a surjection and define f : 2 k χ (q(k), ). Given x R it is clear that f(x) k N q(k)<x 2 k and hence f(x) f(y) for all y (x, ). Let ε (0, ) and choose N N such that 2 N < ε. There exists δ (0, ) such that (x δ, x) (x, x + δ) does not contain q(), q(2),..., q(n). If y (x δ, x) then f(x) f(y) 2 k f(x) k N k N q(k)<y y q(k)<x 2 k f(x) which shows that lim y x f(y) f(x). On the other hand, if y (x, x + δ) then f(x) f(y) 2 k f(x) + k N k N q(k)<y x q(k)<y kn+ 2 k, 2 k > f(x) ε, which implies that lim y x f(y) f(x) provided that x / Q. However, if x q(n) for some n N then f(x) + 2 n f(y) f(x) + 2 n + k N x<q(k)<y 2 k < f(x) + 2 n + ε and hence lim y x f(y) f(x) + 2 n. This shows that Q is the set of discontinuities of f. 3. (a) By the usual differentiation rules F and G are differentiable on R \ {0}. Moreover, by the squeeze theorem. Similarly G (0) 0. F F (h) F (0) h 2 sin(h ) (0) lim lim lim h sin(h ) 0 h 0 h h 0 h h 0 (b) If x R \ {0} then F (x) 2x sin(x ) cos(x ) and hence F (x) 2 x sin(x ) + cos(x ) 2 x +. Therefore F 3 on [, ]. Hence, by the mean value theorem, if a, b [, ] and a < b there exists c (a, b) such that F (b) F (a) F (c) b a 3(b a). It follows that F BV ([, ]) because { } T F () T F ( ) sup F (x k ) F (x k ) n N and x 0, x,..., x n R with x 0 < x < < x n 9

10 { } sup 3(x k x k ) n N and x 0, x,..., x n R with x 0 < x < < x n sup{3x n 3x 0 n N and x 0, x,..., x n R with x 0 < x < < x n } sup{3( ( ))} 6. For each k N {0} define x k : (π(k + 2 )) /2 [0, ]. If n N, then G(x k ) G(x k ) Since k x 2 k ( )k x 2 k ( )k diverges, this implies that G / BV ([, ]). 33. Define an increasing, right continuous function G : R R by lim y x F (y), x < b G(x) : F (b), x b. (x 2 k + x2 k ) x 2 k π(k + 2 ) Then G(b) G(a) F (b) F (a) and G F almost everywhere on (a, b). Moreover, there exists a regular measure µ G on R such that µ G ((a, b]) G(b) G(a) for all a, b R with a < b. Let dµ G dλ + f dm be the Lebesgue-Radon- Nikodym representation of µ G (note that λ 0 by the proof of Lebesgue-Radon-Nikodym), so that and G(x + h) G(x) µ G ((x, x + h]) lim lim h 0 h h 0 m((x, x + h]) f(x) G(x + h) G(x) G(x h) G(x) G(x) G(x h) µ G ((x h, x]) lim lim lim lim h 0 h h 0 h h 0 h h 0 m((x h, x]) f(x) for almost all x R. In particular F (b) F (a) G(b) G(a) µ G ((a, b]) (a,b] f dm b a G dm b a F dm. 35. Since F and G are continuous on [a, b], there exists M (0, ) such that F (x) M and G(x) M for all x [a, b]. Let ε (0, ) and choose δ (0, ) such that F (b k ) F (a k ) < ε 2M and G(b k ) G(a k ) < for every finite collection of disjoint intervals (a, b ), (a 2, b 2 ),..., (a n, b n ) [a, b] with n (b k a k ) < δ. It follows that F G is absolutely continuous on [a, b], because (2) implies that F G(b k ) F G(a k ) ε 2M ( F (b k )G(b k ) F (b k )G(a k ) + F (b k )G(a k ) F (a k )G(a k ) ) F (b k ) G(b k ) G(a k ) + M G(b k ) G(a k ) + 0 F (b k ) F (a k ) G(a k ) F (b k ) F (a k ) M 2πk. (2)

11 < M ε 2M + ε 2M M ε. Therefore, by the fundamental theorem of calculus for Lebesgue integrals, F G is differentiable almost everywhere on [a, b], (F G) L ([a, b], m) and F (b)g(b) F (a)g(a) b a (F G) dm. Since F and G are differentiable almost everywhere on [a, b] (again by the fundamental theorem), it is clear that (F G) F G + GF almost everywhere on [a, b] and hence F (b)g(b) F (a)g(a) b a (F G + GF ) dm. 37. Suppose F is Lipschitz continuous, and choose M (0, ) such that F (y) F (x) M y x for all x, y R. If ε (0, ) and (a, b ), (a 2, b 2 ),..., (a n, b n ) is a finite collection of disjoint intervals with n (b k a k ) < ε M then F (b k ) F (a k ) M b k a k M m (b k a k ) < M ε M ε. Therefore F is absolutely continuous, so it is differentiable almost everywhere. It follows that, for almost every x R, F (x) lim F (y) F (x) y x y x lim F (y) F (x) M y x lim lim M M. y x y x y x y x y x Conversely, suppose F is absolutely continuous and there exists M [0, ) such that F M almost everywhere. If x, y R and x y then F (y) F (x) y x F dm y x F dm y x M dm M(y x) M y x Therefore F is Lipschitz continuous with Lipschitz constant M. 39. Set F 0 : F, and for each k N {0} define a non-negative, increasing, right continuous function G k : R R by lim y a F k (y), x < a G k (x) : lim y x F k (y), x [a, b) F k (b), x b. Fix k N {0}, and note that G k F k almost everywhere on [a, b]. If x [a, b), there is a sequence (x n) n in (x, b) which decreases to x. For each n N it is clear that F k(x n )χ {k} is integrable with respect to the counting measure ν on N; indeed F k (x n )χ {k} dν F k (x n ) F (x n ) <. Since each F k is increasing these functions are dominated by F k(x )χ {k}, and by definition lim n F k (x n )χ {k} G k (x)χ {k} pointwise. Hence, by the dominated convergence theorem G 0 (x) lim F (x n) lim n n F k (x n )χ {k} dν G k (x)χ {k} G k (x). Moreover, if x (, a) then G 0 (x) G 0 (a) G k(a) G k(x) and similarly, if x [b, ) then G 0 (x) F 0 (b) F k(b) G k(x). There exists an outer regular Radon measure µ k on R such that

12 µ k ((s, t]) G k (t) G k (s) for all s, t R with s < t. Let dµ k dλ k + f k dm be the Lebesgue-Radon-Nikodym representation of µ k (note that λ k 0 by the proof of Lebesgue-Radon-Nikodym), so that and G k (x + h) G k (x) µ k ((x, x + h]) lim lim h 0 h h 0 m((x, x + h]) f k(x) G k (x + h) G k (x) G k (x h) G k (x) G k (x) G k (x h) µ k ((x h, x]) lim lim lim lim h 0 h h 0 h h 0 h h 0 m((x h, x]) f k(x) for almost all x R, and hence G k f k almost everywhere. By definition F k 0 almost everywhere. Note that ( ) µ k ((s, t]) µ k ((s, t]) (G k (s) G k (t)) G 0 (s) G 0 (t) µ 0 ((s, t]) for all s, t R with s < t. This implies that µ k µ 0, because both measures are outer regular and every open subset of R is a disjoint union of countably many left-open, right-closed intervals. For each k N choose A k B R such that λ k (A k ) 0 and m(a c k ) 0. Then A k is null with respect to the measure λ k, while its complement Ac k is m-null. Moreover, it is clear that the Borel measure ρ defined by ρ() : f k dm is absolutely continuous with respect to m. Therefore λ 0 λ k and f 0 f k almost everywhere, by the Lebesgue-Radon-Nikodym theorem. In particular, F f 0 f k F k almost everywhere on [a, b]. 40. By construction F n is continuous and F n (R) [0, ] for all n N. Therefore, if n N then G 2 k F k 2 k F k 2 k 2 n kn+ and hence G is the uniform limit of a sequence of continuous functions. This implies that G is continuous. If x, y R kn+ and x < y there exists n N such that [a n, b n ] (x, y), so that F n (x) 0 while F n (y) and hence G(x) F n (x) + k N k n 2 k F k (x) < F n (y) + k N k n 2 k F k (x) F n (y) + k N k n 2 k F k (y) G(y). The complement of the Cantor set C is open, so if x (0, ) \ C then F is constant on a neighbourhood of x, implying that F (x) 0. Since m(c) 0 and F is constant on (, 0) and (, ), this shows that F 0 almost everywhere. If n N, the preimages of (0, ) and (0, ) \ C under the map x x an b n a n have measures b n a n and 2k 3 k (b n a n ) b n a n, which implies that F n 0 almost everywhere by the chain rule. Therefore, by the previous exercise G 2 k F k 0 almost everywhere. 4. (a) If x, y [0, ] and x < y then F (y) m([0, y] A) m([0, x] A) + m((x, y] A) > F (x), so F is strictly increasing on [0, ]. Let ε (0, ). If (a, b ), (a 2, b 2 ),..., (a n, b n ) is a finite collection of disjoint intervals such that n (b k a k ) < ε then F (b k ) F (a k ) m([0, b k ] A) m([0, a k ] A) m((a k, b k ] A) m((a k, b k ]) < ε, so F is absolutely continuous on R. Hence there exists an outer regular Radon measure µ F on R such that µ F ((a, b]) F (b) F (a) b a F dm for all a, b R with a < b. Moreover F (b) F (a) m([0, b] A) m([0, a] A) m((a, b] A) 2 b a χ A dm

13 for all a, b R with a < b. This implies that dµ F F dm χ A dm, since the subset of B R on which these (finite) measures agree is a σ-algebra containing a set which generates B R. By the Lebesgue-Radon-Nikodym theorem F χ A almost everywhere, so F 0 on [0, ] \ A, which has positive measure because m([0, ] A) < m([0, ]). (b) Let ε (0, ). If (a, b ), (a 2, b 2 ),..., (a n, b n ) is a finite collection of disjoint intervals such that n (b k a k ) < ε G(b k ) G(a k ) m([0, b k ] A) m([0, b k ] \ A) m([0, a k ] A) + m([0, a k ] \ A) m((a k, b k ] A) m((a k, b k ] \ A) (m((a k, b k ] A) + m((a k, b k ] \ A)) m((a k, b k ]) < ε, so G is absolutely continuous on R. In particular G is differentiable on some R with m( c ) 0. Moreover b a G dm G(b) G(a) m([0, b] A) m([0, b] \ A) m([0, a] A) + m([0, a] \ A) m((a, b] A) m((a, b] \ A) b a b a χ A dm b a (χ A χ A c) dm χ A c dm for all a, b R such that a < b. This implies that G (χ A χ A c) almost everywhere, by the Lebesgue-Radon- Nikodym theorem applied to each of the measures G dm and (χ A χ A c) dm (which are equal because they are finite and agree on a generating set of B R ). If a, b [0, ], a < b and G is monotone on (a, b), then G 0 or G 0 on (a, b) (because the difference quotient at each point is either non-negative or non-positive). This is a contradiction because χ A χ A c takes on the values and on A and A c respectively, both of which have positive measure and hence G (x) and G (y) for some x, y (a, b). 42. (a) Suppose F is convex and let s, t, s, t (a, b) such that s s < t and s < t t. Then t s t s (0, ] and hence F (t) F ( ) ( ( t s t s t s (t s) + s F t s t + t s ) ) t s t s s t s F (t ) + because if t s t s then t t. Moreover s s t s [0, ), and s s F (s ) F ( s ) s t s (t s) + s F ( s s t s t + t s 0 iff s s, so ) ) s ( s s t s s s t s F (t ) + ( t s t s ) F (s), ( ) s s t F (s). s It follows that F (t) F (s) t s t s F (t ) t s t s F (s) 3

14 and hence t s t s F (t ) t s ( ) t s s ( s t s t s t s F (t ) (t s)(t s (s s)) t s ( F (s ) s s t s F (t ) t s t s ( t s t s + t s s s t s t s ) t s F (t ) F (s ) s s ( ) F (s ) s s t s F (t ) ) t s F (t ) ) F (t ) t s t s F (s ) (t s) t s + s s (t s)(t s ) F (t ) t s t s F (s ) t s t s (F (t ) F (s )) F (t) F (s) t s F (t ) F (s ) t s. (3) Conversely, suppose that (3) holds for all s, t, s, t (a, b) with s s < t and s < t t. Let x, y (a, b) and λ (0, ). Also, set z : λy + ( λ)x. If x < y, then x < z < y so F (z) F (x) z x F (y) F (z), y z which implies that and hence F (z) z x + F (z) y z F (y) y z + F (x) z x ( F (z) z x + ) ( F (y) y z y z + F (x) ) z x ( ) y z + z x ( F (y) (z x)(y z) y z + F (x) ) z x ((z x)f (y) + (y z)f (x)) y x ((λy λx)f (y) + (( λ)y ( λ)x)f (x)) y x λf (y) + ( λ)f (x). We obtain the same result for the case x > y by swapping x with y and replacing λ with λ. Finally, if x y then F (z) F (x) λf (x) + ( λ)f (x) λf (y) + ( λ)f (x). Therefore F is convex. (b) Suppose that F is convex, and let [t, s ] (a, b). There exists s (a, t) and t (s, b). Let ε (0, ) and set { F (t ) F (s } ) F (s) F (t) M : max t s,. t s If (a, b ), (a 2, b 2 ),..., (a n, b n ) [t, s ] is a finite collection of disjoint intervals with n (b k a k ) < ε M then M F (t) F (s) t s F (b k) F (a k ) F (t ) F (s ) b k a k t s M 4

15 for all k {, 2,..., n} and hence F (b k ) F (a k ) M b k a k M (b k a k ) < M ε M ε. Thus F is absolutely continuous on [t, s ], and differentiable on some (a, b). If x 0, y 0 and x 0 < y 0, then for all x (x 0, y 0 ) and y (y 0, b), so for all y (y 0, b) and hence This implies that F is increasing on. F (x) F (x 0 ) x x 0 F (y) F (y 0) y y 0 F (x 0 ) lim x x0 F (x) F (x 0 ) x x 0 F (y) F (y 0) y y 0 F (x 0 ) lim y y0 F (y) F (y 0 ) y y 0 F (y 0 ). Conversely, suppose that F is absolutely continuous and F is increasing on the set (a, b) where F is differentiable. Since F is absolutely continuous, m((a, b) \ ) 0. Let x, y (a, b), λ (0, ) and suppose that x < y. Define z : λy + ( λ)x and T : [x, y] [x, z] by T (t) : λt + ( λ)x. Note that F T is absolutely continuous, because if (a, b ), (a 2, b 2 ),..., (a n, b n ) [x, y] is a finite collection of disjoint intervals, then so is (T (a ), T (b )), (T (a 2 ), T (b 2 )),..., (T (a n ), T (b n )) [x, z], and (T (b k ) T (a k )) (λb k λa k ) λ (b k a k ) < (b k a k ). Since T (t) λt + ( λ)t t for all t [x, y] and F is increasing on, it follows that F (T (y)) F (T (x)) Therefore y x (F T ) dm y x F (T (t))t (t) dt λ y x F (T (t)) dt λ F (z) F (T (y)) F (T (x)) + F (x) λ(f (y) F (x)) + F (x) λf (y) + ( λ)f (x), y x F (t) dt λ(f (y) F (x)). which implies that F is convex by the same argument used in part (a) for the cases x > y and x y. (c) Fix t 0 (a, b) and let t, t 2, t 3, t 4 (a, b) such that t t 2 < t 0 < t 3 t 4. From part (a) F (t 0 ) F (t ) t 0 t F (t 0) F (t 2 ) t 0 t 2 F (t 3) F (t 0 ) t 3 t 0 F (t 4) F (t 0 ) t 4 t 0, which implies that lim t t0 F (t 0 ) F (t) t 0 t and lim t t0 F (t) F (t 0 ) t t 0 exist (by monotone convergence) and F (t 0 ) F (t ) t 0 t F (t 0 ) F (t) F (t) F (t 0 ) lim lim F (t ) F (t 0 ) t t0 t 0 t t t0 t t 0 t t 0 (by taking one limit at a time) for all t (a, t 0 ) and t F (t (t 0, b). Set β : lim 0 ) F (t) t t0 t 0 t and let t (a, b). If t > t 0 then F (t) F (t 0 ) β(t t 0 ), and if t t 0 then F (t) F (t 0 ) 0 β(t t 0 ). Otherwise t < t 0, in which case F (t 0 ) F (t) β(t 0 t) and hence F (t) F (t 0 ) β(t t 0 ). 5

16 (d) Since b g > 0 everywhere, (b g) dµ > 0 and hence g dµ < b dµ b. Similarly a < g dµ, so there exists β R such that F (t) F ( g dµ) β(t g dµ) for all t (a, b). In particular ( ( ) ( )) F g dµ F g dµ + β g g dµ dµ ( ) F g dµ dµ + β g dµ β g dµ dµ ( ) F g dµ + β g dµ β g dµ ( ) F g dµ. 6