T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.


 Reynold Ramsey
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1 Rel Vribles, Fll 2014 Problem set 5 Solution suggestions Exerise 1. Let f be bsolutely ontinuous on [, b] Show tht nd T b (f) P b (f) f (x) dx [f ] +. Conlude tht if f is in AC then it is the differene of two monotone bsolutely ontinuous funtions. In homework 4 exerise 5 we showed tht f (x) dx T b (f) for ll funtions f of bounded vrition. Sine bsolutely ontinuous funtions hve bounded vrition then it suffies to only show the other inequlity. Let {x i } n i0 be prtition of of [, b]. Then f(x i ) f(x i 1 ) xi f (x) dx x i 1 xi x i 1 f (x) dx Tking supremum over ll prtitions of [, b] it follows tht Hene T b (f) T b (f) f (x) dx. f (x) dx. For the seond prt, we similrly see tht for ny prtition {x i } n i0 of [, b] we hve [f(x i ) f(x i 1 )] + [ x i ] + f (x) dx x i 1 1 xi x i 1 [f (x)] + dx f (x) dx. [f (x)] + dx.
2 Tking supremum over ll prtitions of [, b] it follows tht P b (f) [f ] +. For the onverse inequlity note tht f(x) P x (f) N x (f) + f(), so f (x) P x (f) N x (f) lmost everywhere. Sine P x (f) nd N x (f) re nonderesing funtions then P x (f) 0 nd N x (f) 0 lmost everywhere. So [f(x) ] + [P x (f) N x (f) ] + P x (f) lmost everywhere. So we onlude tht [f ] + P x (f) dx P b (f) P (f) P b (f), whih is the inequlity tht we wnted. Thus [f ] + P b (f). Finlly, sine P x (f) nd T x (f) re both integrls they re in prtiulr bsolutely ontinuous. Sine f(x) P x (f) N x (f) + f() 2P x (f) P x (f) N x (f) + f() 2P x (f) T x (f) + f() then f is the differene of two monotone bsolutely ontinuous funtions. Exerise 2. Show tht if f is in AC on [, b] nd if f is never zero there, then g 1 f is lso in AC on [, b]. Sine f is ontinuous funtion on the onneted set [, b] nd never zero, then either f(x) > 0 for ll x [, b] or f(x) < 0 for ll x [, b]. Without loss of generlity we n ssume tht f(x) > 0 for ll x [, b]. Sine [, b] is ompt then f ttins its minimum on [, b], so there exists > 0 so tht f(x) > 0 for ll x [, b]. Fix ε > 0. Sine f is bsolutely ontinuous we find δ > 0 so tht f( i ) f(b i ) < 2 ε 2
3 whenever {( i, b i )} n is disjoint olletion of intervls suh tht This implies tht i b i < δ. 1 g( i ) g(b i ) f( i ) 1 f(b i) f( i ) f(b i ) f( i )f(b i ) 1 f( 2 i ) f(b i ) < ε ε whenever {( i, b i )} n is disjoint olletion of intervls suh tht Hene g 1 f i b i < δ. is bsolutely ontinuous on [, b]. Exerise 3. Show tht there is stritly inresing singulr funtion on [0, 1]. (Rell tht monotone funtion f on [, b] is lled singulr if f 0 lmost everywhere.) We follow the suggestions given by Royden. Lemm 1. Let f be monotone inresing funtion. Then f g + h where g is bsolutely ontinuous nd h is singulr, nd both g nd h re monotone inresing. Proof: Sine f is monotone then f exists lmost everywhere nd is mesurble, nd f 0 sine f is inresing. We define g(x) x f (t) dt for ll x. Now g is bsolutely ontinuous s n integrl nd g is monotone inresing sine f 0. We define h f g. Now 3
4 h f g f f 0 lmost everywhere so h is singulr. To see tht h is monotone inresing, fix x y nd observe tht h(y) h(x) f(y) y f(y) f(x) f(t) dt f(x) + y x f (t) dt. x f (x) dx Sine f is monotone then y f (t) dt f(y) f(x). So we onlude tht h(x) h(y). Finlly, sine f g + h then we re x done. Lemm 2. Let f be nonderesing singulr funtion on [, b]. Then f hs the following property (S): Given ε > 0, δ > 0, there is finite olletion {[y k, x k ]} n of nonoverlpping intervls suh tht x k y k < δ nd (f(x k ) f(y k )) > f(b) f() ε. Proof: Fix ε > 0 nd δ > 0. Denote A {x [, b] : f (x) 0} nd κ. Now if x A then ε b f(x + h) f(x) lim h 0 h f (x) 0, so there exists h x > 0 so tht for ll 0 < h < h x, f(x + h) f(x) h < κ, i.e. f(x + h) f(x) < κh. Now the olletion V {[x, x + h] : x A, h < h x } is Vitli over of A. Sine m(a) < then by Vitli overing theorem there is finite disjoint olletion {[x i, x i + h i ]} n V so tht ( n ) m A \ [x i, x i + h i ] < δ. 4
5 Denote I i [x i, x i + h i ] for ll i. Sine m([, b]) m(a) then b i n l(i i ) b m( i n I i ) m(a \ i n I i ) < δ, so i n l(i i) > b δ. Denote I i I i the open intervl obtined by tking the endpoints of I i wy. Now the omplement of i n I i in [, b] is finite olletion of disjoint losed intervls, denoted by {J i } k, nd dditionlly the possibility exists tht the singleton endpoints {, b} re lso in the omplement in the se of x 1 nd x n + h n b. In this se we tret J 1 {} nd J k {b}. We my ssume tht the intervls re listed in n inresing order, nd we denote J i [ i, b i ] for ll i. Note tht eh intervl J i is n intervl of the form [x in, x in+1 +h in+1] for some i n unless we re deling with the singleton endpoints, nd i k l(j i) < δ. Thus we hve k (f(b i ) f( i )) f(b k ) f( 1 ) + whih finishes the proof. f(b) f() > f(b) f() κ (f( i ) f(b i 1 )) i2 k (f(b i 1 ) f( i )) i2 k i2 h in > f(b) f() κ(b ) f(b) f() ε, Lemm 3. Let f be nonderesing funtion on [, b] with property (S) from Lemm 2. Then f is singulr. Proof: From Lemm 1 it follows tht f g + h where g is bsolutely ontinuous nd h is singulr, nd both g nd h re nonderesing. Fix ε > 0. Sine g is bsolutely ontinuous then there exists δ > 0 so tht (g(y i ) g(x i )) < ε 2 5
6 for ny olletion of disjoint intervls {[x i, y i ]} n with y i x i < δ. Now sine f hs property (S), for the fixed ε > 0 nd δ > 0 there is finite olletion {[ i, b i ]} n of disjoint intervls suh tht i b i < δ nd (f(b i ) f( i )) > f(b) f() ε 2. Sine g(x) x f (t) dt for ll x, then by denoting b 0 nd n+1 b, we hve g(b) g() (g( i+1 ) g(b i )) + i0 i+1 (g(b i ) g( i )) < f (t) dt + ε i0 b i 2 (f( i+1 ) f(b i )) + ε 2 i0 f(b) f() (f(b i ) f( i )) + ε 2 ( < f(b) f() f(b) f() ε ) + ε 2 2 ε. Sine the hoie of ε > 0 ws rbitrry, it follows tht g(b) g() for ll x. Sine g() 0 then g 0. Thus f h nd sine h ws singulr then f is singulr. Lemm 4. Let (f n ) n1 sequene of nonderesing singulr funtions on [, b] so tht the funtion f(x) f n (x) 6 n1
7 is everywhere finite. Show tht f is lso singulr. Let ε > 0 nd δ > 0 be fixed. Sine (f n (b) f n ()) f(b) f() <, n1 then there exists N N suh tht (f n (b) f n ()) < ε 2. nn+1 Denote g(x) N f n(x). Sine eh f n is nonderesing then g is nonderesing, nd sine g (x) N f n(x) 0 for lmost every x, then g is singulr. By Lemm 2 the funtion g hs the property (S). So there exists finite olletion of disjoint intervls {[ i, b i ]} m so tht m b i i < δ nd m (g(b i ) g( i )) > g(b) g() ε 2. For this sme olletion of intervls we lso hve m m ( ) (f(b i ) f( i )) (f n (b i ) f n ( i )) n1 m ( N ) (f n (b i ) f n ( i )) n1 m (g(b i ) g( i )) > g(b) g() ε 2 N (f n (b) f n ()) ε 2 n1 f(b) f() nn+1 > f(b) f() ε 2 ε 2 (f n (b) f n ()) ε 2 f(b) f() ε. 7
8 So we onlude tht f hs the property (S). Sine f is lso nonderesing then by Lemm 3 f is singulr. We now return to our exerise. Sine Q is ountble then the set of ll intervls with rtionl endpoints is ountble. So we n enumerte this set, sy by {[ n, b n ] : n N}, where n, b n Q [0, 1] nd n < b n. We then define f n : [0, 1] R for eh n N by setting ( x f n (x) 2 n n ) C, b n n where C : R R is the Cntor funtion extended ontinuously to the whole rel line by setting C (,0] 0 nd C [1, ) 1. Now note tht f n (x) 0 for ll x n nd f n (x) 2 n for ll x b n, f n(x) 0 lmost everywhere nd f n is nonderesing. It follows tht f(x) : f n (x) n1 is everywhere finite nd thus by Lemm 4 f is singulr on [0, 1]. It lso follows tht f is nonderesing beuse eh f n is. To see tht it is stritly inresing ssume tht x < y. Sine the rtionls re dense in the rels there exists q Q so tht x < q < y. So ny f n ssigned to rtionl intervl with left endpoint t q gives f n (x) 0 nd f n (y) > 0. So it follows tht f(x) < f(y). Hene f is stritly inresing singulr funtion on [0, 1]. Exerise 4. () Let F be AC on [, d] nd let g be stritly inresing nd AC on [, b] with g d. Then F g is AC on [, b]. Fix ε > 0. Sine F is in AC there is κ > 0 so tht F (b i ) F ( i ) < ε whenever {[ i, b i ]} n is disjoint olletion of intervls with b i i < κ. 8
9 Now use κ > 0 in the definition of the bsolute ontinuity of g to find δ > 0 so tht g(y i ) g(x i ) < κ whenever {[x i, y i ]} n is disjoint olletion of intervls with y i x i < δ. So fix disjoint olletion of intervls {[x i, y i ]} n with y i x i < δ. Then sine g is inresing nd ontinuous, the intervls [x i, y i ] mp to intervls under g. More preisely, we hve g[x i, y i ] [g(x i ), g(y i )] for ll i nd the intervls re disjoint s g is stritly inresing. By the bsolute ontinuity of g we lso hve for the length of the intervls tht g(y i ) g(x i ) < κ. So treting the disjoint olletion of intervls {[g(x i ), g(y i )]} n in the definition of the bsolute ontinuity of F we hve tht F (g(y i )) F (g(x i )) < ε. So we hve shown tht there is δ > 0 suh tht for ny disjoint olletion of intervls {[x i, y i ]} n with we hve y i x i < δ. (F g)(y i ) (F g)(x i ) < ε. Hene F g is bsolutely ontinuous. 9
10 (b) Let E {x : g (x) 0}. Then m(g(e)) 0. Fix ε > 0. The proof follows similr rgument s we did in Lemm 2 of Exerise 3. Let κ ε. There now exists δ > 0 so 4(b ) tht A g < ε whenever A [, b] with m(a) < δ. Fix x E. 2 Sine g(x + h) g(x) lim h 0 h 0, there exists h x > 0 so tht for ll 0 < h < h x we hve g(x + h) g(x) h < κ. In other words g(x + h) g(x) < κh for ll 0 < h < h x. We then define V {[x, x + h] : x E, 0 < h < h x }, whih is Vitli over of the set E. Sine m(e) <, by Vitli overing theorem we find finite disjoint subolletion of V, denoted by {[x i, x i + h i ]} n, so tht ( m E \ n ) [x i, x i + h i ] < δ 2. Denote U n [x i, x i + h i ], nd hoose n open set O E \ U with m(o \ (E \ U)) < δ 2. Now so m(o) m(e \ U) m(o \ (E \ U)) < δ 2, m(o) < δ 2 + m(e \ U) < δ 2 + δ 2 δ. We then write O s ountble disjoint union O I k of open intervls I k, so m(g(e \ U)) m(g(o)) m(g(i k )) m ( [min g(x), mx g(x)] ) x I k x I k (mx x I k g(x) min x I k g(x)) 10 g I k O g < ε 2,
11 sine m(o) < δ. Above we lso used the ft tht sine g is ontinuous on the ompt intervls I k then the mximums nd minimums exist. On the other hnd, if t [x i, x i + h i ] then 0 < t x i < h i, so g(t) g(x i ) < κh i. Thus g[x i, x i + h i ] B(g(x i ), κh i ), whih implies tht m(g[x i, x i + h i ]) m(b(g(x i ), κh i )) 2κh i. We then observe tht m(g(e U)) m(g(u)) m(g[x i, x i + h i ]) 2κh i ε 2κ(b ) 2 4(b ) (b ) ε 2. So finlly, m(g(e)) m(g(e \ U)) + m(g(e U)) < ε 2 + ε 2. Sine the hoie of ε > 0 ws rbitrry, then m(g(e)) 0. Exerise 5. Let g be n bsolutely ontinuous monotone funtion on [0, 1] nd E set of mesure zero. Then g(e) hs mesure zero. Fix ε > 0. Sine g is bsolutely ontinuous there exists δ > 0 so tht g(b i ) g( i ) < ε whenever {[ i, b i ]} n is disjoint olletion of intervls with b i i < δ. Tke n open set O E so tht m(o) < δ. Suh set exists sine m(e) 0. We then write O s ountble disjoint union O I k of open intervls I k ( k, b k ). Let m N be fixed. Then m m ( ) b k k l(i k ) l(i k ) m I k m(o) < δ. 11
12 Sine g is monotone nd ontinuous then g(i k ) is the intervl [g( k ), g(b k )] or [g(b k ), g( k )], so by bsolute ontinuity of g we hve m m(g(i k )) m g(b k ) g( k ) < ε. Sine this pplies for ll m N then l(g(i k )) < ε. In prtiulr, it follows tht m(g(e)) m(g(o)) m(g(i k )) < ε. Sine this holds for ll ε > 0, then m(g(e)) 0. Exerise 6. Let g be monotone inresing bsolutely ontinuous funtion on [, b] with g() nd g(b) d. () Show tht for ny open set O [, d], m(o) g 1 (O) g (x) dx. We first write the open set O s ountble disjoint union O I k of open intervls I k ( k, b k ). Sine g is ontinuous then g 1 ( k, b k ) is n open subset of [, b] for ll k, nd sine g is monotone inresing then g 1 ( k, b k ) is n intervl J k (x k, y k ) with g(x k ) k nd g(y k ) b k. Moreover the intervls J k re disjoint sine I k re nd the preimges of disjoint sets re disjoint. Now g 1 (O) is the ountble disjoint union g 1 (O) J k, so g 1 (O) s required. g g (g(y k ) g(x k )) J k m(i k ) m(o), (b k k ) 12
13 (b) Let H {x : g (x) 0}. If E is subset of [, d] with m(e) 0, then F g 1 (E) H hs mesure zero. Let E be subset of [, d] with m(e) 0, nd denote F g 1 (E) H. Assume towrds ontrdition tht m(f ) > 0. Sine g is inresing then g 0 lmost everywhere, so g (x) > 0 for ll x F nd thus F g > 0. Fix n open set O E. Sine F (g 1 (E) H) g 1 (O), then by prt () we hve m(o) g g. g 1 (O) This lower bound holds uniformly for ll open sets O E, so by tking infimum we get m(e) F g > 0. This is ontrdition with the ssumption m(e) 0, so we onlude tht m(f ) 0. F () If E is mesurble subset of [, d], then F g 1 (E) H is mesurble nd m(e) g χ E (g(x))g (x) dx. F Sine E is mesurble then by homework 1 exerise 1 we n write E A N, where A is F σ set nd N hs mesure zero, nd A N. Then F g 1 (E) H g 1 (A N) H (g 1 (A) g 1 (N)) H (g 1 (A) H) (g 1 (N) H). Now H (g ) 1 (R \ {0}) is mesurble sine g is, nd thus g 1 (A) H is mesurble s A ws Borel set. Also, by prt (b) the set g 1 (N) H hs mesure zero nd is thus mesurble. Hene F is mesurble. Note then tht g g χ g 1 (E)g χ E (g)g F g 1 (E) sine χ g 1 (E) χ E (g). So we hve to show tht m(e) equls the bove vlue. We will prove it in severl steps. 13
14 (i) We first ssume tht E is n open set. By prt () we hve m(e) g 1 (E) g χ g 1 (E)g So the sttement is true if E is open. χ E (g)g. (ii) Assume then tht E is G δ set. So there exists nested sequene of open sets {O n } n1 so tht O n+1 O n for ll n nd E n1 O n. Now lim n χ On χ E pointwise nd by dominted onvergene theorem nd prt (i) we hve m(e) lim m(o n ) lim n χ E (g)g, n χ On (g)g so the sttement holds for ll G δ sets. lim χ O n (g)g n (iii) Assume then tht E is ny mesurble set. Then by exerise 1 of problem set 1 we n find G δ set G nd set N with m(n) 0 so tht E N G nd E N. Thus by prt (ii) we hve m(e) m(e N) m(g) g g 1 (E N) H g 1 (E) H g g 1 (E) H χ E (g)g. χ G (g)g g + So the sttement is true for ll mesurble E. g g 1 (G) H g g 1 (N) H (d) If f is nonnegtive mesurble funtion on [, d], then (f g)g is mesurble on [, b] nd f(x) dx f(g(x))g (x) dx. 14
15 Sine g is monotone inresing bsolutely ontinuous funtion nd f is mesurble then (f g) is mesurble. Sine g is mesurble then so is (f g)g. We prove the sttement in severl steps by using the stndrd pproximtion rgument. (i) Assume first tht f χ E is hrteristi funtion of some mesurble set E [, d]. Then by prt () we hve f(x) dx χ E m(e) f(g(x))g (x) dx, χ E (g(x))g (x) dx so the sttement is true for ll hrteristi funtions. (ii) Assume then tht f is simple funtion f n kχ Ek for some onstnts k nd mesurble sets E k [, d]. Then by prt (i) nd linerity of the integrl we hve f(x) dx k χ Ek k k χ Ek (g(x))g (x) dx k χ Ek (g(x))g (x) dx f(g(x))g (x) dx, so the sttement holds for ll simple funtions. χ Ek (iii) Let f then be ny nonnegtive mesurble funtion, nd tke nonderesing sequene (ϕ) n1 of simple funtions so tht ϕ n f for ll n nd ϕ n f pointwise. We use monotone 15
16 onvergene theorem nd prt (ii) to onlude tht f(x) dx lim lim ϕ n(x) dx lim n n n ϕ n (g(x))g (x) dx f(g(x))g (x) dx. So this proves the sttement. ϕ n (x) dx lim ϕ n(g(x))g (x) dx n 16
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