W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying


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1 Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering) Given set E R with λ (E) <, Vitli cover V of E, nd some ε > 0, there re disjoint I 1,..., I n V with λ ( E \ (I 1 I n ) ) < ε. Proof: Strt by picking n open set O E with λ(o) <, nd let W = {I V: I O}. It is esy to check tht W is Vitli cover of E. We now forget bout E for while, nd pick pirwise disjoint intervls I 1, I 2,... in W. We shll do so one by one, strting with I 1, nd we shll do it greedily, trying to mke ech intervl s lrge s possible. Now there my not be ny lrgest size vilble, so we settle for getting within constnt fctor: λ(i k+1 ) > 1 2 sup{ λ(j): J W nd J I j = for j = 1,...,k }. (1) In ordinry prose: At ech step, pick the next intervl disjoint from the other intervls picked this fr, so tht ny other intervl we could hve picked is less thn twice s long. It is conceivble tht this process cnnot go on forever. But if so, we hve lredy covered ll of E with finite, pirwise disjoint set of intervls from W, so we re done. If x E \ (I 1 I k ), then since I 1 I k is closed, there is some δneighbourhood of x tht does not meet I 1 I k, nd so we cn continue with the selection for one more step. If the process continues forever, then we find ( ) λ(i k ) = λ I k λ(o) <, so the sum converges. Pick n lrge enough so tht k=n+1 λ(i k ) < ε, nd let J k be the intervl with the sme center s I k, but five times the length. I clim which in turn implies n E \ I k J k, k=n+1 λ ( E \ n ) I k k=n+1 nd tht is enough to finish the proof. λ(j k ) = 5 k=n+1 λ(i k ) < 5ε, To prove the clim, let x E \ (I 1 I n ). Since W is Vitli cover for E, we cn find some I W with x I nd I I k = for k = 1,..., n. Let m be the smllest nturl number with I I m. Such n m must exist, for otherwise, I would lwys be one of the intervls we could hve picked, nd therefore λ(i k+1 ) > 1 2 λ(i ) for ll k. But this is impossible, since λ(i k ) <. There is more when k = m 1, I is still one of the J s, so we must hve λ(i m ) > 1 2 λ(i ). I I m J m This drwing illustrtes the fct tht, when I m I nd λ(i m ) > 1 2 λ(i ), then J m I. In prticulr, x J m, nd the proof is finlly complete.
2 Dini derivtives These re defined by D + f (y) f (x) f (y) f (x) f (x) = lim, D + f (x) = lim, y x y x y x y x D f (y) f (x) f (y) f (x) f (x) = lim, D f (x) = lim, y x y x y x y x nd clled the upper right, lower right, upper left, ndn lower left Dini derivtives, respectively. The Dini derivtives hve these simple properties: D ± f D ± f, D ± ( f ) = D ± f, D ± g = D f where g (x) = f ( x). Note tht ny of the Dini derivtives cn tke on either of the vlues ±. Our interest in the Dini derivtives stems from the fct tht they lwys exists, nd moreover f hs left derivtive x if nd only if D f (x) = D f (x), right derivtive if nd only if D + f (x) = D + f (x), nd (twosided) derivtive if nd only if ll four Dini derivtives re the sme.
3 Limits to growth for monotone function From now on, < b re rel numbers, nd f : (, b) R is n incresing function. 3 Lemm D + f (x) < nd D f (x) < for λ.e. x (,b). Proof: Let E = { x (,b): D + f (x) = }. The ide of the proof is tht f must hve unbounded growth on this set, if it hs positive mesure. To get this to work, we will first ssume tht f M on (,b). Pick ny (lrge) number m. It follows from the definition tht the set of intervls [x, y] where f (y) f (x) x E, y (x,b), nd > m y x is Vitli cover for E. Pick ny ε > 0, nd let [x k, y k ] be pirwise disjoint intervls of this type for k = 1,..., n with λ (E \ ([x 1, y 1 ] [x n, y n ])) < ε. It follows tht (y 1 x 1 ) + + (y n x n ) > λ (E) ε, nd therefore ( f (y1 ) f (x 1 ) ) + + ( f (y n ) f (x n ) ) > m(λ (E) ε). But becuse f is incresing, the intervls (f (x k ), f (y k )) re pirwise nonoverlpping in [ M, M], so we must finlly hve 2M > m(λ (E) ε). If λ (E) > 0, we cn choose ε < λ (E) nd m lrge enough for this to be contrdiction. If f is unbounded, we pply the bove to the restriction of f to slightly smller intervls ( + n 1,b n 1 ) insted, nd use the fct tht countble union of sets of mesure zero still hs mesure zero. We hve proved tht D + f <.e. To show the sme for D f, pply the first result to the function x f ( x) on ( b, ).
4 Limits to wiggliness for monotone function 4 Lemm D + f (x) = D + f (x) nd D f (x) = D f (x) for λ.e. x (,b). Proof: As in the previous lemm, the second.e. equlity follows from the first by pplying it to x f ( x). So we only prove the first one. Note tht D + f (x) D + f (x) lwys, so we need to show tht the set of x where D + f (x) > D + f (x) hs mesure zero. But for ny such x, we cn lwys find two rtionl numbers r nd s with D + f (x) > s > r > D + f (x). Since the number of rtionl pirs (r, s) with r < s is countble, we only need to show tht the set hs mesure zero. E = { x (,b): D + f (x) > s > r > D + f (x) } (r, s Q fixed) The proof ide is like tht for the previous lemm, only done twice: Use one inequlity to show tht the verge growth rte of f on set of intervls nerly covering E is less thn r, then use the other to find subintervls, still nerly covering E, where the verge growth rte is greter thn s. Then combine the two to get contrdiction, if λ (E) > 0. To get strted, then, we first pick some ε > 0, nd n open set O E with λ(o) < λ (E) + ε. Consider the Vitli cover for E consisting of those intervls [x, y] where x E, x < y, y O, nd f (y) f (x) < r. y x Then let ε > 0, nd use the Vitli covering lemm to pick pirwise disjoint intervls [x k, y k ] of this type, with with λ (E \ ([x 1, y 1 ] [x n, y n ])) < ε. It follows tht ( f (yk ) f (x k ) ) < r (y k x k ). Now let U = (x 1, y n ) (x n, y n ), nd crete yet nother Vitli cover, this time of E U, consisting of those intervls (u, v) where f (v) f (u) u E U, u < v, (u, v) U, nd > s. v u Use the Vitli lemm to lmost cover E U with pirwise disjoint intervls [u j, v j ] of this type, with λ (E U \ ([u 1, v 1 ] [u m, v m ])) < ε. Using resoning tht should be fmilir by now, we get ( f (v j ) f (u j ) ) > s (v j u j ). j =1 But since f is incresing, the intervls [f (u j ), f (v j )] re nonoverlpping. They re lso ech contined in some [f (x k ), f (y k )], so j =1 ( f (v j ) f (u j ) ) j =1 ( f (yk ) f (x k ) ). From the inequlities shown we get m s (v j u j ) < r (y k x k ). j =1 To estimte the right hnd side, note And for the left hnd side, s r n [x k, y k ] O, so (y k x k ) r λ(o) < r ( λ (E) + ε ). m ( m ) (v j u j ) = sλ [u j, v j ] > s ( λ (E U ) ε ) > s ( λ (E) 2ε ), j =1 so now we hve s ( λ (E) 2ε ) < r ( λ (E) + ε ). Recll tht r < s. So if λ (E) > 0, we cn pick ε > 0 smll enough tht the bove inequlity becomes contrdiction.
5 Limited number of corners From now on, < b re rel numbers, nd f : (, b) R is n rbitrry function. 5 Definition. Cll u (,b) strict locl mximum point for f if there is some δ > 0 so tht f (x) < f (u) whenever x (,b) is such tht x u < δ. 6 Lemm The set of strict locl mximum points for f is countble. Proof: Any two u vlues stisfying the definition bove for some δ > 0 must be t lest distnce δ prt, so there is only room for finite number of them in (,b). But ny strict locl mximum must stisfy the definition for some δ {1/n : n N}, so their totl number is countble. 7 Definition. A corner point for f is u (,b) so tht either D f (u) > D + f (u) or D f (u) < D + f (u). To understnd the mening of this definition, consult the illustrtion of Dini derivtives bove, which shows corner of the first kind. Turn the picture upside down to see corner of the second kind. 8 Lemm Any rel vlued function f hs only countble number of corner points. Proof: We only show tht the set of u stisfying the first inequlity is countble. The sme result for the second inequlity will then follow by replcing f by f. If the first inequlity holds, there is some q Q with D f (u) > q > D + f (u). If this holds, then x is strict locl mximum for the function g (x) = f (x) qx, since then D g (u) > 0 > D + g (u). By Lemm 6 only countble number of such u exist for ny q, nd since Q is countble, we re done.
6 Almost everywhere differentibility 9 Theorem Any monotone function is differentible lmost everywhere. Proof: We only need to consider n incresing function f : (,b) R. By Lemm 3 nd 4, f hs finite onesided derivtives lmost everywhere. If f hs onesided derivtives t some point, nd they re different, then f hs corner point there. But Lemm 8 implies tht the corner points hve mesure zero. 10 Exmple. Let f : [0,1] [0,1] be the Cntor function, nd C [0,1] the (stndrd) Cntor set. Then f is loclly constnt on the open set (0,1)\C, so f = 0 there. But λ(c ) = 0, so f = 0 lmost everywhere. In prticulr, 1 f dλ = 0, nd yet f (1) f (0) = 1. 0 This shows tht we cnnot expect the fundmentl theorem of clculus to hold for rbitrry monotone functions more is needed. Wht is hppening here is tht ll of the growth hppens within set of mesure zero.
7 An integrl (in)equlity 11 Proposition Let f : [,b] R be n incresing function. Then f is integrble, nd b f (x)dx f (b) f (). If there is some constnt L so tht f stisfies then equlity holds insted: f (y) f (x) + L(y x) whenever x < y b, b f (x)dx = f (b) f (). By incresing, I men wht some prefer to cll nondecresing. Proof: For simplicity, extend f by setting f (x) = f (b) when x > b. Define f n : [,b] R by f n (x) = n (f (x + n 1 ) f (x) ). Then f n 0, nd f n f.e. Therefore Ftou s lemm gives us b b f (x)dx lim f n (x)dx n b ( = lim n f (x + n 1 ) f (x) ) dx n = lim n n ( = lim n n ( b+n 1 +n 1 b+n 1 b f (b) f (). b f (x)dx f (x)dx n ) f (x)dx +n 1 ) f (x)dx If the stted liner growth condition holds, then f n L, so insted of using Ftou s lemm, we cn use the bounded convergence theorem (BCT), nd the first inequlity in the clcultion bove becomes equlity. Also, lim cn be replce by n ordinry limit. The finl inequlity lso becomes n equlity, since f is now continuous.
8 The first fundmentl theorem of clculus 12 Theorem If f L 1 ([,b]) nd F (x) = f (t)dt for x [,b], then F is differentible.e., nd F = f.e. in [,b]. The proof of this theorem builds on the following 13 Lemm If f L 1 ([,b]) nd f (t)dt = 0 for ll x [,b], then f = 0.e. in [,b]. Proof: Assume tht f > 0 in subset of (,b) of positive mesure. Then this subset hs compct subset K of positive mesure, nd then K f (t)dt > 0. Since b f (t)dt = 0, we must hve (,b)\k f (t)dt < 0. But (,b) \ K is n open set, nd hence is countble disjoint union of open intervls. For t lest one of those intervls, sy (c,d), we must hve (c,d) f (t)dt < 0. But (c,d) f (t)dt = ( d c) f (t)dt = 0, by ssumption nd this contrdiction shows tht f 0.e. Applying this to f, we see tht f 0.e., so f = 0.e. The bove proof is for relvlued f. The extension to complexvlued f by considering the rel nd complex prts seprtely is strightforwrd. Proof of the theorem: Assume first tht f 0. Then F is incresing, so it is differentible.e., nd s Proposition 11 shows tht F (t)dt F (x) for ll x [,b]. Furthermore, if f is bounded, then F stisfies the extr condition of Proposition 11, so the bove inequlity is ctully n equlity. But tht is equivlent to ( F (t) f (t) ) dt = 0 for ll x [,b], nd then the bove lemm shows tht F = f.e. This completes the proof in tht specil cse. If f 0 but f is unbounded, write F n (x) = ( f (t) n ) dt, nd note tht then F (x) F n (x) is the integrl of nonnegtive function, so tht F F n is n incresing function. In prticulr, F F n t ny point where both derivtives exists (i.e.,.e.). Therefore, nd using the result of the previous prgrph, F (t)dt F n (t)dt = ( ) f (t) n dt. Finlly, since f n f when n, we cn use the monotone convergence theorem to conclude tht F (t)dt f (t)dt for ll x [,b]. Just like in the bounded cse, this once more completes the proof, now for the cse when f 0. If f is relvlued integrble function, the theorem redily follows by using the result for the positive nd negtive prts of f. Similrly, if f is complex vlued, using the result for rel nd imginry prts seprtely redily yields the desired result.
9 Absolutely continuous functions Recll tht the Cntor function ψ is continuous nd differentible.e. on [0, 1], whith ψ = 0.e. yet ψ(1) ψ(0), hence it cnnot stisfy ψ(1) = ψ(0) ψ (x)dx, s the second fundmentl theorem of clculus would hve us believe. It turns out tht the remedy is to introduce stronger form of continuity. 14 Definition. A function f on n intervl [, b] is clled bsolutely continuous if for every ε > 0 there is δ > 0 so tht, whenever [ k,b k ] for k = 1,...,n re nonoverlpping intervls with n (b k k ) < δ, we hve n ( f (bk ) f ( k ) ) < ε. 15 Lemm If f L([,b]) nd F (x) = f (t)dt, then F is bsolutely continuous. Proof: Define mesure µ on M by µ(e) = f (x) dx. E [,b] I clim tht for ny ε > 0 there is some δ > 0 so tht λ(e) < δ implies µ(e) < ε. If not, there is some ε > 0 nd, for every n N, set E k M with λ(e k ) < 2 k, yet µ(e k ) ε. The lemm follows from the fct tht, for ny ε > 0, there exists δ > 0 so tht for ny mesurble E [,b] with λ(e) < δ, we hve E f (x) dx < ε. Let F n = k n E k, nd F = n N F n. Then λ(f n ) < 2 1 n, hence λ(f ) = 0. But lso µ(f n ) µ(e n ) ε, hence µ(f ) = lim n µ(f n ) ε (since F n F nd µ(f 1 ) < ). But µ(f ) > 0 while λ(f ) = 0 is clerly impossible, so the stted clim must be true. Now, given ε > 0, pick δ > 0 ccording to the clim bove. If ([ k,b k ]) n is finite sequence of pirwise nonoverlpping intervls, with n (b k k ) < δ let E = n [ k,b k ]. Then λ(e) < δ, nd so µ(e) < ε. Thus F (bk ) F ( k ) = nd the proof is complete. bk k f (t)dt bk k f (t) dt = µ(e) < ε, We shll prove the converse of the bove theorem. An importnt step long the wy is the following: 16 Lemm If f is bsolutely continuous nd differentible.e. on [,b] with f = 0.e., then f is constnt. Lter, we shll see tht bsolutely continuous functions necessrily re differentible.e. (they hve bounded vrition), but for now, we just ssume it. Proof: Let D be the set of points in (,b) in which f is differentible nd the derivtive is zero. Let ε > 0, nd let δ > 0 be s in the definition of bsolute continuity. The intervls [c,d] [,b] for which f (d) f (c) < ε d c form Vitli cover of D, so by the Vitli covering lemm nd the fct tht λ[,b] \ D = 0, there re finite, disjoint sequence of such intervls [c k,d k ] [,b] with k = 1,...,n such tht λ([,b] \ n (c k,d k )) < δ. Order these intervls so tht c 1 < d 1 < c 2 < d 2 < < c n < d n, nd write [ k,b k ] for the gps: So let [ 0,b 0 ] = [,c 1 ], [ 1,b 1 ] = [d 1,c 2 ], nd so on up to [ n,b n ] = [d n,b]. Then n k=0 (b k k ) < δ, so n k=0( f (bk ) f ( k ) ) < ε. Then nd so f (b) f () = f (b) f (d n ) + f (d n ) f (c n ) + f (c n ) f (d 1 ) + f (c 1 ) f () ( = f (bk ) f ( k ) ) ( + f (dk ) f (c k ) ), k=0 f (b) f () f (bk ) f ( k ) + f (dk ) f (c k ) k=0 < ε + ε d k c k (1 + b )ε. Since ε > 0 ws rbitrry, f (b) = f (). The sme rgument with b replced by ny x [,b] shows tht f (x) = f (), nd the proof is therefore complete.
10 The second fundmentl theorem of clculus 17 Theorem If f is bsolutely continuous on [,b], then f is differentible.e. on [,b], f L 1 ([,b]), nd for ll x [,b]. f (x) = f () + f (t)dt Proof: It is shown elsewhere tht f is differentible.e. nd tht f L 1 ([,b]) (see the textbook or perhps I will dd proof here lter). Let g (x) = f (x) f (t)dt. It follows from the first fundmentl theorem of clculus tht g = 0.e., nd from Lemm 15 tht g is bsolutely continuous. Lemm 16 then shows tht g is constnt. The vlue x = 0 shows tht then g (x) = f () for ll x, nd the proof is complete. Note tht I hve not yet lectured on bounded vrition t this point. For now, we tke these fcts for grnted: A complexvlued function hs bounded vrition if nd only if both its rel nd imginry prts do, nd relvlued function hs bounded vrition if nd only if it is the difference between two incresing functions. A function f : [,b] C is sid to hve bounded vrition (BV) if there is some M < so tht f (xk ) f (x k 1 ) M (2) whenever x 0 x 1 x n b. 18 Lemm An bsolutely continuous function on [, b] hs bounded vrition, nd is therefore differentible.e., nd its derivtive is integrble. Proof: Tke δ > 0 corresponding to ε = 1 in the definition of bsolute continuity. Let N N with Nδ > b, nd let δ = (b )/N < δ. Now ssume x 0 x 1 x n b is given. The sum in (2) will not get ny smller if we insert more points into the sequence (x k ), so we my s well ssume tht ech point in [,b] of the form + j δ for j N. Then the sum in (2) splits up into t most N sums, ech involving only x k [ + (j 1)δ, + j δ ] for some j. So in ech of the smller sums x k x k 1 < δ, nd therefore f (xk ) f (x k 1 ) < 1. Therefore, the totl sum is < N, so tht f hs bounded vrition. The sttements on differentibility nd integrbility of the derivtive follows from the fcts bout BV functions mentioned bove, nd the corresponding sttements for incresing functions.
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