Math 702 Mdterm xam Solutons The terms measurable, measure, ntegrable, and almost everywhere (a.e.) n a ucldean space always refer to Lebesgue measure m. Problem. [6 pts] In each case, prove the statement or provde a counterexample: () There s a postve contnuous f 2 L./ such that lm sup x! f.x/ D. TU. Let g W! be the trangular pulse defned by 8 ˆ< x x 0 g.x/ D x 0 x ˆ: 0 otherwse. Then g 0 and g D. The rescaled translaton g n.x/ D ng.n 3.x support n.n n 3 ; n n 3 / and satsfes g n D =n 2. The seres f D nd2 g n n// has defnes a contnuous functon snce ts summands are contnuous wth dsjont supports, and f 2 L./ snce f.x/ dx D g n.x/ dx D n < : 2 nd2 But f.n/ D g n.n/ D n, so lm sup x! f.x/ D. To obtan an example whch s strctly postve, add to ths f any postve contnuous functon n L./ such as =.x 2 /. () If f n W! are ntegrable, f n! f a.e. n, and f n.x/ dx! 0, then f D 0 a.e. n. FALS. The sequence f n D Œ0;n Œ n;0 wth nd2 f n.x/ dx D 0 converges pontwse to the non-zero lmt f D Œ0;/. ;0. omment. The clam would be true f the f n were assumed non-negatve, for then Fatou s lemma would show 0 f.x/ dx lm nf f n.x/ dx D 0 n! whch, because f 0, would mply f D 0 a.e. n.
() very path-connected subset of the plane 2 s measurable. FALS. Take any non-measurable set A and defne D.A / [. f0g/: vdently s path-connected. If were measurable, ts horzontal slce y would be measurable for a.e. y 2. But y D A for every y 0. (v) Gven f W 2!, f the sectons f x ; f y W! are constant functons for a.e. x; y 2, then f D c a.e. for some constant c. TU. Fx any y 0 for whch f y 0 s a constant functon c. Let A be the full measure set of x for whch f x s a constant functon. For any x ; x 2 2 A, f x.y 0 / D f.x ; y 0 / D f y 0.x / D c D f y 0.x 2 / D f.x 2 ; y 0 / D f x2.y 0 /: Ths shows f x D f x2 D c. It follows that f D c on the full measure set A. omment. Observe that we only used the exstence of a sngle y for whch f y s constant. Note also that the clam s false f the sectons f x ; f y are assumed to be a.e. constant even f ths holds for every x; y 2. Smply take f D, where Œ0; 2 s Serpnsk s non-measurable set whch ntersects every horzontal and vertcal lne at most once. Problem 2. [6 pts] Prove that has measure zero f and only f there are open ntervals I ; I 2 ; I 3 ; : : : wth P nd m.i n/ < such that every pont of les n nfntely many of the I n. One drecton s mmedate from Borel-ontell: Suppose fi n g are open ntervals wth P nd m.i n/ <. Then almost every pont of belongs to at most fntely many of the I n. Hence, f every pont of les n nfntely many of the I n, we necessarly have m./ D 0. onversely, suppose m./ D 0. By regularty of Lebesgue measure, for every k there are (dsjont) open ntervals fi j;k g j coverng such that P j m.i j;k/ < 2 k. Take the countable collecton of the I j;k for all j; k and arrange them n a sequence fi n g. Then P n m.i n/ D P k P j m.i j;k/ < P k 2 k D. In partcular, no nterval can show up nfntely many tmes n the sequence fi n g. Snce for every x 2 and every k there s some j such that x 2 I j;k, we see that x belongs to nfntely many of the I n. Problem 3. [0 pts] Suppose.; / s a measure space and f 2 L./. Take an ncreasng sequence fa n g of postve numbers that tend to, and set n D fx 2 W jf.x/j > a n g. Show that lm a n. n / D 0: n!
Let ˆ.t/ D.fx 2 W jf.x/j > tg/ be the dstrbuton functon of f. ecall that (as an applcaton of Fubn-Tonell) we proved ˆ 2 L.0; / and ˆ.t/ dt D jf j d: 0 We prove that tˆ.t/! 0 as t!. Ths wll show n partcular that a nˆ.a n /! 0 as n!, whch s the desred result. Snce ˆ 2 L.0; /, () lm t! t ˆ D lm t! 0 ˆ Œt;/ D 0 by the domnated convergence theorem. Snce ˆ s non-negatve and decreasng, 0 t 2ˆ.t/ t t=2 ˆ for every t > 0. Lettng t! and makng use of () then shows tˆ.t/! 0. Problem 4. [8 pts] Suppose f W!.0; / s ntegrable wth kf k D. For every measurable set wth 0 < m./ <, prove the nequaltes log f.x/ dx m./ log.m.// and t=2 f.x/ p dx m./ p.0 < p < /: Both nequaltes follow from Jensen s nequalty ' f d.' ı f / d for a sutable choce of the convex functon ', takng to be the normalzed Lebesgue measure.=m.// dx on. For the frst nequalty, take '.x/ D log x to obtan log f.x/ dx m./ m./ ˆ log.f.x// dx: Multplyng each sde by m./ and usng f f D then gves log.f.x// dx m./ log f.x/ dx m./ m./ log m./ D m./ log.m.//:
For the second nequalty, take '.x/ D x p : p f.x/ dx m./ f.x/ p dx: m./ Agan, multplyng each sde by m./ and usng f yelds f.x/ p dx m./ f.x/ dx m./ m./ m./ D p m./ p : p Problem 5. [0 pts] Suppose.; / s a measure space, f W! s measurable, and there s a constant c > 0 such that f n d D c for all large n 2 N: Show that f D a.e. for some measurable set. We provde two dfferent proofs. The frst follows a routne path; the second s shorter but a bt trcker. Frst proof. Take the partton D A [ B [ [ D [ nto the dsjont measurable sets A D fx W f.x/ D 0g B D fx W 0 < jf.x/j < g D fx W jf.x/j > g D D fx W f.x/ D D fx W f.x/ D g: It suffces to show that.b/ D. / D.D/ D 0, for then f D on the full measure set A [. By our hypothess, for large n, (2) c D f 2n d D B D D.D/./ g B f 2n d f 2n d: On B, the sequence ff 2n g s postve, monotoncally decreasng, and convergent to 0. By the domnated convergence theorem, f 2n d! 0 as n! : B
On, the sequence ff 2n g s postve, monotoncally ncreasng, and convergent to. By the monotone convergence theorem, f 2n d! d D./. / as n! : Snce f 2n c by (2), we necessarly have. / D 0. Puttng ths nformaton back nto (2) and lettng n! yelds c D.D/./. It also shows B f 2n d D 0 for all large n, whch mples.b/ D 0 snce f 2n > 0 on B. Fnally, use our hypothess agan for large n to wrte c D f 2n d D f 2n d D.D/./: D omparng ths wth c D.D/./ gves.d/ D 0. Second proof. The bnomal expanson! 2n.f 2 f / 2n 2n D. / 2n f 2 f 2n D shows that for large n, D0.f 2 f / 2n d D 2n D0 2n D0 2n!. / f 2n d D c! 2n. / f 2n 2n D0! 2n. / D 0: Snce.f 2 f / 2n 0, t follows that f 2.x/ f.x/ D 0 or f.x/ 2 f0; g for a.e. x 2, whch s what we wanted to prove. omment. The second proof shows that the hypothess can be weakened to c D f d.2n 4n/ for some n 2 N. Problem 6. [0 pts] Suppose d s a measurable set wth m./ D 0. Should there be a homeomorphsm ' W d! d such that './ \ D ;? (You Barely need any fancy tool for ths.) The answer s negatve: There are measure zero sets n d whch meet every homeomorphc mage of themselves. The smplest examples of ths type are obtaned usng Bare s category theorem as follows. Let fx n g be a countable dense set n d and defne V k D [ B.x n ; 2.nk/ /.k D ; 2; 3; : : :/ nd
Then each V k s open and dense, and V V 2 V 3. Snce d s a complete metrc space, Bare s theorem tells us that D T k V k s a dense G ı. Moreover, snce m.v k / m.b.x n ; 2.nk/ // const: 2 d.nk/ const: 2 dk ; nd we have m./ D lm k! m.v k / D 0. Now for every homeomorphsm ' W d! d, '.V k / s open and dense, so './ D T k '.V k/ s also a dense G ı. By another applcaton of Bare s theorem, the ntersecton of two dense G ı s s agan a dense G ı. It follows n partcular that './ \ ;. nd