REAL ANALYSIS I HOMEWORK 1

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1 REAL ANALYSIS I HOMEWORK CİHAN BAHRAN The questons are from Tao s text. Exercse If (x α ) α A s a collecton of numbers x α [0, + ] such that x α <, show that x α = 0 for all but at most countably many α A, even f α A A tself s uncountable. We want to show that the subset A 0 = {α A : x α > 0} of A s countable f the sum s fnte. For every n N, let Then clearly A 0 = A n = {α A : x α > /n}. A n. Now argung contrapostvely, suppose A 0 s uncountable. Then at least one of the A n s should be uncountable, as countable unons of countable sets s countable. Fx such an n. Then for every M, A n has a fnte subset F M of sze M; hence x α x α x α > /n = M/n α A α A n α F M α F M for every M (the frst two nequaltes hold snce x α s are nonnegatve). But by varyng M we can make M/n arbtrarly large (note that n does not depend on M). Hence x α can t be fnte. α A Exercse (Tonell s theorem for seres over arbtrary sets). Let A, B be sets (possbly nfnte or uncountable), and ( ) n A,m B be a doubly nfnte sequence of extended non-negatve reals [0, + ] ndexed by A and B. Show that = =. (n,m) A B n A m B m B n A (Hnt: although not strctly necessary, you may fnd t convenent to frst establsh the fact that f n A x n s fnte, then x n s non-zero for at most countably many n.) Frst, assume that the left hand sde s nfnte. Then for every postve real number R, there s a fnte subset F A B such that (n,m) F > R. Note that the proof of ths nnocent asserton uses the axom of (countable) choce! Ths fact s usually swept under the rug.

2 REAL ANALYSIS I HOMEWORK 2 Note that A has a fnte subset G and B has a fnte subset H such that F G H. Now observe that R < =. (n,m) G H n G m H (n,m) F Note that, for fxed n G, the sum of the collecton ( ) m B s defned as sup. H B s fnte m H Thus for every n G we have and hence m H m B R < n G m B n A m B Snce R > 0 was arbtrary, we conclude that the mddle term n the asserted equalty s also nfnte. In a smlar fashon, the rght hand sde s nfnte. We handled the nfnte case. So now assume that the left hand sde s fnte. Then by Exercse 0.0., the set S = {(n, m) A B : > 0} s countable. Let π A : A B A and π B : A B B to be the projecton maps. Wrte A 0 = π A (S) and B 0 = π B (S) and note that A 0 A and B 0 B are countable sets. Snce S A 0 B 0 and = 0 f (n, m) / S, we get = = (n,m) A B (n,m) S (n,m) A 0 B 0 = n A 0 m B 0 where we used Tonell s theorem for seres n the last equalty. Snce = 0 f m / B, we have = for every n A 0. And A 0 has a smlar property, so m B 0 m B = =. m B 0 m B n A m B n A 0 n A 0 Ths establshes the frst equalty n the fnte sum case. The second equalty s smlar. Exercse.2.4. Let E, F R d be dsjont closed sets, wth at least one of E, F beng compact. Show that dst(e, F ) > 0. Gve a counterexample to show that ths clam fals when the compactness hypothess s dropped. The statement to prove s actually true n the settng of an arbtrary metrc space. Lemma. Let (X, d) be a metrc space and A X. Then A = {x X : dst(x, A) = 0}.

3 REAL ANALYSIS I HOMEWORK 3 Proof. Suppose x X such that dst(x, A) = 0. Then for every ɛ > 0, there exsts a A such that d(x, a) < ɛ, hence B(x, ɛ) A. Snce the collecton {B(x, ɛ) : ɛ > 0} forms a bass of open neghborhoods around x, we conclude that x A. Conversely f x A, then for every ɛ > 0 we have B(x, ɛ) A. Hence there exsts a A (a depends on ɛ) such that d(x, a) < ɛ. Thus by the defnton of nfmum, dst(x, A) < ɛ. As ɛ > 0 was arbtrary, we conclude that dst(x, A) = 0. Proposton 2. Let (X, d) be a metrc space and E, F X be dsjont sets such that E s compact and F s closed. Then dst(e, F ) > 0. Proof. For every x E, snce x / F = F, by Lemma we have dst(x, F ) > 0. In other words, the contnuous functon f : E R x dst(x, F ) satsfes f(e) (0, + ). But E s compact, so f attans a mnmum, say ɛ > 0. Hence for every x E, dst(x, F ) ɛ. Therefore dst(e, F ) ɛ > 0. For a counter-example n the non-compact stuaton, let E = N and F = {n+/n : n N} as subsets of R. Note that for every a, b F a b > /2. So the only convergent sequences n F are eventually constant ones; thus F s closed. Smlarly E s closed. However, for every n N we have n E and n + /n F and hence Thus dst(e, F ) = 0. /n = n (n + /n) = d(n, n + /n) dst(e, F ). Exercse.2.6 Gve an example to show that the reverse statement s false. m (E) = sup m (U) U E,U open Note that For every open set U contaned n E, we have m (U) m (E), that s, m (E) s an upper bound for the set S E = {m (U) : E U-open } of real numbers. Thus by the defnton of sup we obtan sup S E m (E). So a counterexample to the clam must yeld sup S E < m (E). Consder the Lebesgue outer measure for R. Let E = R Q. Snce every nterval contans a ratonal number, the only open set contaned n E s the empty set! So we have S E = {0} whose supremum s (surprsngly) 0. However by subaddtvty we have whch gves m (E) =. = m (R) m (Q) + m (E) = m (E) Exercse.2.7 (Crtera for measurablty) Let E R d. Show that the followng are equvalent: () E s Lebesgue measurable.

4 REAL ANALYSIS I HOMEWORK 4 () (Outer approxmaton by open) For every ε > 0, one can contan E n an open set U wth m (U E) ε. () (Almost open) For every ε > 0, one can fnd an open set U such that m (U E) ε. (v) (Inner approxmaton by closed) For every ε > 0, one can fnd a closed set F contaned n E wth m (E F ) ε. (v) (Almost closed) For every ε > 0, one can fnd a closed set F such that m (F E) ε. (v) (Almost measurable) For every ε > 0, one can fnd a Lebesgue measurable set E ε such that m (E ε E) ε. () was our defnton of measurablty, so () and () are the same statement n our setup. () () s trval because f U contans E then U E = U E. For () (), let ε > 0. Then there exsts an open set V such that m (V E) ε/3. In partcular m (E V ) ε/3 < 2ε/3 by monotoncty. Then by the outer regularty of m, there exsts an open set W contanng E U such that m (W ) < 2ε/3. So U = V W s an open set contanng E and m (U E) = m ((V E) (W E)) m (V E) + m (W E) m (V E) + m (W ) ε/3 + 2ε/3 = ε. For () (v), let ε > 0. Snce E s measurable, so s E c. Therefore by assumpton there exsts an open set U contanng E c such that m (U E c ) ε. Note that F := U c s contaned n E and m (E F ) = m (U E) = m (U E c ) ε. (v) (v) s trval, smlar to () (). (v) (v) holds because closed sets are measurable. For (v) (), let ε > 0. Then there exsts a measurable set D such that m (D E) ɛ/2. And there exsts an open set U contanng D such that m (U D) ɛ/2. Now U E = (U E) (E U) = (D E) ((U D) E) (E U) (D E) (U D) (E D) = (D E) (U D) and hence by subaddtvty and monotoncty we have m (U E) m (D E) + m (U D) ɛ. Exercse.2.9 (Mddle thrds Cantor set). Let I 0 := [0, ] be the unt nterval, let I = [0, /3] [2/3, ] be I 0 wth the nteror of the mddle thrd nterval removed, let I 2 := [0, /9] [2/9, /3] [2/3, 7/9] [8/9, ] be I wth the nteror of the mddle thrd of each of the two ntervals of I removed, and so forth. More formally, wrte I n := a,..., {0,2} [ n 3, n 3 + ].

5 REAL ANALYSIS I HOMEWORK 5 Let C := n= I n be the ntersecton of all the elementary sets I n. Show that C s compact, uncountable, and ull set. By defnton, each I n s the unon of fntely many closed ntervals, so s closed. And snce for any choce of a,... {0, 2} we have n 0 3 n 2 n Ç å 3 = 2 Ç (/3) n+ å = 2 3 /3 Ç = 3 å 2 + = hence I n [0, ]. So beng an ntersecton of closed and bounded sets, C s closed and bounded; hence compact. Let s observe that the famly (I n ) s decreasng. Indeed, gven x I n there exsts a,..., {0, 2} such that Snce we have so x I n. n n 3 x n n 3 n 3, n n 3 x 3 + Consder the set P = {0, 2} N, that s the set of sequences n {0, 2}. Defne a map λ : P C ( ) To see that ths map s well-defned, gven ( ) wth {0, 2} wrte S n = n For every m N we clam that S n I m. We may assume m n. Let b = for n and b = 0 for > m, then we have m b S n = I m. 3 And snce I m s closed, the lmt 3. lm S n = n (whch defntely exsts by comparson wth geometrc seres) les n I m for every m. Hence s n C. 3n

6 REAL ANALYSIS I HOMEWORK 6 Next, we show that λ s njectve. So suppose ( ) and (b n ) are two sequences where, b n {0, 2} such that 3 = b n n 3. n Suppose that the sequences do not concde. So let k be the frst number such that a k b k. WLOG we may assume a k = 2 and b k = 0. 0 = 3 b n n Hence = 2 3 k = b n = 2 3 k + n=k+ n=k+ b n b n n=k+ 2. Cancellng 2 3 k yelds n= 3 = n 3 = 2 whch s nonsense. Thus λ must be njectve and as a result we conclude that C s uncountable as P = {0, 2} N s uncountable. Fnally we deal wth the outer measure of C. We frst show that[ gven two dstnct n n-tuples a,..., and b,..., b n wth, b n {0, 2}, the ntervals 3, n 3 + ] 3 [ n n b and 3, n b 3 + ] are dsjont. Smlar to above, let k be the frst number among,..., n where a k b k and then we may assume a k = 2 and b k = 0. Then n 3 n b 3 = n k =k+ b k + n =k+ = 2 3 k 2 n =k = 2 3 k 2 ( + 3 = k+ + k 2 3 = 2 Ç 3 3 k 3 å k+ + = 3 k + > ) 3 k+ 3

7 thus REAL ANALYSIS I HOMEWORK 7 n 3 > n b n We establshed that the defnng ntervals of I n are dsjont closed ntervals of length. And the number of these ntervals s the number of n-tuples takng values n {0, 2}, 3n whch s 2 n. Thus by countable addtvty of Lebesgue measure, we conclude that m(i n ) = 2n 3. n Fnally, for every n N we have C I n therefore Ç 2 m(c) 3 for every n N. Thus m(c) = 0. Exercse.2.0. Show that the half-open nterval [0, ) cannot be expressed as the countable unon of dsjont closed ntervals. For an addtonal challenge, show that [0, ) cannot be expressed as the countable unon of dsjont closed sets. å n