Exercise Solutions to Real Analysis

Transcription

1 xercse Solutons to Real Analyss Note: References refer to H. L. Royden, Real Analyss xersze 1. Gven any set A any ɛ > 0, there s an open set O such that A O m O m A + ɛ. Soluton 1. If m A =, then there s nothng to prove (take O = R). Suppose now that ɛ > 0 s gven m A <. By defnton, we have, { } m A = nf l(a n, b n ) : A n N (a n, b n ) n N So, by approxmaton for nfma, we know that there exst (a n, b n ) R such that A n N (a n, b n ) l(a n, b n ) m A + ɛ n N So take O = n N (a n, b n ). Then O s open, A O by subaddtvty we have: m O n N l(a n, b n ) m A + ɛ xersze 2. Suppose A, B R m A = 0, then m (A B) = m B Soluton 2. We have from subaddtvty, The result follows. m B m (A B) m A + m B = 0 + m B = m B xersze 3. If A s measurable then, for every ɛ > 0, there s an open set O contanng A such that m(o \ A) < ɛ Soluton 3. Suppose that ɛ > 0 s gven. Then we know by problem 1 that there s an open set O contanng A such that m O < m A + ɛ. In partcular, snce A s measurable O s open (hence also measurable) we may replace m wth m: mo ma < ɛ. On the other h, by measurablty of A the denttes O \ A = O A c, O A = A we get: m(o) = m(a O) + m(a c O) = m(a) + m(o \ A) So ɛ > m(o) m(a) = m(o \ A) xersze 4. If A s measurable then, for every ɛ > 0, there s a closed set C contaned n A such that m(a \ C) < ɛ 1

2 Soluton 4. Snce A s measurable, so s A c. So, gven ɛ > 0, there s O open, contanng A c, such that m(o \ A c ) < ɛ. But then O c s closed s contaned n A. Further, A \ O c = A O = O \ A c. So m(a \ O c ) = m(o \ A c ) < ɛ. xersze 5. Show that f s any measurable set, then ν ν ν + ν ν () Soluton 5. Let A, B be the Hahn decomposton, A postve B negatve. ν + () = ν(a ) 0 ν () = ν(b ) 0. Snce A B are dsjont, we have ν() = ν + () ν () It follows mmedately that ν ν ν + Now, ν = ν + ν ν + + ν = ν + + ν = ν (), snce ν + ν are postve. xersze 6. Show that f ν 1 ν 2 are any two fnte sgned measures, then so s αν 1 + βν 2, where α β are real numbers. Show that αν = α ν ν 1 + ν 2 nu 1 + ν 2 Soluton 6. Set, for convenence, µ = αν 1 +βν 2. It s clear that µ( ) = 0. Further, snce ν 1 ν 2 are fnte measures, µ never admts + or. Now, suppose that {A } s a collecton of dsjont sets. Then we have: µ ( =1A ) = αν 1 ( =1A ) + βν 2 ( =1A ) Now, snce ν 1 ν 2 are sgned measures, ether γν j (A ) =1 converges absolutely, or dverges properly (wth j = 1, 2, γ = α, β respectvely). Now, sum of absolutely convergent seres s agan absolutely convergent. Further, f one s properly dvergent, then so s the sum. In both cases, the sum of the seres can be taken term-by-term. Hence we have: µ( =1A ) = αν 1 (A ) + βν 2 (A ) = (αν 1 (A ) + βν 2 (A )) = µ(a ) =1 =1 Fntude of µ s mmedate by trangle nequalty for real numbers (snce both ν 1 ν 2 are fnte). Now, f ν s or, equalty holds for every nonzero α: αν = α ν. On the other h, snce ν s a fnte real number, we clearly have αν = α ν. Now, assumng that ν 1 + ν 2 s never of the form, f ν 1 + ν 2 =, then clearly ether ν 1 = ± or ν 2 = ±. In ether case, equalty holds: ν 1 + ν 2 = ν 1 + ν 2. If, however, ν 1 + ν 2 s fnte, then both ν 1 ν 2 are fnte. So by trangle nequalty for real numbers we get ν 1 + ν 2 ν 1 + ν 2 =1 =1

3 (of course, above when we speak about the numerc value of ν 1 ν 2, we essentally mean ν 1, ν 2 for a rom measurable set ). xersze 7. We defne ntegraton wth respectt to a sgned measure ν by defnng fdν = fdν + fdν If f M, fdν M ν () Moreover, there s a measurable functon f wth f 1 such that fdν = ν () Soluton 7. Note, that snce ν s a sgned measure, t does not admt both +. In ths case the ntegral f well-defned, snce there doesn t exst a set wth ν + = ν =. So, f f M, fdν = fdν + fdν fdν + + fdν Now, snce ν + ν are both postve measures, the ntegrals wth respect to those measures are defned n the usual sense. Hence all the stard theorems apply. Thus: fdν + + fdν f dν + + f dν M(ν + + ν ) = M ν () Now, let A B be the Hahn decomposton. Defne f(x) = A B. Then we have: fdν = fdν + fdν = A dν + + B dν = Also, t s clear that f 1. ν + (A ) + ν (B ) = ν () xersze 8. (a) Show that the Radon-Nkodym Theorem for a fnte measure µ mples the theorem for a σ-fnte measure µ. (b) Show the unqueness of the functon f n the Radon-Nkodym Theorem. Soluton 8. (a) Suppose that the theorem holds for fnte measures. Suppose now that µ s a σ-fnte measure on. Wrte = of µ-fnte, dsjont sets. Then we know that there exst functons f on such that ν = f for every measurable subset of. Now, defne f = f. Clearly f s nonnegatve measurable. Now, let be a µ-measurable subset of. Let =. form a famly of dsjont measurable sets =. Further, ν = ν = f

4 Snce each f s nonnegatve, f = f = Unqueness of such f s forced by ts exstence (see (b) below). (b) We ll use the decomposton from (a). Observe that each f s unque (a.e. [µ]) on each, for suppose that g s another famly of functons wth f = g But then (f g ) = 0 forcng f = g a.e. on. So, by unqueness of each f on, we conclude that f s also unque (a.e.). xersze 9. Radon-Nkodym dervatves. Let µ, ν, λ be σ-fnte. Show that the Radon-Nkodym dervatve [dν/] has the followng propertes: f (a) If ν µ f s a nonnegatve measurable functon, then [ ] dν fdν = f. (b) (c) If ν µ λ, then (d) If ν µ µ ν, then [ Soluton 9. Let g = [ ] d(ν1 + dν 2 ) = dν ]. [ ] dν = dλ [ ] dν = [ dν [ ] dν1 + ] [ ]. dλ [ ] 1. dν [ ] dν2. Snce fdν = fdν + fdν by defnton, t wll suffce to prove the asserton for fdν + fdν By symmetry snce ν + ν n fact both are postve measures, t wll suffce to prove the asserton only for, say, ν +. The other case wll follow smlarly. Set

5 η = ν +. Suppose that φ s a nonnegatve smple functon φ(x) = a such that φ f. We have: φdη = a η = a g = φg And by defnton we have φg = φg + φg Takng the supremum over all such φ observng that µ + µ, both are fnte, φg s measurable, nonnegatve φg f, we obtan: ( ) sup φg = sup φg + φg = sup φg + sup φg = fg fg + = fg (b) Ths s straghtforward. We have, for every measurable, [ ] [ ] ([ ] dν1 dν2 dν1 (ν 1 + ν 2 ) = + = + So by unqueness, [ ] d(ν1 + dν 2 ) = [ ] dν1 + [ ] dν2. [ dν2 ]) (c) Ths one s also straghtforward: Observe that for each measurable, [ ] dν ν = But by part (a) we have: So by unqueness we obtan: [ ] dν = [ ] dν = dλ [ dν [ dν ] [ ] dλ dλ ] [ ]. dλ (d) Well, snce µ ν, for each measurable we have [ ] µ = dν dν By part (a) we get [ ] [ ] [ ] dν dν = dν dν Snce µ µ from defnton of the ntegral, µ = 1

6 for each measurable, by the unqueness of the Rodon-Nykodm dervatve we obtan [ ] [ ] dν 1 = dν a.e. Hence a.e. [ ] = dν [ ] 1 dν xersze 10. (a) Show that f ν s a sgned measure such that ν µ ν µ, then ν = 0. (b) Show that f ν 1 ν 2 are sgular wth respect to µ, then so s αν 1 + βν 2. (c) Show that f ν 1 ν 2 are absolutely contnuous wth respect to µ, so s αν 1 + βν 2. (d) Prove the ungueness asserton n the Lebesgue decomposton. Soluton 10. (a) Let = A B such that νa = µb = 0. Then clearly ν s zero for every subset of A, snce ν µ s taken to mean ν µ. That s to say, ν A = µ B = 0. Further, snce ν µ for each subset of C of B, µc = 0, we observe that νc = 0. Hence, f s any measurable subset of, wrtng = (A ) (B B), we conclude that nu = 0. (b) There exst decompostons A 1, B 1 A 2, B 2, A, B dsjont, such that = A B ν A = µb = 0. Let = A 1 A 2. Then c = A c 1 A c 2,, c for a decomposton of wth ν 1 = ν 2 = 0 µ c = 0. Hence, (c 1 ν 1 + c 2 ν 2 ) = µ c = 0 (c) Let be a µ-measurable subset of such that µ = 0. Then by hypothess ν 1 = ν 2 = 0. Hence (αν 1 + βν 2 ) = αν 1 + βν 2 = 0. (d) Suppose, by the Lebesgue decomposton, ν = ν 0 + ν 1 ν = ν 0 + ν 1. Then clearly 0 = (ν 0 ν 0 ) + (ν 1 ν 1 ). Observe that by parts (b) (c), (ν 0 ν 0 ) µ (ν 1 ν 1 ) µ. Hence there exsts a decomposton A, B such that (ν 1 ν 1 )A = µb = 0. So for each subset of B, by absolute contnuty, (ν 0 ν 0 ) = 0. Hence (ν 1 ν 1 ) = 0, so that ν 1 = ν 1. On the other h, for each subset of A, (ν 1 ν 1 ) = 0, so that (ν 0 ν 0 ) = 0, hence ν 0 = ν 0. Ths shows unqueness of ν 0 ν 1. xersze 11. Use the followng example to show that the hypothess n the Radon- Nkodym Theorem that µ s σ-fnte cannot be omtted. Let = [0, 1], B the class of Lebesgue measurable subsets of, take ν to be Lebesgue measure µ to be the countng measure on B. Then ν s fnte absolutely contnuous wth respect to µ, but there s no functon f such that ν = f for all B. At what pont does the proof of Theorem 23 break down for ths example? Soluton 11. Suppose that f s any nonnegatve µ-measurable functons on [0, 1]. Now, gven [0, 1], we see mmedately that f = α R + (the extended postve real numbers) where α s zero f f s zero (but ths s a pathologcal case whch we exclude mmedately); otherwse:

7 () α s fnte provded that s a fnte set f s bounded. But f s a fnte set, ν = 0, forcng f to be dentcally zero on every fnte subset of [0, 1], hence on [0, 1]. We already excluded ths case. () α s nfnte, then s nfnte or s fnte but f s not bounded on t. In the former case, ν s stll fnte, n the latter case, ν s zero. We see that Radon-Nykodm fals n ths case. Well, observe that the theorem s frst proved for µ fnte, then extended to the σ-fnte case. However, the extenson s vald only when can be decomposed nto a countable collecton of µ-fnte sets (see problem 4). However, [0, 1] cannot be decomposed nto a countable collecton of µ-fnte sets. In a sense, by soluton to problem 4, the extenson wll break down snce an uncountable sum s always properly dvergent (not to menton that we don t know how to take an ntegral of an uncountable sum, or to nterchange the ntegral the lmt n ths case). xersze 12. (a) Let L 1 (, m) be the metrc space wth dstance d(f, g) = tan 1 f g dm Defne convergence n measure as before. Prove that convergence n measure concdes wth convergence n ths metrc space. (b) do the same for f g d(f, g) = 1 + f g dm ( prove that t s a metrc, too). xtra: Show that ether one of these metrc spaces s complete. Soluton 12. (a) In general, convergence n measure only mples convergence of a subsequence a.e. But n ths case, we show that f f n f 0, then d(f n, f) 0. Observe frst that f φ(x) = x tan 1 (x), then φ (x) = x. Hence φ (x) > 0 2 for all x. Thus φ(x) s strctly ncreasng. Further, φ(0) = 0. Ths shows that φ(x) 0 for all x, hence x tan 1 (x) for all x. Usng ths, we have: d(f n, f) = tan 1 f n f dm f n f dm = f n f Thus, snce f n f 0, we must have d(f n, f) 0. f g 1+ f g (b) If we set d(f, g) = dm, only the trangle nequalty requres attenton (the other two propertes of a metrc are obvous). Set φ(x) = x 1+x. Then φ 1 (x) = (1+x) whch s clearly everywhere (on the nonnegatve axs) postve. 2 Hence φ s an ncreasng functon. So, φ( a + b ) φ( a + b ) = a + b a 1+ a + b 1+ b 1+ a + b = φ( a ) + φ( b ). Substtutng f g for a g h for b, we get φ( f h ) φ( f g ) + φ( g h ). Therefore, φ( f h )dm φ( f g )+φ( g h )dm = φ( f g )dm+ φ( g h )dm

8 So d(f, h) d(f, g) + d(g, h). Now, observe agan that f n f d(f n, f). Ths s obvous. Now usng the argument from part (a), we see that convergence n measure mples convergence n the metrc d. We show now completeness wth respect to the measure d(f, g) = f g 1+ f g. Suppose that f n s a Cauchy sequence. Set f = lmf n. Now, gven ɛ > 0, we know that there exsts N such that for all n, m N, we have d(f n, f m ) < ɛ/2. So, f n f m dm < ɛ/2 1 + f n f m But then, by Fatou s lemma, we get: f n f 1 + f n f dm = f n f m lm m 1 + f n f m dm lm m f n f m dm ɛ/2 < ɛ 1 + f n f m for all n N. Ths shows that d(f n, f) < ɛ for all n N. Hence f n converges to f. Further, snce f = lmf n, each f n s n L 1 (, m), we know that f s also n L 1 (, m). Ths establshes the result. xersze 13. Let ν : B R (no nfntes allowed!) be a sgned measure. Prove that sup A B ν(a) <, nf A B ν(a) >. Soluton 13. Suppose that for each natural m > 0 there exsts a measurable set A m wth ν(a m ) > m. Now, set B 1 = A 1 B n>1 = A n \ (A 1 A n 1 ). Clearly B form a collecton of dsjont sets, n fact, A n = n k=1 B k. Ths shows that, m < ν(a m ) = ν( m k=1b k ) Therefore, m < ν(b k ) = ν( k=1b k ) = k=1a k k=1 for all natural m > 0. Hence ν( k=1 A k) =. But ths cannot be by hypothess! Now, gven that ν s a measure, by problem 2, we see that ν s also a measure. Further, snce ν does not admt ±, nether does ν. Now apply the above result to ν. We see that sup A B ν(a) <. Hence nf A B ν(a) >. xersze 14. Prove that there exst A 1 A 2 so that ν(a 1 ) = sup A B ν(a) ν(a 2 ) = nf A B ν(a). Soluton 14. Frst of all, observe that snce ν( ) = 0, the nfmum s at most 0 the supremum s at least 0. Well, set S = sup A B ν(a) I = nf A B ν(a). If S =, then for each natural m > 0 there s a measurable A m wth ν(a m ) > m. Now apply the constructon n the frst part of problem 9 above to obtan a set A wth ν(a) =. Smlarly, f I =, then consder ν, whose supremum s. Apply the argument above to get a set A such that ν(a) =. Then ν(a) =. Suppose now that S, I are fnte real numbers. Then by approxmaton for the

9 suprema, for each natural n > 0, there exsts A n such that S 1/n ν(a n ) S. Consder B as n the frst part of problem 9. Set A = k=1 B k. Then clearly (see frst part of problem 9) ν(a) S 1/n for all n. But then ν(a) S. So ν(a) = S. Now consder ν apply the constructon above to get a set A such that ν(a) = I ( I s the supremum n ths case). But then ν(a) = I. xersze 15. xtend the Radon-Nkodym Theorem to the case of sgned measures. Soluton 15. xersze 16. Complex Measures. (a) Show that eacy complex measure ν may be expressed as ν = µ 1 µ 2 +µ 3 +µ 4, where µ are fnte measures. (b) Show that for each complex measure ν there s a measure µ a complexvalued measurable functon φ wth φ = 1 such that for each set n B, ν = φ Soluton 16. Lemma 0.1. Let (, B, µ) be a fnte measure space g an ntegrable functon such that for some constant M, gφ M φ p for all smple functons φ. Then g L q. xersze 17. Prove Lemma 0.1. Soluton 17. xersze 18. Show that Lemma 0.1 remans true f we only assume µ to be σ-fnte. Soluton 18. Lemma 0.2. Let n be a sequence of dsjont measurable sets, for each n let f n be a functon n L p (1 p < ) that vanshes outsde n. Set f = n=1 f n. Then f L p f only f f n p <. In ths case f = f n n L p ; that s, f f p 0 xersze 19. Prove Lemma 0.2 Soluton 19. =1 f p = f n p xersze 20. For g L q, let F be the lnear functonal on L p defned by F (f) = fg Show that F = g q. n=1

10 Soluton 20. xersze 21. Let µ be the countng measure on a countable set. Show that L p (µ) = l p. Soluton 21. xersze 22. Let µ 1, µ 2 be defned on the σ-algebra of Lebesgue measurable subsets of R by µ 1 (A) = n A 1 2 µ n 2 (A) = n A 1/n. Here n s taken to run through the natural numbers. xhbt a Hahn decomposton for 3µ 1 µ 2. Soluton 22. We show frst that µ 1 µ 2 are measures, µ 1 fnte µ 2 σ-fnte, whch wll establsh that 3µ 1 µ 2 s a sgned, σ-fnte measure. Well, notce that snce n runs over the set of natural numbers, we may restrct our attenton only to subsets A of (0, ). Clearly both are zero on the empty set. σ-addtvty follows from defnton of nfnte seres the fact that all terms are postve. Further, snce µ 1 (0, ) s smply a geometrc seres, t s fnte. Settng (0, ) = n [n 1, n], we see that µ 2 s σ-fnte. Now, observe that for all n 4, 3/2 n < 1/n. Conversely, for all n < 4, 3/2 n > 1/n. Hence we see that for every subset of of (0, 4), 3µ 1 () µ 2 () 0. On the other h, for every subset of [4, ), 3µ 1 () µ 2 () 0. Therefore we have the followng Hahn decomposton: A = (0, 4) B = [4, ). xersze 23. Fnd the Lebesgue decomposton for µ 1 w.r.t. µ 2 for µ 2 w.r.t. µ 1. Fnd correspondng Radon-Nkodym dervatves for the a.c. parts. Soluton 23. Set ν 0 = µ 1 ν 1 = 0. Let A = N B = (0, ) \ N. Then clearly ν 1 µ 2, snce ν 1 (A) = µ 2 (B) = 0. Now, f µ 2 () = 0, then N =. But n ths case µ 1 () = 0. Hence we see that ν 0 = µ 1 µ 2. Hence hour Lebesgue decomposton of µ 1 w.r.t. µ 2 s: µ 1 = ν 0 + ν 1 Repeat the process as above: set ν 0 = µ 2 ν 1 = 0. Then the Lebesgue decomposton of µ 2 w.r.t. µ 1 s µ 2 = ν 0 + ν 1 Snce µ 1 s the absolutely contnuous part of the Lebesgue decomposton w.r.t. µ 2, vce versa, we must fnd the Rodon-Nkodym dervatves f, g, such that: µ 1 () = f 2 Well, defne Then we have: f 2 = k=1 µ 2 () = f(x) = (k 1,k] k=1 g 1 k 2 k (k 1,k] f 2 = k=1 k 2 k µ 2( (k 1, k])

11 Observe that µ 2 ( (k 1, k]) = 0 f (k 1, k] = 1/k otherwse. In whch case, from above we obtan: { k 0 f (k 1, k] = 2 k µ 2( (k 1, k]) = 1/2 k otherwse Hence, Smlarly, settng one obtans xersze 24. Let µ 2 () = g(x) = k=1 µ 1 () = f 2 2 k k (k 1,k] g 1 1 µ 1 () = 1 + x 2 dx + 5δ 1 + 2δ π µ 2 () = dx + πδ 1 + 2δ e [0,1] Fnd the Lebesgue decomposton for µ 1 w.r.t. µ 2 for µ 2 w.r.t. µ 1. Fnd correspondng Radon-Nkodym dervatves for the a.c. parts. Soluton 24. Set ν 0 () = [0,1] dx + πδ 1 ν 1 () = 2δ e Set also A = R\{e} B = {e}. Observe that A B = R ν 1 (A) = µ 1 (B) = 0. So ν 1 µ 1. Further, suppose that µ 1 () = 0. Then we see that x 2 dx = 0 1, π /. But ths means that m() = 0 (where m s the Lebesgue measure) πδ 1 () = 0. Hence ν 0 () = 0, showng that ν 1 µ 1. It s clear that µ 2 = ν 0 + ν 1. For the other case, take ν 0 () = [0,1] x 2 dx + 5δ 1() 1 ν 1 () = [0,1] 1 + x 2 dx + 2δ π() c It s easy to see that µ 1 = ν 0 + ν 1 that ν 0 µ 2. Now, take A = [0, 1] {e} B = A c. Then we observe that ν 1 (A) = µ 2 (B) = 0, so that ν 1 µ 2. Ths gves the Lebesgue decomposton.

12 Now we fnd the Rodon-Nkodym dervatves n each case. So we must fnd measurable f, g such that: ν 0 () = f 1 ν 0 () = To ths end the followng result wll be useful. g 2 Proposton 0.3. Suppose that µ 1, µ 2,..., µ n are postve measures on a measure space (B, ) a 1, a 2,..., a n are nonnegatve constants. Suppose further that f s a µ 1 -measurable, nonnegatve functon. Then () If λ = a 1 µ 1 + a 2 µ 2 + a n µ n, g s a λ-measurable functons, then gdλ = a 1 g 1 + a 2 g a n g n () If ν() = f 1 g s a ν-measurable functon, then gdν = gf Proof. The second asserton () was proved as part of a prevous homework (p. 268 #22(b)). The frst asserton () follows easly by frst consderng g to be a smple functons, n whch case the result holds. Then extendng the result by the Monotone Convergence Theorem, by consderng a sequence φ n of smple functons that ncrease to g. Now take f = (1 + x 2 ) [0,1) + π 5 {1} g = x 2 [0,1) + 5 π {1} Then by Proposton 0.3, t follows that ν 0 () = g 1 ν 0 () = f 2 xersze 25. Show that the product of two sem-rngs s a sem-rng, but the product of two rngs s not necessarly a rng. Soluton 25. xersze 26. Assume that s a sequence of dsjont measurable sets =. Then for any set A we have µ (A ) = µ (A )

13 Soluton 26. Ths s true when n = 1. Suppose the result holds for some n. Consder n + 1 measurable, dsjont sets 1,... n, n+1. Now we have, by measurablty of n+1 the fact that all are dsjont: µ (A [ n+1 ] ) = µ (A [ n+1 ] n+1 ) + µ (A [ n+1 ] n+1 ) = µ (A [ n ]) + µ (A n+1 ) And by nducton hypothess from above we obtan: µ (A [ n+1 ] n n+1 ) = µ (A ) + µ (A n+1 ) = µ (A ) Now, After takng the lmt we obtan Hence µ (A ) µ (A [ n ]) = µ (A ) µ (A ) = µ (A ) µ (A ) n µ (A ) xersze 27. Let A be a collecton of sets whch s closed under fnte unons fnte ntersectons; an algebra of sets, for example. (a) Show that A σ s closed under countable unons fnte ntersectons. (b) Show that each set A σδ s the ntersecton of a decreasng sequence of sets n A σ. Soluton 27. (a) Suppose that A s a countable unon of sets n A σ. Then by defnton each A s n turn a countable unon of sets n A, say A = j Bj. But then [ A = j Bj ] s a countable unon of sets Bj, hence s n A,j N σ. Suppose that A 1, A 2,..., A n s a fnte sequence of sets n A σ. Then agan, each s n turn a countable unon of sets from A, say A = Bj. Now we have: [ A = j Bj] [ = j=1 n Bj ] Now, snce A s by hypothess closed under fnte ntersectons, n B j s n A. Hence above we end up wth a countable unon of sets from A, whch s n A σ. Ths completes the proof. (b) Suppose that B A σδ. But then B s a countable ntersecton of sets A n A σ. Let B 1 = A 1 for n 2, B n = n A. Then by part (a), B A σ for each B = j B j wth B n+1 B n. Ths completes the proof.

14 xersze 28. Let µ be a fnte measure on an algebra A, µ the nduced outer measure. Show that a set s measurable f only f for each ɛ > 0 there s a set A A σ, A, such that µ ( \ A) < ɛ. Soluton 28. By A δ we underst the collecton of all countable ntersectons of sets from A. Proposton 0.4. If µ s a fnte measure A, B are µ-measurable sets wth B A, then µ (A \ B) = µ (A) µ (B). Proof. Gven that A, B are measurable, µ (A) = µ (A B) + µ (A B c ). Snce B A, we have µ (A) = µ (B) + µ (A \ B) = µ (A \ B) = µ (A) µ (B). ( = ) Snce s measurable, so s c. By approxmaton for the nfmum, there exsts a cover of c of sets A n A such that (set A = n A n ) ( ) µ ( c ) + ɛ > n Snce c A, we know that A c. Further, A c = [ n A n ] c = n A c A δ µ(a n ) µ (A) Now, by Proposton 0.4, µ () = µ () µ ( c ) µ (A c ) = µ () µ (A), where s our total space. In addton from ( ) above we get: µ (A) µ ( c ) < ɛ hence µ () µ () + µ (A) µ ( c ) < ɛ Hence µ () µ (A c ) < ɛ And agan from Proposton 1, snce certanly A c s measurable, we get µ ( \ A c ) = µ () µ (A c ) < ɛ Ths completes the proof n one drecton. ( =) Suppose that gven, for all ɛ > 0 there exsts A A δ such that µ ( \ A) < ɛ For convenence let us set C = \ A. Now, for any set B we have: µ ( ) + µ (B c ) µ (B ) + µ (B A c ) snce A mples c A c. Further, µ (B ) + µ (B A c ) = µ ([B ] [B C]) + µ (B A c ) µ (B ) + µ (B C) + µ (B A c ) µ ([B ]) + ɛ + µ (B A c ) Snce A s measurable, we have so we obtan from above µ (B) = µ (B A) + µ (B A c ) µ ( ) + µ (B c ) µ (B) + ɛ Nether the left nor the rght h sde depends on ɛ. Hence µ ( ) + µ (B c ) µ (B)

15 From whch we deduce that s measurable. Ths completes the proof n the other drecton.