Real Analysis HW 10 Solutions


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1 Rel Anlysis HW 10 Solutions Problem 47: Show tht funtion f is bsolutely ontinuous on [, b if nd only if for eh ɛ > 0, there is δ > 0 suh tht for every finite disjoint olletion {( k, b k )} n of open intervls in (, b), [f(b k ) f( k ) < ɛ, if [b k k < δ. Solution: Clerly, by the tringle inequlity the usul definition of bsolute ontinuity implies this one. It remins to show the onverse. To see this, let {( k, b k )} n be olletion of disjoint open intervls in [, b, we denote by the invervls {( + k, b+ k )}n+ the intervls suh tht f(b + k ) f(+ k ) 0 nd {( k, b k )}n the intervls suh tht f(b k ) f( k ) < 0, nd let δ + nd δ be suh tht n + n + n+ [f(b + k ) f(+ k ) = f(b + k ) f(+ k ) < ɛ/2, if [b + k + k < δ+ nd n [f(b k ) f( k ) = n Summing the two inequlities bove it follows tht f(b k ) f( k ) < ɛ n f(b k ) f( k ) < ɛ/2, if [b k k < δ. if [b k k < min{δ +, δ }. Problem 48: The Cntor Lebesgue funtion ϕ is ontinuous nd inresing on [0, 1. Conlude from Theorem 10 tht ϕ is not bsolutely ontinuous on [0, 1. Compre this resoning with tht proposed in Problem 40. Solution: The Cntor Lebesgue funtion hs the property tht ϕ (x) = 0.e. nd ϕ(1) ϕ(0) = 1. If it were ontinuous, this would imply tht 0 = ϕ = ϕ(1) ϕ(0) = 1, [0,1 1
2 whih is lerly ontrdition. This differs from the resoning proposed in problem 40 sine we re using here hrteriztion of bsolute ontinuity in terms of the Fundmentl theorem of lulus insted of the definition of bsolute ontinuity. Problem 49: Let f be ontinuous on [, b nd differentible lmost everywhere on (, b). Show tht [ lim Diff 1/nf f = f(b) f() if nd only if [ = lim Diff 1/n f. Solution: Sine f is ontinuous nd differentible.e. on (, b), we hve tht nd [ ( lim Diff 1/n f = lim Av1/n f(b) Av 1/n f() ) = f(b) f(), [ lim Diff 1/nf = for lmost ll x. The result then follows immeditely. f Problem 52: Let f nd g be bsolutely ontinuous on [, b. Show tht f g = f(b)g(b) f()g() f g. Solution: Sine g is bsolutely ontinuous {Diff 1/n g} is uniformly integrble nd therefore {(Diff 1/n g) f} is uniformly integrble. Thus f g = lim however by diret omputtion we see tht [ b f Diff 1/n g = n f(x)g(x + 1/n)dx = n 1/n 1/n f Diff 1/n g. (f(x) f(x + 1/n))g(x + 1/n)dx + n n f(x + 1/n)g(x + 1/n)dx 1/n = Av 1/n (f g)(b) Av 1/n (f g)() b 1/n f(x + 1/n)g(x + 1/n)dx Diff 1/n f g. f(x + 1/n)g(x + 1/n)dx 2
3 Sine f nd g re ontinuous, Av 1/n (f g)(b) f(b)g(b), Av(f g)() f()g(), nd f is bsolutly ontinuous so {Diff 1/n f g} is uniformly integrble, therefore Hene we hve our result. Diff 1/n f g f g. Problem 55: Let f be of bounded vrition on [, b nd define v(x) = T V (f [,x ) for ll x [, b. (i) Show tht f v.e. on [, b, nd infer from this tht f T V (f). (ii) Show tht the bove is n equlity if nd only if f is bsolutely ontinuous on [, b. (iii) Compre prts (i) nd (ii) with Corollries 4 nd 12, repetively. Solution: (i) Suppose we tke prtition P = {u, v}, with u, v [, b therefore f(v) f(u) = V (f, P ) T V (f [u,v ) = T (f [,v ) T (f [,u ) f(v) f(u) v u T (f [,v) T (f [,u ), v u sine f is of bounded vrition, nd T V (f [,x ) is inresing, the limits s u v exists pointwise.e., so f v.e. However, f v v(b) v() = T V (f). (ii) If f is bsolutely ontinuous, then we know tht for ny (u, v) [, b v u f = f(v) f(u). Also f is of bounded vrition, therefore for ny prtition P of [, b, V (f, P ) = f(b k ) f( k ) = 3 k k f k k f = f.
4 Therefore nd so by the previous inequlity, T V (f) f T V (f) = To show the overse, suppose tht we hve equlity. We n see from the work in prt (i) tht this must imply tht f. v = v(b) v(). Sine v is stritly inresing, this mens tht v is bsolutely ontinuous on [, b. Let δ > 0 be suh tht if {( k, b k )} is olletion of disjoint open intervls then T V (f [,bk ) T V (f [,k ) = T V (f [k,b k ) < ɛ whenever n [b k k < δ. By the definition of T V (f [k,b k ) hve Therefore f(b k ) f( k ) T V (f [k,b k ). f(b k ) f( k ) whenever n [b k k < δ. T V (f [k,b k ) < ɛ (iii) If f is inresing s in Corollries 4 nd 12, then we know tht totl vrition is T V (f) = f(b) f(), nd f is positive, Prt (i) sys tht f f(b) f() whih is equivlent to Corollry 4, nd prt (ii) sys tht f is bsolutely ontinuous if nd only if whih is equivlent to Corollry 12. f = f(b) f() Problem 56: Let g be stritly inresing nd bsolutely ontinuous on [, b. 4
5 (i) Show tht for ny open subset O of (, b), m(g(o)) = (ii) Show tht for ny G δ subset of (, b), m(g()) = O g (x)dx. g (x)dx. (iii) Show tht for ny subset of [, b tht hs mesure 0, its imge g() lso hs mesure 0, so tht m(g()) = 0 = g (x)dx. (iv) Show tht for ny mesurble subset A of [, b, m(g(a)) = g (x)dx. (v) Let = g(), nd d = g(b). Show tht for ny simple funtion φ on [, d, d φ(y)dy = A φ(g(x))g (x)dx. (vi) Show tht for ny nonnegtive integrble funtion f over [, d. d f(y)dy = f(g(x))g (x)dx. (vii) Show tht prt (i) follows from (vi) in the se tht f is the hrteristi funtion of O nd the omposition is defined. Solution: (i) Write O ountble olletion of disjoint open intervls, O = ( k, b k ), sine g is inresing nd ontinuous, we n lso write g(o) = (g( k), g(b k )), where {(g( k ), g(b k ))} is lso ountble olletion of disjoint open intervls. Thus we see m(g(o)) = [g(b k ) g( k ) whih by the bsolute ontinuity of g gives, m(g(o)) = 5 [k k g = O g.
6 (ii) Write G δ sets s the intersetion of ountble olletion of open sets = O k. Sine for ny n, n = n O k is open nd desending, we hve m(g( n )) = g. n Sine g is stritly inresing nd ontinuous, we know g( n ) is lso open nd desending, therefore, by ontinuity of mesure nd ontinuity of integrtion m(g()) = g. (iii) Problem 40. (iv) Sine A is mesurble, we my write it s A = F, union of G δ set nd F set of mesure 0. Sine g( F ) = g() g(f ), nd g(f ) is mesure 0 nd g() is G δ, m(g(a)) = m(g()) = g = g. (v)* We first show this for n inditor funtion χ of mesurble set [, d. Tht is, we would like to show tht m() = (χ g) g. Note, the diffiulty here is tht we do not know the mesurbility of χ g χ g 1 (), sine the inverse of stritly inresing bsolutely ontinuous funtion my not mp mesurble sets to mesurble sets. This n be seen in ounterexmple by Silvi Sptru where n bsolutely ontinuous stritly inresing funtion f is onstruted so tht f = 0.e. on ft ntor set B, nd f mps B to set of mesure 0. Hene g 1 behves like the CntorLebesgue funtion, mpping set of mesure 0 to set of positive mesure. In prtiulr this mens tht there f 1 mps mesure zero set to nonmesurble set. However, it turns out tht while χ g my not be mesurble, (χ g) g will lwys be mesurble. We will show this by pproximtion with mesurble funtions. Denote h := (χ g) g nd note tht sine g is ontinuous, then for ny open set O nd ompt K, (χ O g) nd (χ K g) re both mesurble funtions. In prtiulr this implies by (iv) tht m(o) = (χ O g) g, nd m(k) = A (χ K g) g. We now hoose {O n } deresing sequene of open sets ontining, nd {K n } n inresing sequene of ompt sets ontined in so tht, m(o n ) m() nd m(k n ) m(). 6
7 Define φ n := (χ Kn g) g nd ψ n := (χ On g) g. We see tht for every n, φ 1... φ n h ψ n... ψ 1. Sine g 0.e., {φ n } nd {ψ n } re monotone sequenes of funtions bounded by g. Thus they hve pointwise limits.e. In prtiulr sine g is integrble, dominted onvergene gives Sine ψ n φ n 0 we onlude lim (ψ n φ n ) = m() m() = 0. lim (ψ n φ n ) = 0 pointwise. Therefore lim h φ n lim (ψ n φ n ) = 0, pointwise. So h is mesurble, being the pointwise limit of mesurble funtions φ n. We onlude by monotone (or dominted) onvergene tht m() = lim (χ Kn g) g = (χ g) g. Now suppose tht ϕ is simple funtion over (, d) given in nonil form by ϕ = s k χ k. where { k } is finite disjoint olletion of mesurble sets suh tht n k = (, d). We find g(b) ϕ = s k m( k ) = s k (χ k g) g = (ϕ g) g. g() (vi) If f is nonnegtive integrble funtion over [, d there is monotone sequene of simple funtions ϕ n tht onverge to f p.w. Therefore by the previous problem d ϕ n = so by the monotone onvergene theorem, d f = (ϕ n g) g (f g) g. (vii) Choose f to be the hrteristi funtion of g(o), where O (, b). Then m(g(o)) = (χ g(o) g)g = χ O g = g O 7
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