Ma 5051 — Real Variables and Functional AnalysisSolutions for Take-Home MidtermProf. Sawyer — Washington UniversityLet (X, M, µ) be a measure space. RecallRAf(x)dµ =RIA(x)f(x)dµ for A ∈ Mand f ∈ L+, where IA(x) is the indicator function of A.1. (a) The problem is to show that the integrand is dominated by an integrablefunction. Since 1 − e−x=Rx0e−ydy ≤Rx0dy = x, it follows that 1 − x ≤ e−x,1 − (1/n)x ≤ e−x/n, and¡1 − (1/n)x¢n≤ e−x. Thus the integrand is dominated byxke−x, which is integrable on (0, ∞). Then by the dominated convergence theoremlimn→∞Zn0xk¡1 − n−1x¢ndx =Z∞0limn→∞I(0,n](x) xk¡1 − n−1x¢ndx=Z∞0xke−xdx = k!(b) The substitution x → x/√n changes the integral toI(n) =Z√n01 + x2¡1 + (1/n)x2¢ndxBy the binomial theorem (see also the model solutions for Problem 4a of HW5),µ1 +x2n¶n≥ 1 + nx2n+n(n − 1)2µx2n¶2≥ 1 +13x4for x > 0 and n ≥ 3. Thus the integrand of I(n) is dominated by 3(1 + x2)/(3+ x4),which is integrable on (0, ∞). Then by the dominated convergence theoremlimn→∞I(n) =Z∞0limn→∞I(0,√n](x)1 + x2¡1 + (1/n)x2¢ndx=Z∞01 + x2ex2dx =Z∞0e−x2dx +Z∞012x³2xe−x2´dx=32Z∞0e−x2dx =34√π2. The integral isI(a) =Z∞0e−y2sin(ay) dy =Z∞0e−y2∞Xn=0(−1)na2n+1y2n+1(2n + 1)!dyMa 5051– Real Variables and Functional Analysis– Take-Home Midterm. . . . . . . . . 2The partial sums of the integrand of I(a) are dominated for fixed a > 0 bye−y2∞Xn=0a2n+1y2n+1(2n + 1)!≤ e−y2∞Xn=0anynn!= e−y2eaywhere the second series is formed from the first by adding in the even termsa2ny2n/(2n)!. Since the last expression above is integrable on (0, ∞), dominatedconvergence allows us to interchange the integral and sum in I(a) and concludeI(a) =∞Xn=0(−1)na2n+1(2n + 1)!Z∞0e−y2y2n+1dy =∞Xn=0(−1)na2n+1(2n + 1)!12Z∞0e−xxndx=∞Xn=0(−1)na2n+1n!2(2n + 1)!=a2∞Xn=0(−a2/2)n(2n + 1)(2n − 1) . . . (3)(1)since 2nn! = (2n)(2n − 2) . . . 2.One can use complex-variable techniques to show I(a) =R∞0e−y2sin(ay) dy =e−a2/4Ra/20ex2dx, but that is not required.See the Appendix for two derivations of this identity using complex-variablemethods, one based on analytic continuation and one using Cauchy’s theorem.3. Let A = lim infn→∞Rfn(x) dµ. Then there exists a sequence nk↑ ∞ such thatlimk→∞Rfnkdµ = A. Since fnk→ f in measure, there exists a further subsequence{fnkj} such that limj→∞fnkj(x) = f(x) a.e. µ. Since fn(x) ≥ 0, this impliesZf(x) dµ ≤ lim infj→∞Zfnkj(x) dµ = limk→∞Zfnk(x) dµ = lim infn→∞Zfn(x) dµandRf(x) dµ ≤ lim infn→∞Rfn(x) dµ.4. Since µ is a finite measure on X = [0, 1], by the dominated convergence theoremlimn→∞Z10Z10e−n(x−y)2dµ(x)dµ(y) = I =Z10Z10J(x, y)dµ(x)dµ(y)where J(x, x) = 1 and J(x, y) = 0 for y 6= x. The inner integral in I above, as afunction of y, isZ10J(x, y)dµ(x) =Z10J(x, y)(dm(x) + 2dδa(x)) = 2J(a, y)Ma 5051– Real Variables and Functional Analysis– Take-Home Midterm. . . . . . . . . 3since, for fixed y, J(x, y) = 0 (Lebesgue) a.e. Thus the double integral isI =Z102J(a, y)dµ(y) =Z102J(a, y)(dm(y) + 2dδa(y)) = 4J(a, a) = 45. (a) Show that Gf= {(x, y) : 0 ≤ y < f (x) } is measurable in the productσ-algebra (that is, Gf∈ M ⊗ B(R+)).Proof I of (a) : By Proposition 2.10, there exist simple functions fn(x) suchthat 0 ≤ fn(x) ↑ f(x) for all x. If h(x) =Pmj=1cjIAj(x) is a nonnegative simplefunction with disjoint Aj∈ M, then Gh=Smj=1¡Aj× [0, cj)¢is a finite disjointunion of measurable rectangles and so is product measurable. Since Gfn↑ Gfif0 ≤ fn(x) ↑ f(x) for all x, the set Gfis also product measurable.Proof II of (a): The functions F1(x, y) = f (x) and F2(x, y) = y are bothmeasurable in the product σ-algebra, since in each case {(x, y) : Fj(x, y) ≤ λ } is ameasurable rectangle. Thus F (x, y) = F1(x, y) −F2(x, y) = f(x) −y is also productmeasurable, by Proposition 2.6 (page 45) in the text. Finally, Gf= {(x, y) :F (x, y) > 0 }.(b) By the Fubini-Tonelli theorem for the product measure of µ on (X, M)and Lebesgue measure m on R+= [0, ∞),(µ × m)(Gf) =ZXZ∞0IGf(x, y) dm(y)dµ(x) =Z∞0ZXIGf(x, y) dµ(x)dm(y)For fixed x, the section (Gf)x= {y : y < f(x) } = [0, f (x)), so that the innerintegral of the first integral above isR∞0IGf(x, y) dm(y) = f (x). Thus the firstiterated integral equalsRXf(x) dµ.For fixed y, the section (Gf)y= {x : f (x) > y }, so thatRXIGf(x, y) dm(y) =F (y) = µ({x : f(x) > y }) and the second integral above isR∞0F (y) dy. Thus thetwo expressions are the same, and are also the same as (µ × m)(Gf).6. It is sufficient to assume f( x) ≥ 0 and g(y) ≥ 0, since otherwise we can writef = f+− f−and g = g+− g−. Define simple functions fn(x), gm(y) such that0 ≤ fn(x) ↑ f(x) for all x and 0 ≤ gm(y) ↑ g(y) for all y.(a) If fn(x) =Pm1j=1ajIAj(x) and gn(y) =Pm2k=1bjIBj(y) for fixed n anddisjoint sets {Aj} in X and {Bk} in Y , then {Aj× Bk} are disjoint subsets ofX × Y and{(x, y) : fn(x)gn(y) ≤ λ } =[[{Aj× Bk: ajbk≤ λ }is a finite union of measurable rectangles for each λ. Thus the functions fn(x)gn(y)are product measurable.Ma 5051– Real Variables and Functional Analysis– Take-Home Midterm. . . . . . . . . 4Since 0 ≤ fn(x)gn(y) ↑ h(x, y) = f(x)g(y) for all x and y, it follows thath(x, y) is product measurable. (You can also argue directly from Proposition 2.6 asin Proof II of Problem 5a.)(b) Proof I. Since f ∈ L1(µ) and g ∈ L1(β), the sets X1= {x : f(x) > 0 }and Y1= {y : g(y) > 0 } are σ-finite. The function h(x, y) = 0 for (x, y) /∈ X1×Y1.By the definition of integrals as the supremum of simple functions underneath,RX×Yh(x, y)dν =RX1×Y1h(x, y)dν. Thus part (b) follows from Tonelli’s theoremrestricted to X1× Y1.Proof II. By definition, (µ×β)(A×B) = µ(A)β(B) for A ∈ M1and B ∈ M2.If f1(x) =Pnj=1cjIAj(x) for disjoint Aj∈ M1and g1(y) =Pmk=1dkIBk(y) fordisjoint Bk∈ M2, thenZf1(x)g1(y) d(µ × β) =nXj=1mXk=1cjdk(µ × β)(Aj× Bk)=nXj=1cjµ(Aj)ÃmXk=1dkβ(Bk)!=µZf1(x) dµ¶µZg1(y) dβ¶Thus the identity is true for nonnegative simple functions. Since 0 ≤ fn(x)gn(y) ↑h(x, y) = f(x, y) for all x and y, it follows from the increasing limits theorem thatit is true for h(x, y) as well.7. (a) “If f is Borel measurable on Rn, then so is f ◦ T ” This is true even if T isnot invertible, since, since by the definition of a measurable function on
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