Kenneth Kuttler

20.4. EXERCISES 531

6. Explain why for each t > 0,x→ e−tx is a function in L1 (R) and∫

∞

0 e−txdx = 1t . Thus∫ R

0sin(t)

t dt =∫ R

0∫

∞

0 sin(t)e−txdxdt Now explain why you can change the order ofintegration in the above iterated integral. Then compute what you get. Next pass toa limit as R→ ∞ and show

∫∞

0sin(t)

t dt = 12 π

7. Let f (y) = g(y) = |y|−1/2on (−1,0)∪(0,1) and f (y) = g(y) = 0 off (−1,0)∪(0,1).Find x where

∫R f (x− y)g(y)dy makes sense.

8. Let Ei be a Borel set in R. Show that ∏ni=1 Ei is a Borel set in Rn.

9. Let {an} be an increasing sequence of numbers in (0,1) which converges to 1. Let gnbe a nonnegative function which equals zero outside (an,an+1) such that

∫gndx = 1.

Now for (x,y) ∈ [0,1)× [0,1) define f (x,y) ≡ ∑∞k=1 gn (y)(gn (x)−gn+1 (x)) . Ex-

plain why this is actually a finite sum for each such (x,y) so there are no conver-gence questions in the infinite sum. Explain why f is a continuous function on[0,1)× [0,1). You can extend f to equal zero off [0,1)× [0,1) if you like. Show theiterated integrals exist but are not equal. In fact, show∫ 1

∫ 1

0f (x,y)dydx = 1 ̸= 0 =

∫ 1

0f (x,y)dxdy.

Does this example contradict the Fubini theorem, Corollary 10.14.11 on Page 307?Explain why or why not.

10. Let f : [a,b]→ R be Rieman integrable. Thus f is a bounded function and by Dar-boux’s theorem, there exists a unique number between all the upper sums and lowersums of f , this number being the Riemann integral. Show that f is Lebesgue mea-surable and

∫ ba f (x)dx =

∫[a,b] f dm where the second integral in the above is the

Lebesgue integral taken with respect to one dimensional Lebesgue measure and thefirst is the ordinary Riemann integral.

11. Let (Ω,F ,µ) be a σ finite measure space and let f : Ω→ [0,∞) be measurable. Alsolet φ : [0,∞)→ R be increasing with φ (0) = 0 and φ a C1 function. Show that∫

Ω

φ ◦ f dµ =∫

∞

0φ′ (t)µ ([ f > t])dt.

Hint: This can be done using the following steps. Let tni = i2−n. Show that

X[ f>t] (ω) = limn→∞

∞

∑i=0

X[ f>tni+1]

(ω)X[tni ,t

ni+1)

(t)

Now this is a countable sum of F ×B ([0,∞)) measurable functions and so it fol-lows that (t,ω)→X[ f>t] (ω) is F ×B ([0,∞)) measurable. Consequently, so isX[ f>t] (ω)φ (t) . Note that it is important in the argument to have f > t. Now ob-serve ∫

Ω

φ ◦ f dµ =∫

Ω

∫ f (ω)

0φ′ (t)dtdµ =

∫Ω

∫∞

0X[ f>t] (ω)φ

′ (t)dtdµ

Use Fubini’s theorem. For your information, this does not require the measure spaceto be σ finite. You can use a different argument which ties in to the first definition ofthe Lebesgue integral. The function t→ µ ([ f > t]) is called the distribution function.

20.4.10.11.EXERCISES 531Explain why for each t > 0,x — e~” is a function in L! (IR) and fy’ e~dx = +. Thus*R sin(t)OFdt = Se Jo sin(t)e~ “dxdt Now explain why you can change the order ofintegration in the above iterated Oe 1. Then compute what you get. Next pass toa limit as R + co and show Jy” Odt=Let f(y) = (x)= sen 0) and f(y) =g(y) =0off (-1,0)U(0, 1).Find x where fp f (x —y) g (y) dy makes sense.Let E; be a Borel set in R. Show that J]/_, E; is a Borel set in R”.Let {a,,} be an increasing sequence of numbers in (0,1) which converges to 1. Let g,be a nonnegative function which equals zero outside (a,,an41) such that [ g,dx = 1.Now for (x,y) € [0,1) x [0,1) define f (x,y) = YZ) 8n(¥) (Gn (X) — Bn4i (x)). Ex-plain why this is actually a finite sum for each such (x,y) so there are no conver-gence questions in the infinite sum. Explain why f is a continuous function on(0,1) x [0, 1). You can extend f to equal zero off [0,1) x [0, 1) if you like. Show theiterated integrals exist but are not equal. In fact, show[ [ feaar= 1 40=[ [ rosyaray.Does this example contradict the Fubini theorem, Corollary 10.14.11 on Page 307?Explain why or why not.Let f : [a,b] + R be Rieman integrable. Thus f is a bounded function and by Dar-boux’s theorem, there exists a unique number between all the upper sums and lowersums of f, this number being the Riemann integral. Show that f is Lebesgue mea-surable and f iH Ff (x)dx = fia. fdm where the second integral in the above is theLebesgue integral taken with respect to one dimensional Lebesgue measure and thefirst is the ordinary Riemann integral.Let (Q,.F,) bea o finite measure space and let f : Q — [0,°°) be measurable. Alsolet @ : [0,0c) — R be increasing with ¢ (0) = 0 and @ aC! function. Show that[eerau= [0 u(r > adarHint: This can be done using the following steps. Let ¢’ = i2~”. Show thatXi psi) (@) = lim y 4 (ptt) (@) Zin yn) (t)nyo ivi+lNow this is a countable sum of .¥ x @([0,-c)) measurable functions and so it fol-lows that (t,@) + Zips, (@) is F x A([0,cc)) measurable. Consequently, so is2 si) (@) (t). Note that it is important in the argument to have f > 1. Now ob-serve[ve fdu = Ls (t)dtdu = Lf ZH pon (@) 0! (t)dtduUse Fubini’s theorem. For your information, this does not require the measure spaceto be o finite. You can use a different argument which ties in to the first definition ofthe Lebesgue integral. The function t — yt ([,f > t]) is called the distribution function.