8.8. EXERCISES 215

Now explain why this converges to 0 on the right. This will involve the dominatedconvergence theorem. Conclude that

∫f XEdmn = 0 for every bounded measurable

set E. Show that this implies that∫

f XEdmn = 0 for every measurable E. Explainwhy this requires f = 0 a.e. The result which gets used over and over in all of this isthe dominated convergence theorem.

20. Let F (x) =(∫ x

0 e−t2dt)2

, so

F ′ (x) = 2e−x2(∫ x

0e−t2

)= 2xe−x2

(∫ 1

0e−(ux)2

du).

Now integrate by parts to get the following.

F (x) = e(x)+1+∫ x

0e−t2

∫ 1

0

(−2tu2e−t2u2

)dudt, lim

x→∞e(x) = 0

Now change the order of integration in this integral to get

F (x) = e(x)+1−∫ 1

0u2∫ x

02te−t2(1+u2)dtdu.

Modifying e(x) as needed, obtain

F (x) = e(x)+1−∫ 1

0

u2

1+u2 = e(x)+∫ 1

0

11+u2 du = e(x)+

π

4

Show∫

0 e−t2dt =

√π

2 . Justify all the steps in the above using whatever theorems areapplicable.

21. The Dini derivates are as follows. In these formulas, f is a real valued functiondefined on R and ∆ f (+) will be f (x+h)− f (x)

h for h > 0 and ∆ f (−) will be f (x)− f (x−h)h

for h > 0.

D+ f (x) ≡ lim suph→0+

∆ f (+) ,D+ f (x)≡ lim infh→0+

∆ f (+)

D− f (x) ≡ lim suph→0+

∆ f (−) ,D− f (x)≡ lim infh→0+

∆ f (−)

Thus when these are all equal, the function has a derivative. Now suppose f is anincreasing function. Let

Nab ={

x : D+ f (x)> b > a > D+ f (x)},a≥ 0

Let V be an open set which contains Nab∩ (−r,r)≡ Nrab such that

m(V \ (Nab∩ (−r,r)))< ε

Then explain why there exist disjoint intervals [ai,bi] such that

m(Nrab \∪i [ai,bi]) = m(Nr

ab \∪i (ai,bi)) = 0

andf (bi)− f (ai)≤ am(ai,bi)

8.8. EXERCISES 21520.21.Now explain why this converges to 0 on the right. This will involve the dominatedconvergence theorem. Conclude that { f 2¢dm, = 0 for every bounded measurableset E. Show that this implies that [ f 2zdm, = 0 for every measurable E. Explainwhy this requires f = 0 a.e. The result which gets used over and over in all of this isthe dominated convergence theorem.x2 2Let F (x) = (ie dr) , SOx 1F'(x)=2e™ (/ «*) =2xe" (/ eau) .Jo 0Now integrate by parts to get the following.x 2 fl > 22F(x) =e(s) +14 [ et [ (—21e-P*) duat, tim e(x) =00 0XooNow change the order of integration in this integral to getF (x)= ett fu > [ate *( (1) dtdu.Modifying e (x) as needed, obtain1 I 1F (x) =e(x)+1- [ Taw =e(x)+ F Tq tt =e +aShow fo. et dt= ve Justify all the steps in the above using whatever theorems areapplicable.The Dini derivates are as follows. In these formulas, f is a real valued functiondefined on R and Af (+) will be &"-) for p> 0 and Af (—) will be S-fo-)for h > 0.D* f(x) = lim sup Af(+),D,f(x) =lim inf Af (+)h—>0+ h>0+Df (x) = lim sup Af(—),D_f (x) =lim inf Af(-)h0+ h>0+Thus when these are all equal, the function has a derivative. Now suppose f is anincreasing function. LetNap = {x:D* f (x) >b>a> Df (x)$,a>0Let V be an open set which contains Nay (—r,r) = N/, such thatm(V\ (Na A(—1,r))) <€Then explain why there exist disjoint intervals [a;,b;| such thatm (Naw \ Ui lai, bil) = m (Nap \ Ui (ai, bi) = 0andFf (bi) — f (ai) < am (ai, bi)