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Mathematics LibreTexts

8.6.E: Problems on Integrability and Convergence Theorems

( \newcommand{\kernel}{\mathrm{null}\,}\)

Exercise 8.6.E.1

Fill in the missing details in the proofs of this section.

Exercise 8.6.E.2

(i) Show that if f:SE is bounded and m-measurable on A, with mA<, then f is m -integrable on A( Theorem 2) and
Af=cmA,


where inf f[A]csupf[A].
(ii) Prove that if f also has the Darboux property on A, then
(x0A)c=f(x0).

[Hint: Take g=1 in Theorem 3.]
(iii) What results if A=[a,b] and m= Lebesgue measure?

Exercise 8.6.E.3

Prove Theorem 4 assuming that the fn are measurable on A and that
(k)Afk>


instead of fn0.
[Hint: As {fn}, show that
(nk)Afn>.

If
(n)Afn=,

then
Af=limAfn=.

Otherwise,
(nk)|Afn|<;

so fn is integrable. (Why?) By Corollary 1 in §5, assume |fn|<. (Why?) Apply Theorem 4 to hn=fnfk(nk), considering two cases:
Ah< and Ah=.]

Exercise 8.6.E.4

Show that if fnf (pointwise) on AM, there are M-measurable maps Fnfn and Ff on A, with FnF (pointwise) on A, such that
AF=¯Af and AFn=¯Afn.


[Hint: By Lemma 2 of §5, fix measurable maps hf and hnfn with the same integrals. Let
Fn=infkn(hhk),n=1,2,,

and F=supnFnh. (Why?) Proceed.]

Exercise 8.6.E.5

For AM and any (even nonmeasurable) functions f,fn:SE, prove the following.
(i) If fnf( a.e. ) on A, then
¯Afn¯Af,


provided
(n)¯Afn>.

(ii) If fnf( a.e. ) on A, then
_Afn_Af,

provided
(n)_Afn<.

[Hint: Replace f,fn by F,Fn as in Problem 4. Then apply Problem 3 to Fn; thus obtain (i). For (ii), use (i) and Theorem 1(e) in §5. (All is orthodox; why?)]

Exercise 8.6.E.6

Show by examples that
(i) the conditions
¯Afn> and _Afn<


in Problem 5 are essential; and
(ii) Problem 5(i) fails for lower integrals. What about 5(ii)?
[Hints: (i) Let A=(0,1)E1,m= Lebesgue measure, fn= on (0,1n),fn=1 elsewhere.
(ii) Let M={E1,},mE1=1,m=0,fn=1 on (n,n),fn=0 elsewhere. If f=1 on A=E1, then fnf, but not
_Afn_Af.

Explain!]

Exercise 8.6.E.7

Given fn:SE and AM, let
gn=infknfk and hn=supknfk(n=1,2,).


Prove that
(i) ¯Alim_fnlim__Afn provided (n)¯Agn>; and
(ii) _A¯limfn¯lim_Afnprovided(n)_Ahn<.
[Hint: Apply Problem 5 to gn and hn.]
(iii) Give examples for which
¯Alim_fn¯limA¯Afn and _A¯limfnlim__Afn.

(See Note 2).

Exercise 8.6.E.8

Let fn0 on AM and fnf( a.e. ) on A. Let AX,XM.
Prove the following.
(i) If
¯Afn¯Af<,


then
¯Xfn¯Xf.

(ii) This fails for sign-changing fn.
[Hints: If (i) fails, then
lim_X¯Xfn<¯Xf or lim_X¯Xfn>¯Xf.

Find a subsequence of
{¯Xfn} or {¯AXfn}

contradicting Lemma 2.
(ii) Let m= Lebesgue measure; A=(0,1),X=(0,12),
fn={n on (0,12n],n on (112n,1[.

Exercise 8.6.E.9

9. (i) Show that if f and g are m-measurable and nonnegative on A, then
(a,b0)A(af+bg)=aAf+bAg.


(ii) If, in addition, Af< or Ag<, this formula holds for any a,bE1.
[Hint: Proceed as in Theorem 1.]

Exercise 8.6.E.10

10. If
f=n=1fn,


with all fn measurable and nonnegative on A, then
Af=n=1Afn.

[Hint: Apply Theorem 4 to the maps
gn=nk=1fkf.

Use Problem 9.]

Exercise 8.6.E.11

If
q=n=1A|fn|<


and the fn are m-measurable on A, then
n=1|fn|<(a.e.) on A

and f=n=1fn is m-integrable on A, with
Af=n=1Afn.

[Hint: Let g=n=1|fn|. By Problem 10,
Ag=n=1A|fn|=q<;

so g<(a.e.) on A. (Why?) Apply Theorem 5 and Note 1 to the maps
gn=nk=1fk;

note that |gn|g.]

Exercise 8.6.E.12

(Convergence in measure; see Problem 11(ii) of §3).
(i) Prove Riesz' theorem: If fnf in measure on AS, there is a subsequence {fnk} such that fnkf (almost uniformly), hence (a.e.), on A.
[Outline: Taking
σk=δk=2k,


pick, step by step, naturals
n1<n2<<nk<

and sets DkM such that (k)
mDk<2k

and
ρ(fnk,f)<2k

on ADk. (Explain!) Let
En=k=nDk,

mEn<21n.( Why?) Show that
(n)(k>n)ρ(fnk,f)<21n

on AEn. Use Problem 11 in §3.]
(ii) For maps fn:SE and g:SE1 deduce that if
fnf

in measure on A and
(n)|fn|g( a.e. ) on A,

then
|f|g( a.e. ) on A.

[ Hint: fnkf(a.e.) on A.]

Exercise 8.6.E.13

Continuing Problem 12( ii ), let
fnf


in measure on AM(fn:SE) and
(n)|fn|g(a.e.) on A,

with
¯Ag<.

Prove that
limn¯A|fnf|=0.

Does
¯Afn¯Af?

[Outline: From Corollary 1 of §5, infer that g=0 on AC, where
C=k=1Ck(disjoint),

mCk<. (We may assume gM-measurable on A. Why?) Also,
>Ag=ACg+Cg=0+k=1Ckg;

so the series converges. Hence
(ε>0)(p)Agε<pk=1Ckg=Hg,

where
H=pk=1CkM

and mH<. As |fnf|2g( a.e. ), we get
 (1) _A|fnf|¯A|fnf|¯H|fnf|+AH2g<¯H|fnf|+2ε.

(Explain!)
As mH<, we can fix σ>0 with
σmH<ε.

Also, by Theorem 6, fix δ such that
2Xg<ε

whenever AX,XM and mX<δ.
As fnf in measure on H, we find M-sets DnH such that
(n>n0)mDn<δ

and
|fnf|<σ on An=HDn.

(We may use the standard metric, as |f| and |fn|< a.e. Why?) Thus from (1), we get
¯A|fnf|¯H|fnf|+2ε=¯An|fnf|+¯Dn|fnf|+2ε<¯An|fnf|+3εσmH+3ε<4ε

for n>n0. (Explain!) Hence
lim¯A|fnf|=0.

See also Problem 7 in §5 and Note 1 of §6 (for measurable functions) as regards
lim¯Afn]

Exercise 8.6.E.14

Do Problem 12 in §3 (Lebesgue-Egorov theorems) for T=E, assuming
(n)|fn|g(a.e.) on A,


with
Ag<

(instead of mA<).
[Hint: With Hi(k) as before, it suffices that
limim(AHi(k))=0.

(Why?) Verify that
(n)ρ(fn,f)=|fnf|2g(a.e.) on A,

and
(i,k)AHi(k)A(2g1k)Q(mQ=0).

Infer that
(i,k)m(AHi(k))<.

Now, as (k)Hi(k) (why?), right continuity applies.]


8.6.E: Problems on Integrability and Convergence Theorems is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.

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