8.6.E: Problems on Integrability and Convergence Theorems
( \newcommand{\kernel}{\mathrm{null}\,}\)
Fill in the missing details in the proofs of this section.
(i) Show that if f:S→E∗ is bounded and m-measurable on A, with mA<∞, then f is m -integrable on A( Theorem 2) and
∫Af=c⋅mA,
where inf f[A]≤c≤supf[A].
(ii) Prove that if f also has the Darboux property on A, then
(∃x0∈A)c=f(x0).
[Hint: Take g=1 in Theorem 3.]
(iii) What results if A=[a,b] and m= Lebesgue measure?
Prove Theorem 4 assuming that the fn are measurable on A and that
(∃k)∫Afk>−∞
instead of fn≥0.
[Hint: As {fn}↑, show that
(∀n≥k)∫Afn>−∞.
If
(∃n)∫Afn=∞,
then
∫Af=lim∫Afn=∞.
Otherwise,
(∀n≥k)|∫Afn|<∞;
so fn is integrable. (Why?) By Corollary 1 in §5, assume |fn|<∞. (Why?) Apply Theorem 4 to hn=fn−fk(n≥k), considering two cases:
∫Ah<∞ and ∫Ah=∞.]
Show that if fn↗f (pointwise) on A∈M, there are M-measurable maps Fn≥fn and F≥f on A, with Fn↗F (pointwise) on A, such that
∫AF=¯∫Af and ∫AFn=¯∫Afn.
[Hint: By Lemma 2 of §5, fix measurable maps h≥f and hn≥fn with the same integrals. Let
Fn=infk≥n(h∧hk),n=1,2,…,
and F=supnFn≤h. (Why?) Proceed.]
For A∈M and any (even nonmeasurable) functions f,fn:S→E∗, prove the following.
(i) If fn↗f( a.e. ) on A, then
¯∫Afn↗¯∫Af,
provided
(∃n)¯∫Afn>−∞.
(ii) If fn↘f( 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?)]
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 fn→f, but not
∫_Afn→∫_Af.
Explain!]
Given fn:S→E∗ and A∈M, let
gn=infk≥nfk and hn=supk≥nfk(n=1,2,…).
Prove that
(i) ¯∫Alim_fn≤lim_∫_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¯limfn≠lim_∫_Afn.
(See Note 2).
Let fn≥0 on A∈M and fn→f( a.e. ) on A. Let A⊇X,X∈M.
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 {¯∫A−Xfn}
contradicting Lemma 2.
(ii) Let m= Lebesgue measure; A=(0,1),X=(0,12),
fn={n on (0,12n],−n on (1−12n,1[.
⇒9. (i) Show that if f and g are m-measurable and nonnegative on A, then
(∀a,b≥0)∫A(af+bg)=a∫Af+b∫Ag.
(ii) If, in addition, ∫Af<∞ or ∫Ag<∞, this formula holds for any a,b∈E1.
[Hint: Proceed as in Theorem 1.]
⇒10. If
f=∞∑n=1fn,
with all fn measurable and nonnegative on A, then
∫Af=∞∑n=1∫Afn.
[Hint: Apply Theorem 4 to the maps
gn=n∑k=1fk↗f.
Use Problem 9.]
If
q=∞∑n=1∫A|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=1∫Afn.
[Hint: Let g=∑∞n=1|fn|. By Problem 10,
∫Ag=∞∑n=1∫A|fn|=q<∞;
so g<∞(a.e.) on A. (Why?) Apply Theorem 5 and Note 1 to the maps
gn=n∑k=1fk;
note that |gn|≤g.]
(Convergence in measure; see Problem 11(ii) of §3).
(i) Prove Riesz' theorem: If fn→f in measure on A⊆S, there is a subsequence {fnk} such that fnk→f (almost uniformly), hence (a.e.), on A.
[Outline: Taking
σk=δk=2−k,
pick, step by step, naturals
n1<n2<⋯<nk<⋯
and sets Dk∈M such that (∀k)
mDk<2−k
and
ρ′(fnk,f)<2−k
on A−Dk. (Explain!) Let
En=∞⋃k=nDk,
mEn<21−n.( Why?) Show that
(∀n)(∀k>n)ρ′(fnk,f)<21−n
on A−En. Use Problem 11 in §3.]
(ii) For maps fn:S→E and g:S→E1 deduce that if
fn→f
in measure on A and
(∀n)|fn|≤g( a.e. ) on A,
then
|f|≤g( a.e. ) on A.
[ Hint: fnk→f(a.e.) on A.]
Continuing Problem 12( ii ), let
fn→f
in measure on A∈M(fn:S→E) and
(∀n)|fn|≤g(a.e.) on A,
with
¯∫Ag<∞.
Prove that
limn→∞¯∫A|fn−f|=0.
Does
¯∫Afn→¯∫Af?
[Outline: From Corollary 1 of §5, infer that g=0 on A−C, where
C=∞⋃k=1Ck(disjoint),
mCk<∞. (We may assume gM-measurable on A. Why?) Also,
∞>∫Ag=∫A−Cg+∫Cg=0+∞∑k=1∫Ckg;
so the series converges. Hence
(∀ε>0)(∃p)∫Ag−ε<p∑k=1∫Ckg=∫Hg,
where
H=p⋃k=1Ck∈M
and mH<∞. As |fn−f|≤2g( a.e. ), we get
(1) ∫_A|fn−f|≤¯∫A|fn−f|≤¯∫H|fn−f|+∫A−H2g<¯∫H|fn−f|+2ε.
(Explain!)
As mH<∞, we can fix σ>0 with
σ⋅mH<ε.
Also, by Theorem 6, fix δ such that
2∫Xg<ε
whenever A⊇X,X∈M and mX<δ.
As fn→f in measure on H, we find M-sets Dn⊆H such that
(∀n>n0)mDn<δ
and
|fn−f|<σ on An=H−Dn.
(We may use the standard metric, as |f| and |fn|<∞ a.e. Why?) Thus from (1), we get
¯∫A|fn−f|≤¯∫H|fn−f|+2ε=¯∫An|fn−f|+¯∫Dn|fn−f|+2ε<¯∫An|fn−f|+3ε≤σ⋅mH+3ε<4ε
for n>n0. (Explain!) Hence
lim¯∫A|fn−f|=0.
See also Problem 7 in §5 and Note 1 of §6 (for measurable functions) as regards
lim¯∫Afn⋅]
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
limi→∞m(A−Hi(k))=0.
(Why?) Verify that
(∀n)ρ′(fn,f)=|fn−f|≤2g(a.e.) on A,
and
(∀i,k)A−Hi(k)⊆A(2g≥1k)∪Q(mQ=0).
Infer that
(∀i,k)m(A−Hi(k))<∞.
Now, as (∀k)Hi(k)↘∅ (why?), right continuity applies.]