7.5.E: Problems on Premeasures and Related Topics
( \newcommand{\kernel}{\mathrm{null}\,}\)
Fill in the missing details in the proofs, notes, and examples of this section.
Describe m∗ on 2S induced by a premeasure μ:C→E∗ such that each of the following hold.
(a) C={S,∅},μS=1.
(b) C={S,∅,and all singletons};μS=∞,μ{x}=1.
(c) C as in (b), with S uncountable; μS=1, and μX=0 otherwise.
(d) C={all proper subsets of S};μX=1 when ∅⊂X⊂S;μ∅=0.
Show that the premeasures
v′:C′→[0,∞]
induce one and the same (Lebesgue) outer measure m∗ in En, with v′=v (volume, as in §2):
(a) C′={open intervals};
(b) C′={half-open intervals};
(c) C′={closed intervals};
(d) C′=Cσ;
(e) C′={open sets};
(f) C′={half-open cubes};
[Hints: (a) Let m′ be the v′-induced outer measure; let C={all intervals}. As C′⊆C,m′A≥m∗A. (Why?) Also,
(∀ε>0)(∃{Bk}⊆C)A⊆⋃kBk and ∑vBk≤m∗A+ε.
(Why?) By Lemma 1 in §2,
(∃{Ck}⊆C′)Bk⊆Ck and vBk+ε2k>v′Ck.
Deduce that m∗A≥m′A,m∗=m′. Similarly for (b) and (c). For (d), use Corollary 1 and Note 3 in §1. For (e), use Lemma 2 in §2. For (f), use Problem 2 in §2.]
Do Problem 3(a)-(c), with m∗ replaced by the Jordan outer content c∗ (Note 6).
Do Problem 3, with v and m∗ replaced by the LS premeasure and outer measure. (Use Problem 7 in §4.)
Show that a set A⊆En is bounded iff its outer Jordan content is finite.
Find a set A⊆E1 such that
(i) its Lebesgue outer measure is 0 (m∗A=0), while its Jordan outer content c∗A=∞;
(ii) m∗A=0,c∗A=1 (see Corollary 6 in §2).
Let
μ1,μ2:C→[0,∞]
be two premeasures in S and let m∗1 and m∗2 be the outer measures induced by them.
Prove that if m∗1=m∗2 on C, then m∗1=m∗2 on all of 2S.
With the notation of Definition 3 and Note 6, prove the following.
(i) If A⊆B⊆S and m∗B=0, then m∗A=0; similarly for c∗.
[Hint: Use monotonicity.]
(ii) The set family
{X⊆S|c∗A=0}
is a hereditary set ring, i.e., a ring R such that
(∀B∈R)(∀A⊆B)A∈R.
(iii) The set family
{X⊆S|m∗X=0}
is a hereditary σ-ring.
(iv) So also is
H={those X⊆S that have basic coverings};
thus H is the hereditary σ-ring generated by C (see Problem 14 in §3).
Continuing Problem 8(iv), prove that if μ is σ-finite (Definition 4), so is m∗ when restricted to H.
Show, moreover, that if C is a semiring, then each X∈H has a basic covering {Yn}, with m∗Yn<∞ and with all Yn disjoint.
[Hint: Show that
X⊆∞⋃n=1∞⋃k=1Bnk
for some sets Bnk∈C, with μBnk<∞. Then use Note 4 in §5 and Corollary 1 of §1.]
Show that if
s:C→E∗
is σ-finite and additive on C, a semiring, then the σ-ring R generated by C equals the σ-ring R′ generated by
C′={X∈C||sX|<∞}
(cf. Problem 6 in §4).
[Hint: By σ-finiteness,
(∀X∈C)(∃{An}⊆C||sAn|<∞)X⊆⋃nAn;
so
X=⋃n(X∩An),X∩An∈C′.
(Use Lemma 3 in §4.)
Thus (∀X∈C)X is a countable union of C′-sets; so C⊆R′. Deduce R⊆R′. Proceed.]
With all as in Theorem 3, prove that if A has basic coverings, then
(∃B∈Aδ)A⊆B and m∗A=m∗B.
[Hint: By formula (4),
(∀n∈N)(∃Xn∈A|A⊆Xn)m∗A≤mXn≤m∗A+1n.
(Explain!) Set
B=∞⋂n=1Xn∈Aδ.
Proceed. For Aδ, see Definition 2(b) in §3.]
Let (S,C,μ) and m∗ be as in Definition 3. Show that if C is a σ-field in S, then
(∀A⊆S)(∃B∈C)A⊆B and m∗A=μB.
[Hint: Use Problem 11 and Note 3.]
⇒∗ Show that if
s:C→E
is σ-finite and σ-additive on C, a semiring, then s has at most one σ-additive extension to the σ-ring R generated by C.
(Note that s is automatically σ-finite if it is finite, e.g., complex or vector valued.)
[Outline: Let
s′,s′′:R→E
be two σ-additive extensions of s. By Problem 10, R is also generated by
C′={X∈C||sX|<∞}.
Now set
R∗={X∈R|s′X=s′′X}.
Show that R∗ satisfies properties (i)-(iii) of Theorem 3 in §3, with C replaced by C′; so R=R∗.]
Let m∗n(n=1,2,…) be outer measures in S such that
(∀X⊆S)(∀n)m∗nX≤m∗n+1X.
Set
μ∗=limn→∞m∗n.
Show that μ∗ is an outer measure in S (see Note 5).
An outer measure m∗ in a metric space (S,ρ) is said to have the Carathéodory property (CP) iff
m∗(X∪Y)≥m∗X+m∗Y
whenever ρ(X,Y)>0, where
ρ(X,Y)=inf{ρ(x,y)|x∈X,y∈Y}.
For such m∗, prove that
m∗(⋃kXk)=∑km∗Xk
if {Xk}⊆2S and
ρ(Xi,Xk)>0(i≠k).
[Hint: For finite unions, use the CP, subadditivity, and induction. Deduce that
(∀n)n∑k=1m∗Xk≤m∗∞⋃k=1Xk.
Let n→∞. Proceed.]
Let (S,C,μ) and m∗ be as in Definition 3, with ρ a metric for S. Let μn be the restriction of μ to the family Cn of all X∈C of diameter
dX≤1n.
Let m∗n be the μn-induced outer measure in S.
Prove that
(i) {m∗n}↑ as in Problem 14;
(ii) the outer measure
μ∗=limn→∞m∗n
has the CP (see Problem 15), and
μ∗≥m∗ on 2S.
[Outline: Let ρ(X,Y)>ε>0(X,Y⊆S).
If for some n,X∪Y has no basic covering from Cn, then
μ∗(X∪Y)≥m∗n(X∪Y)=∞≥μ∗X+μ∗Y,
and the CP follows. (Explain!)
Thus assume
(∀n>1ε)(∀k)(∃Bnk∈Cn)X∪Y⊆∞⋃k=1Bnk.
One can choose the Bnk so that
∞∑k=1μBnk≤m∗n(X∪Y)+ε.
(Why?) As
dBnk≤1n<ε,
some Bnk cover X only, others Y only. (Why?) Deduce that
(∀n>1ε)m∗nX+m∗nY≤∞∑k=1μnBnk≤m∗n(X∪Y)+ε.
Let ε→0 and then n→∞.
Also, m∗≤m∗n≤μ∗. (Why?)]
Continuing Problem 16, suppose that
(∀ε>0)(∀n,k)(∀B∈C)(∃Bnk∈Cn)
B⊆∞⋃k=1Bnk and μB+ε≥∞∑k=1μBnk.
Show that
m∗=limn→∞μ∗n=μ∗,
so m∗ itself has the CP.
[Hints: It suffices to prove that m∗A≥μ∗A if m∗A<∞. (Why?)
Now, given ε>0,A has a covering
{Bi}⊆c
such that
m∗A+ε≥∑μBi.
(Why?) By assumption,
(∀n)Bi⊆∞⋃k=1Bink∈Cn and μBi+ε2i≥∞∑k=1μBink.
Deduce that
m∗A+ε>∑μBi≥∞∑i=1(∞∑k=1μBink−ε2i)=∑i,kμBink−ε≥m∗nA−ε.
Let ε→0; then n→∞.]
Using Problem 17, show that the Lebesgue and Lebesgue-Stieltjes outer measures have the CP.