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Mathematics LibreTexts

7.4.E: Problems on Set Functions

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Exercise 7.4.E.1

Prove Theorem 2 in detail for semirings.
[Hint: We know that
XnXn1=mni=1Yni (disjoint)
for some YniC, so
¯s(XnXn1)=mni=1sYni,
with ¯s as in Theorem 1.]

Exercise 7.4.E.2

Let s be additive on M, a ring. Prove that s is also σ-additive provided s is either
(i) left continuous, or
(ii) finite on M and right-continuous at ; i.e.,
limnsXn=0
when Xn (XnM).
[Hint: Let
A=nAn (disjoint),A,AnM.
Set
Xn=nk=1Ak,Yn=AXn.
Verify that Xn,YnM,XnA,Yn.
In case (i),
sA=limsXn=k=1sAk.
(Why?)
For (ii), use the Yn.]

Exercise 7.4.E.3

Let
M={all intervals in the rational field RE1}.
Let
sX=ba
if a,b are the endpoints of XM (a,bR,ab). Prove that
(i) M is a semiring;
(ii) s is continuous;
(iii) s is additive but not σ-additive; thus Problem 2 fails for semirings.
[Hint: R is countable. Thus each XM is a countable union of singletons {x}=[x,x]; hence sX=0 if s were σ-additive.]

Exercise 7.4.E.3

Let N= {naturals}. Let
M={all finite subsets of N and their complements in N}.
If XM, let sX=0 if X is finite, and sX= otherwise. Show that
(i) M is a set field;
(ii) s is right continuous and additive, but not σ-additive.
Thus Problem 2 (ii) fails if s is not finite.

Exercise 7.4.E.4

Let
C={finite and infinite intervals in E1}.
If a,b are the endpoints of an interval X (a,bE,a<b), set
sX={ba,a<b,0,a=b.
Show that s is σ-additive on C, a semiring.
Let
Xn=(n,);
so sXn=n= and Xn. (Verify!) Yet
limsXn=s.
Does this contradict Theorem 2?

Exercise 7.4.E.5

Fill in the missing proof details in Theorem 1.

Exercise 7.4.E.6

Let s be additive on M. Prove the following.
(i) If M is a ring or semiring, so is
N={XM||sX|<}
if N.
(ii) If M is generated by a set family C, with |s|< on C, then |s|< on M.
[Hint: Use Problem 16 in §3.]

Exercise 7.4.E.7

(Lebesgue-Stieltjes set functions.) Let α and sα be as in Example (d). Prove the following.
(i) sα0 on C iff α on E1 (see Theorem 2 in Chapter 4, §5).
(ii) sα{p}=sα[p,p]=0 iff α is continuous at p.
(iii) sα is additive.
[Hint: If
A=ni=1Ai (disjoint),
the intervals Ai1,Ai must be adjacent. For two such intervals, consider all cases like
(a,b](b,c),[a,b)[b,c], etc.
Then use induction on n.]
(iv) If α is right continuous at a and b, then
sα(a,b]=α(b)α(b).
If α is continuous at a and b, then
sα[a,b]=sα(a,b]=sα[a,b)=sα(a,b).
(v) If α on E1, then sα satisfies Lemma 1 and Corollary 2 in §1 (same proof), as well as Lemma 1, Theorem 1, Corollaries 1-4, and Note 3 in §2 (everything except Corollaries 5 and 6).
[Hint: Use (i) and (iii). For Lemma 1 in §2, take first a half-open B=(a,b]; use the definition of a right-side limit along with Theorems 1 and 2 in Chapter 4, §5, to prove
(ε>0)(c>b)0α(c)α(b+)<ε;
then set C=(a,c). Similarly for B=[a,b), etc. and for the closed interval AB.]
(vi) If α(x)=x then sα=v, the volume (or length) function in E1.

Exercise 7.4.E.8

Construct LS set functions (Example (d)), with α (see Problem 7(v)), so that
(i) sα[0,1]sα[1,2];
(ii) sαE1=1 (after extending sα to Cσsetsin\(E1);
(ii') sαE1=c for a fixed c(0,);
(iii) sα{0}=1 and sα[0,1]>sα(0,1].
Describe sα if α(x)=[x] (the integral part of x).
[Hint: See Figure 16 in Chapter 4, §1.]

Exercise 7.4.E.9

For an arbitrary α:E1E1, define σα:CE1 by
σα[a,b]=σα(a,b]=σσ[a,b)=σα(a,b)=α(b)α(a)
(the original Stieltjes method). Prove that σα is additive but not σ-additive unless α is continuous (for Theorem 2 fails).


7.4.E: Problems on Set Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.

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