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STAT 709: Exercise
Exercise 1.12 Let ν and λ be two measures on a σ-field F on Ω such that ν(A) = λ(A) for any A ∈ C, where C ⊂ F is a collection having the property that if A and B are in C, then so is A ∩ B. Assume that there are Ai ∈ C, i = 1, 2 , ..., such that ∪Ai = Ω and ν(Ai) < ∞ for all i. Show that ν(A) = λ(A) for any A ∈ σ(C), where σ(C) is the smallest σ-field containing C. Note. Solving this problem requires knowing properties of measures (Shao, 2003, §1.1.1). The technique used in solving this exercise is called the “good sets principle”. All sets in C have property A and we want to show that all sets in σ(C) also have property A. Let G be the collection of all sets having property A (good sets). Then, all we need to show is that G is a σ-field. Solution. Define G = {A ∈ F : ν(A) = λ(A)}. Since C ⊂ G, σ(C) ⊂ G if G is a σ-field. Hence, the result follows if we can show that G is a σ-field. (i) Since both ν and λ are measures, 0 = ν(∅) = λ(∅) and, thus, the empty set ∅ ∈ G. (ii) For any B ∈ F, by the inclusion and exclusion formula,
ν
( (^) n ⋃
i=
Ai ∩ B
)
∑
1 ≤i≤n
ν(Ai ∩ B) −
∑
1 ≤i previous expression. Thus, Bc^ ∈ G whenever B ∈ G. (iii) Suppose that Bi ∈ G, i = 1, 2 , .... Note that
ν(B 1 ∪ B 2 ) = ν(B 1 ) + ν(B 1 c ∩ B 2 ) = λ(B 1 ) + λ(B 1 c ∩ B 2 ) = λ(B 1 ∪ B 2 ),
since B 1 c ∩B 2 ∈ G. Thus, B 1 ∪B 2 ∈ G. This shows that for any n, ∪ni=1Bi ∈ G. By the continuity property of measures,
ν
( (^) ∞ ⋃ i=
Bi
) = lim n→∞ ν
( (^) n ⋃ i=
Bi
) = lim n→∞ λ
( (^) n ⋃ i=
Bi
) = λ
( (^) ∞ ⋃ i=
Bi
) .
Hence, ∪Bi ∈ G.
Exercise 1.22. Let ν be a measure on a σ-field F on Ω and f and g be Borel functions with respect to F. Show that (i) if
∫ f dν exists and a ∈ R, then
∫ (af )dν exists and is equal to a
∫ f dν; (ii) if both
∫ f dν and
∫ gdν exist and
∫ f dν +
∫ ∫ gdν^ is well defined, then (f + g)dν exists and is equal to
∫ f dν +
∫ gdν. Note. For integrals in calculus, properties such as
∫ (af )dν = a
∫ ∫ f dν^ and (f + g)dν
∫ f dν +
∫ gdν are obvious. However, the proof of them are com- plicated for integrals defined on general measure spaces. As shown in this exercise, the proof often has to be broken into several steps: simple functions, nonnegative functions, and then general functions. Solution. (i) If a = 0, then
∫ (af )dν =
∫ 0 dν = 0 = a
∫ f dν. Suppose that a > 0 and f ≥ 0. By definition, there exists a sequence of nonnegative simple functions sn such that sn ≤ f and limn
∫ sndν =
∫ f dν. Then asn ≤ af and limn
∫ asndν = a limn
∫ sndν = a
∫ ∫ f dν.^ This shows (af )dν ≥ a
∫ f dν. Let b = a−^1 and consider the function h = b−^1 f. From what we have shown,
∫ f dν =
∫ (bh)dν ≥ b
∫ hdν = a−^1
∫ ∫ (af^ )dν.^ Hence (af )dν = a
∫ f dν. For a > 0 and general f , the result follows by considering af = af+ −af−. For a < 0, the result follows by considering af = |a|f− − |a|f+. (ii) Consider the case where f ≥ 0 and g ≥ 0. If both f and g are simple functions, the result is obvious. Let sn, tn, and rn be simple functions such that 0 ≤ sn ≤ f , limn
∫ sndν =
∫ f dν, 0 ≤ tn ≤ g, limn
∫ tndν =
∫ gdν, 0 ≤ rn ≤ f + g, and limn
∫ rndν =
∫ (f + g)dν. Then sn + tn is simple, 0 ≤ sn + tn ≤ f + g, and ∫ f dν +
∫ gdν = lim n
∫ sndν + lim n
∫ tndν
= limn
∫ (sn + tn)dν,
Suppose now that
∫ f−dν = ∞. Then
∫ f+dν < ∞ since
∫ ∫ f dν^ exists. Since f dν +
∫ gdν is well defined, we must have
∫ g+dν < ∞. Since (f + g)+ ≤ f+ + g+,
∫ (f + g)+dν < ∞. Thus,
∫ (f + g)−dν = ∞ and
∫ (f + g)dν = −∞. On the other hand, we also have
∫ f dν +
∫ gdν = −∞. Similarly, we can prove the case where
∫ f+dν = ∞ and
∫ f−dν < ∞.
Exercise 1.30. Let F be a cumulative distribution function on the real line R and a ∈ R. Show that ∫ [F (x + a) − F (x)]dx = a.
Solution. For a ≥ 0, ∫ [F (x + a) − F (x)]dx =
∫ ∫ I(x,x+a](y)dF (y)dx.
Since I(x,x+a](y) ≥ 0, by Fubini’s theorem, the above integral is equal to
∫ ∫ I(y−a,y](x)dxdF (y) =
∫ adF (y) = a.
The proof for the case of a < 0 is similar.
Exercise 1.31. Let F and G be two cumulative distribution functions on the real line. Show that if F and G have no common points of discontinuity in the interval [a, b], then ∫
(a,b]
G(x)dF (x) = F (b)G(b) − F (a)G(a) −
∫
(a,b]
F (x)dG(x).
Solution. Let PF and PG be the probability measures corresponding to F and G, respectively, and let P = PF × PG be the product measure (see Shao, 2003, §1.1.1). Consider the following three Borel sets in R^2 : A = {(x, y) : x ≤ y, a < y ≤ b}, B = {(x, y) : y ≤ x, a < x ≤ b}, and C = {(x, y) : a < x ≤ b, x = y}. Since F and G have no common points of discontinuity, P (C) = 0. Then,
F (b)G(b) − F (a)G(a) = P ((−∞, b]×(−∞, b]) − P ((−∞, a]×(−∞, a]) = P (A) + P (B) − P (C) = P (A) + P (B) =
∫
A
dP +
∫
B
dP
∫
(a,b]
∫
(−∞,y]
dPF dPG +
∫
(a,b]
∫
(−∞,x]
dPGdPF
∫
(a,b]
F (y)dPG +
∫
(a,b]
G(x)dPF
=
∫
(a,b]
F (y)dG(y) +
∫
(a,b]
G(x)dF (x)
∫
(a,b]
F (x)dG(x) +
∫
(a,b]
G(x)dF (x),
where the fifth equality follows from Fubini’s theorem.
