Math 515
Professor Lieberman
September 7, 2004
HOMEWORK #2 SOLUTIONS
Chapter 3
4. To show that n is countably additive, let hEj i be a sequence of disjoint sets. We consider
three possibilities. (A complete proof requires a lot more set theory than I want to deal
with in this course.) First, only finitely many sets are nonempty and the nonempty sets are
all finite. Then the number of elements in the union of these sets is equal to the sum of the
number of elements in each set because the sets
P are disjoint. Second, there is an infinite
set in our list. Then the union is infinite, and
nEj = ∞. Finally, ifP
there are infinitely
many nonempty sets, then the union has infinitely many elements and
nEj = ∞, again.
If E is a set and y ∈ R, then the function f : E → E + y defined by f (x) = x + y is
one-to-one and onto, so E and E + y have the same number of elements, so n is translation
invariant.
Finally, n is clearly defined for all sets of real numbers (or any other kind of set).
7. Let E be a set and y ∈ R. If hIn i is a sequence of open intervals which covers E, then hIn +yi
is a sequence of open sets which covers E +y and `(In ) = `(In +y), so m∗ E ≥ m∗ (E +y). On
the other hand, if hIn i is a sequence of open intervals which covers E + y, then hIn + (−y)i
is a sequence of open sets which covers E, so m∗ E ≤ m∗ (E + y). Combining these two
inequalities yields that m∗ is translation invariant.
Chapter 11
7. First, we show that if E ∈ B0 , then Ẽ ∈ B0 . So write E = A ∪ B with B ∈ B and A a
subset of C ∈ B such that µC = 0. Now set A0 = Ẽ ∩ (C ∪ B). Then Ẽ = A0 ∪ B̃ and
A0 ⊂ C, so Ẽ ∈ B0 .
Next, if hEn i is a sequence of sets in B0 , then there are sequences hAn i and hBn i such
that S
E n = An ∪ S
Bn , Bn ∈ B,
of some Cn ∈ B with µCn = 0. Now set
S and An is a subset
S
A = An , B = Bn , C = Cn , and E = PEn . Clearly, E = A ∪ B and, because B is a
σ-algebra, B ∈ B and C ∈ B. Finally µC ≤ µCn = 0, so µC = 0, and therefore E ∈ B0 .
To define µ0 and to show that it’s a measure, suppose E ∈ B0 can be written as E =
A1 ∪ B1 and E = A2 ∪ B2 with Bi ∈ B and Ai a subset of some Ci ∈ B for i = 1, 2. Then
B1 ⊂ B2 ∪ (B1 ∩ B̃2 ), so
µB1 ≤ µB2 + µ(B1 ∩ B̃2 ).
In addition
B1 ∩ B̃2 ⊂ E ∩ B̃2 ⊂ C2 ,
so µ(B1 ∩ B̃2 ) ≤ µC2 = 0. It follows that µB1 ≤ µB2 . By symmetry, we also have
µB2 ≤ µB1 and therefore µB1 = µB2 . Therefore, however we write E as E = A ∪ B
with B ∈ B and A a subset of C ∈ B such that µC = 0, we get the same number for
µB, so we take this number to be µ0 E. Then µ0 is defined and nonnegative on B0 and
µ0 (∅) = µ(∅) = 0. Finally, if hEn i is a sequence of disjoint measurable set, then we write
En = An ∪ Bn with Bn ∈ B and An a subset of Cn ∈ B such that µCn = 0. We also write
1
2
S
S
S
S
A = An , B = Bn , C = Cn , and E = En and we note that E = A ∪ B, B ∈ B,
A ⊂ C, and C ∈ B. Moreover, hBn i is a sequence of disjoint sets in B. Then
X
X
µ0 (E) = µ(B) =
µ(Bn ) =
µ0 (En ),
and therefore µ0 is a measure.
Finally, if µ0 E = 0 and if F ⊂ E, then we write E = A ∪ B with B ∈ B and A a subset
of C ∈ B such that µC = 0. Since µ0 E = 0, we see that µB = 0 and hence E ⊂ C 0 with
C 0 = B ∪ C 0 , and µC 0 = 0. It follows that F ⊂ C 0 , so F ∈ B0 and µ0 F = 0.
Chapter 12
1. Suppose µ∗ (E) = 0. Then, for any A ⊂ X, we have
µ∗ (A ∩ E) + µ∗ (A ∩ Ẽ) ≤ µ∗ (E) + µ∗ (A) = µ∗ (A),
so E ∈ B. Hence, if E ∈ B and µ̄(E) = 0, then µ∗ (E) = 0 and any subset F of E also
satisfies µ∗ (F ) = 0, so F ∈ B and µ̄F = 0.