Math 516
Professor Lieberman
March 7, 2005
HOMEWORK #7 SOLUTIONS
Chapter 9
46. Let A be the set of all functions of the form
n
X
F (x, y) =
gi (x)hi (y)
i=1
with gi and hi continuous. Then each F ∈ A is continuous since it’s a sum of products of
continuous functions. To check that A is an algebra, note first that it’s a linear space. In
addition, if
m
X
G(x, y) =
g̃i (x)h̃i (y),
i=1
then
F (x, y)G(x, y) =
m
X
nĝk (x)ĥk (y)
k=1
with
ĝk (x) = gi (x)g̃j (x),
ĥk (x) = hi (x)h̃j (x)
for k = n(j − 1) + i.
To see that A separates points, let (x, y) 6= (x0 , y 0 ). If x 6= x0 , let g : X → R be a
continuous function such that g(x) 6= g(x0 ) and take n = 1, g1 = g, and h1 ≡ 1. If x = x0 ,
let h : Y → R be a continuous function such that h(y) 6= h(y 0 ) and take n = 1, g1 ≡ 1, and
h1 = h.
P
Finally, A contains the constants because c = 1i=1 g1 (x)h1 (y) with g1 ≡ c and h1 ≡ 1.
Therefore the Stone-Weierstrass Theorem implies the result.
Chapter 11
25. Define measures µn on the class of Lebesgue measurable sets by µn (E) = m(E ∩ (n, ∞))
(where m is Lebesgue measure). It’s easy to check that each µn is a measure and that hµn i
is decreasing. If we set A = [0, ∞) and Ai = [i − 1, i) for i = 1, . . . , then µn (Ai ) = 0 if
n > i, so µ(Ai ) = 0 for each i. However, µn A = ∞ for all n, so µA = ∞. Since A = ∪Ai ,
it follows that
∞
X
µA = ∞ =
6 0=
µAi ,
i=1
and therefore µ isn’t a measure because it isn’t countably additive.
27. (a) Let X be the interval [0, 1]. Then A = [0, 1], B = ∅ and A0 = [0, 1), B 0 = {1} are two
distinct Hahn decompositions.
(b) Let {A, B} and {A0 , B 0 } be two Hahn decompositions. Then A ∼ A0 ⊂ A, so it’s a
positive set, and A ∼ A0 ⊂ B 0 , so it’s also a negative set. Hence A ∼ A0 is a null set. A
similar argument shows that A0 ∼ A, B ∼ B 0 , and B 0 ∼ B are all null sets and therefore
A∆A0 and B∆B 0 are null sets by Lemma 51.
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