Chapter 7 Infinite product spaces

7.3. PRODUCT MEASURE SPACES 9
Theorem 7.15 Let S be a semialgebra and let µ : S ! [0, 1] be a set function. Suppose
the properties hold:
(1) Finite additivity: A, Ai 2 S ,A= S n i=1 Ai disjoint ) µ(A) =Â n i=1 µ(Ai),
(2) Countable subadditivity: A, Ai 2 S ,A= S • i=1 Ai disjoint ) µ(A) apple  • i=1 µ(Ai).
Then there exists a unique extension ¯µ : S ! [0, 1] of µ, which is a measure on S .
Proof. Note that (1) immediately implies that µ(∆) =0. The proof comes in four
steps:
Definition of ¯µ: For A 2 S , then A = S n i=1 Si for some Si 2 S . Then we let
¯µ(A) = n i=1 µ(Si). This definition does not depend on the representation of A;
if A = S m j=1 Ti for some Tj 2 S , then the fact that S is a p-system and finite
additivity of µ ensure that
n
 µ(Si) =
i=1
n ⇣ [ m ⌘
 µ Si \ Tj =
i=1 j=1
n m
 Â
i=1 j=1
µ(Si \ Tj) =
m
 µ(Tj).
j=1
Finite additivity of ¯µ: By the very definition, ¯µ is also finitely additive, because if
An is given as the disjoint union S
i Si,n, then S
n An = S
i,n Si,n and thus ¯µ( S
n An) =
Âi,n µ(Si,n) =Ân ¯µ(An). (All indices are finite.)
Countable subadditivity of ¯µ: We claim that ¯µ is countably subadditive, whenever
the disjoint union lies in S . Let Ai 2 S and A = S
k 1 Ak 2 S . Writing Ak =
S
n Sj—note the efficient way to index the Sj’s contributing to A kapplejn Ak is also in S
because S is an algebra. Since we already showed that ¯µ is finitely additive, we
have
n
¯µ(A) = ¯µ(Bn)+  ¯µ(A k)
n
 ¯µ(A k).
k=1
k=1
Letting n ! •, the opposite inequality follows.