Banach Lattices and the Dual of C(X)

6 NOTES FOR MA 110A
for all f ∈ C(X).
Proof. First, without loss of generality, we can suppose µ = 1. We
start by noting µ(f+) = µ+(f+) − µ−(f+) ≤ µ+(f+) and µ(f+) ≥
−µ−(f+) so |µ(f+)| ≤ |µ|(f+). Similarly, |µ(f−)| ≤ |µ|(f−), so
|µ(f)| ≤ |f| d|µ| (3.5)
As in von Neumann’s proof of the RN theorem, we have
|µ(f)| ≤ |f| 2 1/2 d|µ|
since d|µ| = µ = 1, so there is g ∈ L2 with
µ(f) = g(x)f(x) d|µ|(x) (3.6)
Originally, this is true for f ∈ C(X), but by use of the dominated
convergence theorem, we can define µ(f) for f any bounded Baire
function and both (3.6) and (3.5) hold.
Now let ε > 0 and let f be the characteristic function of {x | g(x) >
1 + ε} ≡ Aε. Then (3.6) says
and (3.5) says
µ(f) ≥ (1 + ε)|µ|(Aε)
|µ(f)| ≤ |µ|(Aε)
It follows that |µ|(Aε) = 0. Taking ε = 1/2 n , we see g(x) ≤ 1 for a.e.
x. Similarly, g(x) ≥ −1, so
so
|g(x)| ≤ 1 a.e. x (3.7)
From (3.6), we have for f ∈ C(X),
|µ(f)| ≤ |g| d|µ| f∞
(3.8)
1 = µ ≤ |g| d|µ| (3.9)
Since |µ(X)| = 1 and (3.7) holds, we see |g(x)| = 1 for a.e. x.
Proof of the Hahn Decomposition Theorem. Define
so
A = {x | g(x) = +1}
X \ A = {x | g(x) = −1}