Let X be a compact Hausdorff
space, C(X) the algebra of all continuous complex valued functions on X, and let A
be a supnorm algebra on X, that is, A is a uniformly closed algebra of continuous
complex valued functions on X that contains the constants and separates the points.
If ϕ is a complex homomorphism of A then let M(ϕ) be the set of all positive,
regular, Borel measures on X that represent ϕ. If μ is a finite, (complex), regular,
Borel measure on X then we write μ ⊥ A if ∫
f dμ = 0 for all f ∈ A. Let
ϕ be a complex homomorphism of A and m ∈ M(ϕ), then we say that m
satisfies the Riesz theorem if whenever μ is a finite, (complex), regular, Borel
measure on X and μ ⊥ A then μ_{a} ⊥ A and μ_{s} ⊥ A where μ = μ_{a} + μ_{s} is the
Lebesgue decomposition of μ with respect to m. It is quite easy to see that if
m ∈ M(ϕ) and m satisfies the Riesz theorem then for all ρ ∈ M(ϕ) we have ρ is
absolutely continuous with respect to m. We will show that this condition is also
sufficient. This is done by means of a theorem which says that if F ⊆ X is a
compact G_{δ} such that m(F) = 0 for all m ∈ M(ϕ) then there exists a sequence
f_{n} in A such that f_{n}≦ 1 on X, ϕ(f_{n}) → 1, and f_{n} → 0 uniformly on
F.
