Bochner's theorem : Let $G$ be a locally compact Abelian group. Then for any $ \phi \in \ P(G) $ there is a unique positive Radon measure $ \ μ \in \ $ M ($ \widehat{G} $) such that $ \phi (x) = \int_{ \widehat{G} } \beta(x) ~ \mathrm{d}{μ(\beta)} $
Uniqueness: Suppose $ \int_{ \widehat{G} } \beta(x) ~ \mathrm{d}{μ(\beta)} = 0 $ for all $ x \in \ G $ with some (not necesseary positive) Radon measure $μ$. Then it holds
$$ 0=\int_{ {G} } \int_{ \widehat{G} } f(x) \beta(x) ~ \mathrm{d}{μ(\beta) d(x) }. \quad \ (1) $$ for all $ f \in { {L^{1}}(G) } $.
${ \widehat{{L^{1}}(G)} }$ is dense in $C0(\widehat{G})$ and hence this implies $ μ = 0 \quad \ (2) $. suppose that $ μ_1 $and $ μ_2 $ are two positive Radon measure that satisfied in theorem.let $μ= μ_1- μ_2 $.by (2) we have $ μ_1= μ_2 \quad \ (3)$
My question:I do not understand why $ μ = 0$? is (3) correct?
I will be grateful if you help me.