Checking only for null measure that the fourier transform characterises measure

Question

Let $X$ be a separable Fréchet space and let $B(X)$ denote its Borel sigma algebra. If $\mu$ and $\nu$ are Borel measures on $X$ with $\mu'=\nu'$ then $\mu=\nu$. Here $\mu'$ is the characteristic function of $\mu$.

The proof proceeds like this:

1. Pick a cylinder set $C$ with with a base in $B(\mathbb{R}^d)$

2. Let $F\subset X^*$ be spanned by $\{f_1,\dots,f_d\}$

3. Suppose $\mu'=0$ and denote $\mu_F$ for the image measure under $P_F(x)=(f_1(x),\dots,f_d(x))$

4. Shows $\mu'_F=0$ using image measure and adjoint map
5. $\mu_F=0$ implies $\mu$ restricted to $E(X,F)$ is 0. $E(X)$ is the sigma algebra generated by the cylindrical sets in $X$

I don't see why showing this for the zero measure is enough and why $E(X,F)$ is smaller than $B(X)$

Answer 1

Measures here are not necessarily positive measures. $\mu -\nu$ is a measure whose charateristic function is $0$ and the arguement shows that $\mu -\nu=0$ so $\mu =\nu$.