Let $\mu$ be a finite measure on the sigma algebra $\mathcal{B}$ of the borel set (over the real line).
The Fourier-Transform of $\mu$ is defined to be the function $\hat{\mu} \; : \; \mathbb{R} \to \mathbb{C}$ such that
$\hat{\mu}(t) := \int_{\mathbb{R}}{e^{itx} d\mu(x)}$
Let $\mu,\nu$ be two finite measure on $\mathcal{B}$. I want to show that if $\hat{\mu} = \hat{\nu}$ then $\mu = \nu$.
To prove it if you wish you can use the following lemma
Let $\mu,\nu$ be two finite measure on $\mathcal{B}$ such that
$\mu(\, [a,b) \, ) = \nu( \, [a , b) \, )$ for any $a < b$
then $\mu = \nu$