How similar are the measures if their Fourier transform coincide? Let $\mu$ and $\nu$ be finite Borel nonnegative measures exponentially decreasing at infinity, i.e. there exists $A > 0$ such that
$$
   \int\limits_{0}^{\infty} e^{Ax} \mu(dx) < \infty, \;\;\; \int\limits_{0}^{\infty} e^{Ax} \nu(dx) < \infty
$$
and let supports $\mathrm{supp}(\mu)$ and $\mathrm{supp}(\nu)$ lie in $\mathbb{R}_{+} = [0,\infty)$ and 
$$
   \int\limits_{0}^{\infty} e^{-p x}\mu(dx) = \int\limits_{0}^{\infty} e^{-px} \nu(dx), \;\;\; p \geq 0
$$
Functions on the LHS and on the RHS of this equation may be continued analitically to $\mathbb{R}_{+} + i\mathbb{R}$. Hence Fourier transforms of $\mu$ and $\nu$ are well defined and coincide. But what can we say in this case about $\mu$ and $\nu$? How similar they are? Are they equal in general?
 A: The conditions imply that the identity
$$
\int_0^\infty e^{-px} \, d\mu(x) = \int_0^\infty e^{-px} \, d\nu(x)
$$
holds for all complex $p$ with $\mathrm{Re}\, p>-A$, so adding up the two identities for purely imaginary $p$ with positive and negative imaginary part we indeed get equality of the Fourier transforms
$$
\int_{-\infty}^\infty e^{-itx}\, d\mu(x) = \int_{-\infty}^\infty e^{-itx} \, d\nu(x)
$$
for all $t \in \mathbb{R}$. Now to show that the measures are actually the same is a somewhat technical standard proof, found in many textbooks. First you use the Weierstrass approximation theorem to show that every continuous periodic function can be uniformly approximated by trigonometric polynomials, so $\mu$- and $\nu$-integrals of continuous periodic functions are the same. Then you can approximate compactly supported continuous continuous functions by periodic functions with periods $x_n \to \infty$, which implies that $\mu$- and $\nu$-integrals agree on compactly supported continuous functions. Lastly you can approximate characteristic functions of intervals by compactly supported continuous functions and you get that $\mu$ and $\nu$ agree on intervals, which shows $\mu = \nu$. 
