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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?

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What do you mean by exponentially decreasing? –  Davide Giraudo Nov 18 '12 at 14:07
    
@DavideGiraudo I've explained in my post –  Nimza Nov 18 '12 at 14:15
    
If you just have finite measures, then indeed the identity can only be continued to the right half-plane. The equality of Fourier transforms would be the same identity on the boundary of this domain, so it would not automatically hold, unless you can show that this analytic function extends continuously to the boundary. However, with your exponential decay condition you can actually extend the identity analytically to the halfplane $\mathrm{Re} z > -A$, and then the Fourier transforms are actually the same, and so the measures are, too. –  Lukas Geyer Nov 18 '12 at 14:19
    
@LukasGeyer why measures are equal if their Fourier transforms are actually the same? (It is my initial question in the post!) –  Nimza Nov 18 '12 at 14:21

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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 $\mu = \nu$.

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