It is quite easy to show that if $\mu$ is positive finite Borel measure on, say $[0,1]$, and for all $n \in \mathbb{N}$


$$\int\_{[0,1]} e^{-nx}\mu(dx)=0$$


holds true, then $\mu=0$. Does this still hold if $\mu$ is signed finite Borel measure?


For positive measure, by applying Holder's inequality it can be shown that $\int\_{[0,1]} e^{-rx} \mu(dx)=0$ holds for all positive rational $r$, hence by DCT holds for all $r\geq 0$. By taking differentiation under the integral sign and set $t=0$ we find $$\int\_{[0,1]}f(x)\mu(dx)=0$$ holds for all polynomial $f(\cdot)$ and hence for continuous $f(\cdot)$ by Weistrass's approximation theorem. Now Lusin's theorem for Radon measure implies the above equality also holds for measurable $f(\cdot)$, especially for indicating functions so we reach the conclusion.


But for signed measure, it seems Holder's inequality does not work out so I tried to show
$\int\_{[0,1]}e^{-nx}\mu\_+(dx)=0$ and $\int\_{[0,1]}e^{-nx}\mu\_{-}(dx)=0$ where $\mu=\mu\_+-\mu\_-$ is the Hahn decomposition. But it did not work out so well.


[Possible solution]Below is one possible solution I gave.


Let $T:\mathbb{R}\supset A \to B\subset \mathbb{R}$ be diffeomorphism,($\mathcal{C}^1$-diffeomorphism seems to be sufficient but smoothness isn't really what matters here), then
$$\int\_A f(x) \mu(dx)=\int\_B f\circ T^{-1} (y) (\mu\circ T^{-1})(dy)$$
holds for all $f$ that is $\mu$-measurable and $\mu\circ T^{-1}$ is the image measure induced naturally by $T$. Now let $T={x \mapsto e^{-nx}}$, and $A=[0,1]$, then $B=[e^{-1},1]$, and
$$\int\_{[e^{-1},1]}x^n \tilde{\mu}(dx)=0$$
where $\tilde{\mu}=\mu\circ T^{-1}$. Now the conclusion follows immediately by similar argument as stated above.



On vanishing signed measure
This is an old Theorem of M. Lerch, 1903. For more info see Theorem 6.2, Chap.II, $\S 6$ of



> 
> D.V. Widder: The Laplace Transform, Princeton University Press 1941 (or Dover 2010).
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