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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.

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  • $\begingroup$ We can use Stone-Weierstass theorem in order to prove that $\int_{[0,1]}f(x)\mu(dx)=0$ holds for all $f$ in the closure of the vector space generated by $x\mapsto e^{-nx}$, $n\in\mathbb N$. We thus have for all $f$ continuous that $\int_{[0,1]}f(x)d\mu^+(x)=\int_{[0,1]}f(x)d\mu^-(x)$ and we can conclude. $\endgroup$
    – Davide Giraudo
    Aug 7, 2013 at 11:52
  • $\begingroup$ @DavideGiraudo Do I understand any continuous function can be 'approximate' in some sense by the linear combination of the family $\{x \mapsto e^{-nx}\}_{n \in \mathbb{N}}$? $\endgroup$
    – zzr
    Aug 7, 2013 at 12:12
  • $\begingroup$ Yes (actually uniformly), that what I mean. $\endgroup$
    – Davide Giraudo
    Aug 7, 2013 at 12:17
  • $\begingroup$ exponentials can be uniformly approximated by polynomials but I am not very sure if the converse holds true? Say,how can $x\mapsto x^2$ be approximated by exponential family $\{x \mapsto e^{-nx}\}$? If the converse holds then it is clear that $\{x \mapsto e^{-nx}\}$ can be applied G-S process to form an CONS in $L^2$ but I am not sure yet. $\endgroup$
    – zzr
    Aug 7, 2013 at 12:31
  • $\begingroup$ @RoyHan: it is a consequence of the quoted theorem. But you can also use the classic Weierstrass approximation theorem (uniform approximation by polynomials) applied to the continuous function $f(-\log t )$ on $[1/e, 1]$. If $P$ is uniformly close to $f\circ(-\log)$ on $[1/e, 1]$, then $P(e^{-x})$ is uniformly close to $f(x)$ on $[0,1]$. $\endgroup$
    – Pietro Majer
    Aug 7, 2013 at 12:37

1 Answer 1

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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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