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Let $\mu,\nu$ be finite Borel measures on $\mathbb R$. Assume that there is an open interval $(a,b)$ on which the Laplace transforms exist and coincide:

$$ \int_{-\infty}^\infty e^{-tx}\,d\mu(x) = \int_{-\infty}^\infty e^{-tx}\,d\nu(x) <\infty $$ for all $t\in(a,b)$. Does this imply that $\mu = \nu$?

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    $\begingroup$ Yes. WLOG, $(a,b)=(-2\delta,2\delta)$. Then we have $e^{-\delta|x|}$ integrable with respect to $\mu$ and $\nu$, so we can write the LT of $\mu-\nu$ for complex arguments in the vertical strip around the imaginary axis and it will be analytic there. Being $0$ on $(-\delta,\delta)$, it has to be zero in the whole strip. But on the imaginary axis, it is just the Fourier transform, which determines a finite signed measure on $\mathbb R$ uniquely. $\endgroup$
    – fedja
    Commented Mar 22, 2023 at 13:04

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I suppose that $\mu$ and $\nu$ are real measures and "exist" means absolute convergence. Then both transforms can be extended to bounded analytic functions in the strip $\{ t+i\sigma:a<t<b\}$ and coincide in this trip by uniqueness theorem for analytic functions. But then the restrictions on the line $t=t_0$ for some $t_0\in(a,b)$ can be considered as Fourier transforms of $e^{-t_0x}d\mu(x)$ and $e^{-t_0x}\nu(x)$, therefore by the uniqueness theorem of Fourier transform, these measures coincide, and thus $\mu=\nu$.

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  • $\begingroup$ Thanks for the nice answer! Do you know of a reference for this result? $\endgroup$
    – Lau
    Commented Mar 22, 2023 at 19:45

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