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 $\mu = \nu$?



Injectivity of two sided Laplace transform
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 $\mu=\nu$.


