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Mar 22, 2023 at 19:44 vote accept Lau
Mar 22, 2023 at 13:28 answer added Alexandre Eremenko timeline score: 2
Mar 22, 2023 at 13:04 comment added fedja 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.
Mar 22, 2023 at 11:59 history edited Martin Sleziak CC BY-SA 4.0
minor typos
Mar 22, 2023 at 11:57 history asked Lau CC BY-SA 4.0