Lerch's uniqueness theorem

Remarks

By linearity it suffices to show that $h = f - g$ vanishes almost everywhere when its transform vanishes for $s > s_0$. Let $\varphi(t) = \int_0^t e^{-s_0 \tau} h(\tau) \, d\tau$; it is continuous and bounded, and integration by parts turns $H(s_0 + n) = 0$, $n = 1, 2, \ldots$, into $\int_0^\infty e^{-n t} \varphi(t) \, dt = 0$. The substitution $x = e^{-t}$ turns these into $\int_0^1 x^{n-1} \psi(x) \, dx = 0$ for a continuous function $\psi$, so all moments of $\psi$ vanish. By the Weierstrass approximation theorem $\psi = 0$, hence $\varphi = 0$ and $h = 0$ almost everywhere. The proof shows that equality at the points $s_0 + n$, $n = 1, 2, \ldots$, is already enough. Without continuity the conclusion "almost everywhere" cannot be improved: changing $f$ at one point does not change $F$.

Going deeper

For a student

This theorem is what makes the inverse Laplace transform well defined, up to changes on null sets. It justifies the table method: once $F$ matches a known transform, the original function is determined. It does not by itself give a formula for $f$; that is the job of the Bromwich integral and of Post's inversion formula.