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  "body": "\n## Statement\n\nLet $f, g \\colon [0, \\infty) \\to \\C$ be [[Lebesgue integrable|DWJF7ABPR1]] on each interval $[0, T]$.\nAssume that the Laplace integrals of $f$ and $g$ converge at some real number $s_0$.\nLet $F$ and $G$ be their [[Laplace transforms|05XT5AVJBK]].\nIf $F(s) = G(s)$ for every real $s \u003e s_0$, then $f = g$ almost everywhere on $[0, \\infty)$.\nIf in addition $f$ and $g$ are continuous, then $f = g$ everywhere on $[0, \\infty)$.\n\n## Remarks\n\nBy linearity it suffices to show that $h = f - g$ vanishes almost everywhere when its transform vanishes for $s \u003e s_0$.\nLet $\\varphi(t) = \\int_0^t e^{-s_0 \\tau} h(\\tau) \\, d\\tau$; it is continuous and bounded, and integration by parts turns $H(s_0 + n) = 0$, $n = 1, 2, \\ldots$, into $\\int_0^\\infty e^{-n t} \\varphi(t) \\, dt = 0$.\nThe 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.\nBy the Weierstrass approximation theorem $\\psi = 0$, hence $\\varphi = 0$ and $h = 0$ almost everywhere.\nThe proof shows that equality at the points $s_0 + n$, $n = 1, 2, \\ldots$, is already enough.\nWithout continuity the conclusion \"almost everywhere\" cannot be improved: changing $f$ at one point does not change $F$.\n\n## Explanations\n\n### L1 human\n\nTwo functions with the same Laplace transform are essentially the same function.\nSo, when one finds a function with a given transform, one has found the only reasonable one.\n\n### L2 human\n\nThis theorem is what makes the [[inverse Laplace transform|07WBTX1MHA]] well defined, up to changes on null sets.\nIt justifies the table method: once $F$ matches a known transform, the original function is determined.\nIt does not by itself give a formula for $f$; that is the job of the [[Bromwich integral|5KNAXNHRY6]] and of [[Post's inversion formula|D6QKPBS2EN]].\n",
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