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  "body": "\n## Definition\n\nLet $(E, \\mathcal{E})$ be a measurable space and let $(P_t)_{t \\ge 0}$ be a sub-Markov transition semigroup on $E$.\nSo the $P_t$ are kernels as for [[Markov kernels|2AVNDP8BJP]] but with $P_t(x, E) \\le 1$, and $P_s P_t = P_{s+t}$.\nAssume that $(t, x) \\mapsto P_t(x, A)$ is measurable on $[0, \\infty) \\times E$ for each $A \\in \\mathcal{E}$.\nThe *potential operator* (or potential kernel) of $(P_t)$ is the kernel $U$ given by\n$$U(x, A) = \\int_0^\\infty P_t(x, A) \\, dt \\in [0, \\infty], \\qquad x \\in E, \\ A \\in \\mathcal{E}.$$\nFor measurable $f \\colon E \\to [0, \\infty]$ it acts by $U f(x) = \\int_0^\\infty P_t f(x) \\, dt$.\n\n## Remarks\n\nIf $(X_t)$ is a Markov process with semigroup $(P_t)$ and law $P^x$ when started at $x$, then $U f(x) = E^x \\int_0^\\infty f(X_t) \\, dt$.\nSo $U(x, A)$ is the expected total time that the process spends in $A$.\nFor each measurable $f \\ge 0$, the function $U f$ is [[excessive|46436KE7NT]].\nExample: for [[Brownian motion|22735FGJV6]] in $\\R^3$, the Gaussian integral $\\int_0^\\infty (2\\pi t)^{-3/2} e^{-r^2/2t} \\, dt = 1/(2\\pi r)$ gives\n$$U(x, A) = \\int_A \\frac{dy}{2\\pi \\abs{x - y}}.$$\nNon-example of finiteness: for Brownian motion in $\\R^2$, $U(x, A) = \\infty$ for every nonempty open $A$, so $U$ is not a useful kernel there; the process is recurrent.\n$U$ is the limit as $\\alpha \\downarrow 0$ of the [[resolvent|J5YNNKSB9P]] $U_\\alpha = \\int_0^\\infty e^{-\\alpha t} P_t \\, dt$.\nWhen $L$ is the [[infinitesimal generator|VPA41PWRR1]] and $U f$ is finite and in the domain of $L$, one has $-L U f = f$.\nIn discrete time the analogue is $U = \\sum_{n \\ge 0} P^n$, the Green kernel of a Markov chain.\n",
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