Potential operator of a Markov process

Remarks

If $(X_t)$ is a Markov process with semigroup $(P_t)$ and law $P^x$ when started at $x$, then $U f(x) = E^x \int_0^\infty f(X_t) \, dt$. So $U(x, A)$ is the expected total time that the process spends in $A$. For each measurable $f \ge 0$, the function $U f$ is excessive. Example: for Brownian motion in $\R^3$, the Gaussian integral $\int_0^\infty (2\pi t)^{-3/2} e^{-r^2/2t} \, dt = 1/(2\pi r)$ gives $$U(x, A) = \int_A \frac{dy}{2\pi \abs{x - y}}.$$ Non-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. $U$ is the limit as $\alpha \downarrow 0$ of the resolvent $U_\alpha = \int_0^\infty e^{-\alpha t} P_t \, dt$. When $L$ is the infinitesimal generator and $U f$ is finite and in the domain of $L$, one has $-L U f = f$. In discrete time the analogue is $U = \sum_{n \ge 0} P^n$, the Green kernel of a Markov chain.