Polar sets and Brownian hitting

Remarks

Proof idea: the hitting probability $h(x) = P^x(T_K < \infty)$ is harmonic off $K$ and is the equilibrium potential of $K$, up to regularization. For $d \ge 3$ it is the potential of the equilibrium measure of $K$. It vanishes identically exactly when the Newtonian capacity of $K$ is $0$. For $d = 2$ a similar argument, with logarithmic capacity and Brownian motion killed on leaving a large disc, applies. Polar sets are exactly the sets of capacity zero, which gives the statement. Example: a single point is polar for $d \ge 2$, so planar Brownian motion started away from $0$ never hits $0$. A countable union of polar sets is polar, so planar Brownian motion almost surely avoids all points with rational coordinates, except possibly its starting point at time $0$. A line segment in $\R^2$ is not polar, so Brownian motion hits it with positive probability. In dimension $1$ the picture is different: Brownian motion hits every point almost surely, and only the empty set is polar.

Going deeper

For a student

This is the probabilistic description of the exceptional sets of potential theory. Statements that hold "except on a polar set", such as the boundary behaviour of Perron solutions, become statements that hold almost surely for Brownian paths. It explains why polar sets are removable for bounded harmonic functions: Brownian motion never feels them. In the language of Hunt processes, a set is polar for a process when the process almost surely never enters it, and this definition generalizes the classical one.