Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 6 / b / Solution

The assertion is false without a boundedness or uniform integrability condition. Take the upper half-planeand $u(x,y)=y$. This function is harmonic and continuous on $\overline{D}$, with boundary value $f=0$. The second coordinate of planar Brownian motion is one-dimensional Brownian motion, so its first hitting time $\tau$ of zero is finite almost surely. NeverthelessThe stopped local martingale $u(B_{t\wedge \tau })$ is not uniformly integrable, which is exactly why optional stopping fails in the limit.