Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 201 / 5 / b

Translate the centre of the ball to zero. Choose an inner radius $\rho$ strictly smaller than the ball's radius. If the starting point already lies in the ball there is nothing to prove; otherwise write $d=|z|>\rho$ and choose $R>d$. Let $H_{\rho },H_R$ be the first hits of the two circles.

The Brownian exit time from the annulus $\rho <|z|<R$ is finite almost surely: it is no later than exit of the first coordinate from $(-R,R)$, which is finite by part (a). The function $\log |z|$ is harmonic on this annulus. The Itô formula and optional stopping theorem apply to its bounded stopped values, yieldingSolving gives the planar Brownian annulus hitting probabilityAs $R\to \infty$ this tends to one, while the event in the formula is contained in $\{H_{\rho }<\infty \}$. Thus the inner circle is hit almost surely. That circle lies strictly inside the original open ball, proving the recurrence of planar Brownian motion in the form requested:A probability-one intersection over balls with rational centres and positive rational radii makes this conclusion simultaneous for all nonempty open balls, since every such ball contains one from that countable family.