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On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log |B_t|$, so $M$ is a continuous local martingale.

To show that it is not a martingale, compute its expectation at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho =|W_t|$, its angle is uniform. The angular average of a logarithmic potential isFor $\rho <1$ this follows by averaging the real part of the power series for $\log (1+\rho e^{i\theta })$; for $\rho >1$ factor out $\rho$ and apply the same argument to ${\rho }^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.

The radius has density $(\rho /t)e^{-{\rho }^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and ${\int }_0^{1}r|\log r|dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yieldsSince $M_0=0$, this contradicts preservation of expectation by a martingale. Thus the logarithmic radius of planar Brownian motion is a strict local martingale. It is not a nonnegative local martingale; its negative values allow its expectation to increase.