 Solution

By part (a), translated by $a$, starting with $B_0\sim {\nu }_{a,R}$ givesPart (c) says that the same hitting law from ${\nu }_{0,R}$ tends in total variation distance to ${\nu }_{a,r_1}$. On the other hand, every path from the circle of radius $R>r_2$ to the target disc must first hit the circle of radius $r_2$. Part (a) and the Strong Markov property show that its position there has law ${\nu }_{0,r_2}$ and thatindependently of $R$. Letting $R\to \infty$ in part (c) proveswhen $B_0\sim {\nu }_{0,r_2}$.