 Solution

For a bounded measurable $f$, the dual estimate for total variation distance givesThe right-hand side tends to zero, so in particular the integrals converge for every bounded continuous $f$. Thus weak convergence of random variables follows.

The converse fails. On $\mathbb{R}$, let ${\mu }_n={\delta }_{1/n}$ and $\mu ={\delta }_0$. Continuity gives $f(1/n)\to f(0)$, so ${\mu }_n$ converges weakly to $\mu$. However, for $A=\{0\}$,for every $n$, so there is no convergence in total variation distance.