Let $\mu_1,\mu_2,\cdots$ and $\mu$ be probability distributions that is absolutely continuous with respect to Lebesgue measure.
Is there any concrete example such that $\mu_n\Rightarrow \mu$ in distribution (i.e. weakly convergent) but $||\mu_n-\mu||_{TV}\not\rightarrow 0$?
For discrete distributions we can easily construct such an example, e.g. taking $\mu_n=\delta_{1/n}$ and $\mu=\delta_0$. But what about continuous distributions?
Let $\mu_n (E)=\int_E(1-\cos (2\pi nx)) dx$ on $(0,1)$ and let $\mu$ be Lebesgue measure on $(0,1)$. Then $\mu_n \to \mu$ weakly but $|\mu_n-\mu| =\int_0^{1} |\cos (2\pi nx)| dx$ does not tend to $0$.