Product of two sequences of random variables converging in distribution The following is an exercise in Prof Tao's lecture notes on probability theory. 
We assume that a sequence of random variables $\xi_n \rightarrow \xi$ in distribution and also $\nu_n \rightarrow \nu$ in distribution as well. 
(i) If $\nu$ equals a constant almost surely, then prove that the product $\xi_n \nu_n$ converges to $\xi \nu$ in distribution. 
(ii) Find a counterexample to (i) in the case $\nu$ does not equal a constant a.s. 
Any help, hints will be greatly appreciated. 
 A: (i) We have to show that $\xi_n(\nu_n-\nu)$ goes to $0$ in probability. Take $\varepsilon\gt 0$; then for each positive $R$, 
$$\mu\{|\xi_n(\nu_n-\nu)|\gt\varepsilon\}\leqslant \mu\{|\xi_n|\geqslant R\}+\mu\{|\nu_n-\nu|\geqslant \varepsilon/R\}.$$
We then have by portmanteau theorem that for each $R$, 
$$\limsup_{n\to\infty}\mu\{|\xi_n(\nu_n-\nu)|\gt\varepsilon\}\leqslant\mu\{|\xi|\geqslant R\},$$
and we conclude, as $R$ was arbitrary. 
(ii) Take $\xi_n$ and $\nu_n$ discrete random variable with disjoint support and which converge to a continuous distribution (for example, we can use Riemann sums). 
A: It is worth noting that there can't be any examples for (ii) with any type of convergence:


*

*$v_n$ does not converge to a constant almost surely by task description

*if $v_n$ was to converge in probability to a constant, then, by Slutsky's theorem, the product $\xi_n v_n$ also converged in distribution

*if $v_n$ was to converge in distribution to a constant, then it also converged in probability. Continue with the second point from here. 

