Product of probability measures $\mu_{n}\nu_{n}\implies\mu\nu$ if and only if $\mu_{n}\implies\mu$ and $\nu_{n}\implies\nu$

Question

I'm trying to prove the next:

For probability measures on the line $\mu_{n}\times \nu_{n}\implies\mu\times\nu$ if and only if $\mu_{n}\implies\mu$ and $\nu_{n}\implies\nu.$

Here $\implies$ denotes convergence on distribution.

My idea is utilizing property of $\displaystyle\lim_{x\rightarrow\infty}\mu_{n}((-\infty,x])=1$ on the limit of the product $\mu_{n}\nu_{n}$ to conclude the first direction. For the second implication, I'm not sure if it's possible to apply a kind of " product of limits " to get "limit of a product." I think this idea will work on the intersection of the points of continuity of the distribution functions, but I don't get it clear.

Any kind of help is thanked in advanced.

Answer 1

Are you familiar with characteristic functions. $\mu_n \times \nu_n$ converges in distribution to $ \mu \times \nu $ if and only if the characteristic function converge at every point and this simply says $\phi_n (t,s) \to \phi (t,s)$ where $\phi_n (t,s)$ and $ \phi (t,s)$ are the joint characteristic functions of $(\mu_n,\nu_n)$ and $\mu,\nu)$ respectively. But $\phi (t,s) = \int e^{itx} d\mu(x) \int e^{itx} d\nu(x)$ by Fubini's Theorem and a similar formula holds for $\phi_n (t,s)$. The result now is quite obvious. ( In one direction you will have to put $t=0$ and $s=0$).