$X_n$ converges in distr. to $0$, $Y_n$ converges in prob. to $Y$. Show that $g(X_n,Y_n)$ converges in probability to $g(0,Y)$.

Question

Suppose that $X_n$ converges in distribution to $0$ and $Y_n$ converges in probability to $Y$. Let $g:\mathbb{R}^2 \to \mathbb{R}$ be such that $g(x,y)$ is a continuous function of $y$ for all $x$, and $g(x,y)$ is continuous at $x=0$ for all $y$. Show that $g(X_n,Y_n)$ converges in probability to $g(0,Y)$.

Unfortunately I don't have a whole lot of work to show for this one, I am stumped here. I thought perhaps Fubini's Theorem might come into play. So far, I have been trying to prove it using the straight up definition of convergence in probability but I've barely gone anywhere with that. Any help here would be much appreciated.

Answer 1

1. $X_n \to 0$ in distribution implies $X_n \to 0$ in probability. So $(X_n, Y_n) \to (0, Y)$ in probability. So could extract a subsequence $(X_{n_k}, Y_{n_k}) \to (0,Y)$ a.s.
2. Notice that $g(x,y)$ (viewed as a function of two variables) is continuous at $(0,y)$ for all $y$. So $g(X_{n_k}, Y_{n_k}) \to g(0,Y)$ a.s. This implies the result.