TITLE: Conditions for $\lim_{n\to\infty} f_n(X_n) = f(X)$ where $X_n, X$ is random variables
QUESTION [1 upvotes]: Let $(X_n)_{n\in\mathbb{N}}$ be a sequence of real valued random variables converging to random variable $X$ with probability one. Let $f_n:\mathbb{R}\to\mathbb{R}$ be a sequence of continuous functions converging to $f$. What conditions are required for
$$
\lim_{n\to\infty} f_n(X_n) = f(\lim_{n\to\infty} X) = f(X)
$$
to hold? Is it sufficient that $f_n$ converges pointwise to $f$?

REPLY [2 votes]: For this result you need uniformly convergence of the $f_n$.
Let's show this first:
So assume $f_n \to f$ uniformly hence it holds $$\parallel f_n - f\parallel_\infty \to 0$$ and $f$ is continuous as the uniform limit of continuous functions so it follows:
$$\begin{align*}|f_n(X_n) - f(X)| &= |f_n(X_n) - f(X_n) + f(X_n) - f(X)| \\&\le |f_n(X_n) - f(X_n)| + |f(X_n) - f(X)| \\&\le \parallel f_n - f\parallel_\infty + |f(X_n) - f(X)|\end{align*}$$
And by taking $\lim_\limits{n\to\infty}$ on both sides we get $$\lim_{n\to\infty} |f_n(X_n) - f(X)| = 0$$ what's equivalent to $$f_n(X_n) \to f(X)$$
That we need uniform convergence you can see if you consider: $$f_n(x) = -(nx-1)\cdot 1_\left[0,\frac{1}{n}\right]$$ and $$X_n \equiv \frac{1}{n}$$
Then $$f_n(x) \to f(x) = \begin{cases} 1, &x=0 \\ 0 & x\not=0\end{cases}$$ point-wise but not uniformly and $$X_n \to X \equiv 0$$ but $$\lim_{n\to\infty} f_n(X_n) = \lim_{n\to\infty} 0 = 0 \not= 1 = f(0) = f(X)$$
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