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 $f$?

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    $\begingroup$ Continuity of $f$ is sufficient. $\endgroup$
    – Sarastro
    Commented Nov 21, 2017 at 13:03
  • $\begingroup$ Thank you Sarastro. You have answered my original question. How about my new/modified question where we have a sequence of converging functions $f_n$? $\endgroup$ Commented Nov 21, 2017 at 14:02

1 Answer 1


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 $$\lim_{n\to\infty} f_n(X_n) = \lim_{n\to\infty} 0 = 0 \not= 1 = f(0) = f(X)$$

  • $\begingroup$ Thank you very much Gono. I agree that uniform convergence is sufficient but I don't think it is necessary. I think that as long as $f$ is continuous then pointwise convergence of $f_n$ is enough. From your second inequality, if $f$ is continuous and $f_n$ converges to it pointwise, then $|f_n(X_n) - f(X_n)| \to 0$ with probability one. No? $\endgroup$ Commented Nov 22, 2017 at 3:24
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    $\begingroup$ @iwanttolearn: Pointwise convergence is definitely NOT enough. Consider the sequence of functions $f_n(x)=\max\{0,1-|nx-1|\}$. This converges pointwise to $0$ which is continuous, however $f_n(1/n)=1$ even though $1/n \to 0$. In fact, uniform convergence is both necessary and sufficient. $\endgroup$
    – shalop
    Commented Nov 22, 2017 at 6:58
  • $\begingroup$ @Shalop: I see. Thank you. $\endgroup$ Commented Nov 22, 2017 at 9:03
  • $\begingroup$ I think compact convergence (uniform convergence on all compact sets) is sufficient. We know that on any sample path $w$ from the set of non-zero probability paths, $X_n(w)$ converges to $X(w)$ and so the set $(X_n(w))_{n\in\mathbb{N}}$ is bounded. Therefore $|f_n(X_n(w)) - f(X_n(w))| \to 0$. Thoughts? $\endgroup$ Commented Nov 25, 2017 at 7:05
  • $\begingroup$ If you do the proof $\omega$-wise this should work. But your statement "on any sample path $\omega$ from the set of non-zero probability paths" is not very useful… there are sequences of r.v. where each path has zero probability… but actually this doesn't matter if you do your proof $\omega$-wise on the set where the almost sure convergence $X_n \to X$ holds. $\endgroup$
    – Gono
    Commented Nov 25, 2017 at 19:54

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