# If $X_n$ converges to $X$ in probability, then for $f$ continuous, then $f(X_n)$ converges in probability to $f(X)$

The following is an exercise in the book Measure Theory and Probability, by Athreya and Lahire.

Let $$X_n$$ converge to $$X$$ in probability. If $$f$$ is continuous, then $$f(X_n)$$ converges in probability to $$f(X)$$

This seems like a simple exercise, but I haven’t been able to solve it. This is what I have tried so for.

Assuming that $$f$$ is uniformly continuous, then for $$\epsilon > 0 \ \exists \delta >0 : \mid X_n(w) - X(w) \mid < \delta \implies \mid f(X_n(w)) - f(X(w)) \mid < \epsilon$$

Hence, we know that $$P(\mid X_n - X\mid < \delta ) \leq P(\mid f(X_n) - f(X) \mid < \epsilon )$$, and since $$X_n \rightarrow_p X$$, then $$1 = \lim_{n \to \infty } P(\mid X_n - X\mid < \delta ) \leq \lim_{n\to \infty}P(\mid f(X_n) - f(X) \mid < \epsilon )$$

Now, is the above solution correct? And how does one prove if $$f$$ is not uniformly continuous?

• You know that $X_n \to X$ in probability iff for every subsequence $(n_k)$ there exists sub-subsequence $(n_{k_m})$ such that $X_{n_{k_m}} \to X$ almost surely? It is the easiest way of proving your statement – Dominik Kutek Aug 4 '20 at 12:57
• I haven’t been introduced to this theorem. Do you have a reference to it’s proof? – Davi Barreira Aug 4 '20 at 13:12
• I've put an answer with the short proof. I thought a little bit about your approach. It is valid for uniformly continuous functions, indeed. However I don't see how to prove it for $f$ only continuous in that way. If you know anything about convergence in distribution and tightness, then you could for any $\varepsilon > 0$ find such $M$ that $\mathbb P(|X_n| > M) < \varepsilon, \mathbb P(|X| > M) < \varepsilon$ for every $n \in \mathbb N$. On $[-M,M]$ continuous function is uniformly continuous, so you can bound it there as you did above and on the rest there is no problem due to tightness – Dominik Kutek Aug 4 '20 at 13:40

Theorem: Let $$(X_n)$$ be a sequence of random variables. Then $$X_n \to X$$ in probability, iff for every subsequence $$(n_k)$$ there exists sub-subsequence $$(n_{k_m})$$ such that $$X_{n_{k_m}} \to X$$ almost surely.

Firstly lemma: If $$X_n \to X$$ in probability, then there exists subsequence $$(n_k)$$ such that $$X_{n_k} \to X$$ almost surely.

Proof of lemma: By convergence in probability, we have a subsequence $$(n_k)$$ such that $$\mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) \le \frac{1}{k^2}$$. Hence $$\sum_{k=1}^\infty \mathbb P(|X_{n_k} - X| \ge \frac{1}{k^2}) < \infty$$ so by borel cantelli, almost surely we have $$|X_{n_k} - X| < \frac{1}{k^2}$$ starting from some $$k>K$$, so $$X_{n_k} \to X$$ almost surely.

Proof of theorem.

By Lemma we have => direction (since for any subsequence $$(n_k)$$ we have $$X_{n_k} \to X$$ in probability, too).

"<=" Assume contrary, that there $$X_n \not \to X$$ in probability. Hence by definition, there exists $$\varepsilon >0, \delta >0$$ and subsequence $$(n_k)$$ such that $$\mathbb P(|X_{n_k} - X| > \varepsilon) > \delta$$ for every $$k \in \mathbb N$$. But from that subsequence $$(n_k)$$ we cannot choose any sub-subsequence $$(n_{k_m})$$ such that $$X_{n_{k_m}} \to X$$ almost surely (because $$X_{n_{k_m}} \not \to X$$ in probability and this is necessary condition).

Having theorem we can proceed as follows in your question:

Let $$f$$ be continuous and $$X_n \to X$$ in probability. We want to show that $$Y_n := f(X_n) \to f(X) =: Y$$ in probability. Take any subsequence $$(n_k)$$. We know that there exists sub-subsequence $$(n_{k_m})$$ such that $$X_{n_{k_m}} \to X$$ almost surely. Hence $$Y_{n_{k_m}} \to Y$$ almost surely (cause continuous functions of pointwise convergent sequences are pointwise convergent). Since $$(n_k)$$ was arbitrary, by theorem we know that $$Y_n \to Y$$ in probability.