If $X_n \to X$ in probability and $f$ is continuous, then $f(X_n) \to f(X)$ in probability without using subsequences Suppose that $\{X_n:n \in \mathbb{N}\}$, $X$ are real-valued random variables on a probability space $(\Omega, \mathcal{F}, P)$ and $f: \mathbb{R} \to \mathbb{R}$ is a continuous function. Show that if $X_n \to X$ in probability, then $f(X_n) \to f(X)$ in probability. One should use the definition of convergence in probability (i.e. for every $\epsilon > 0, \ P(|f(X_n) - f(X)| > \epsilon) \to 0$ as $n \to \infty$). The hint is to use the fact that any continuous function on a compact subset of $\mathbb{R}$ is uniformly continuous. More specifically, the goal is not to use the kind of argument that relies on subsequences of $X_n$ and $f(X_n)$.
My attempt so far:
First fix $\omega \in \Omega$ and choose arbitrary $\epsilon >0$. Next, define $A_{\omega} := \{ X_n(\omega) : |X_n(\omega) - X(\omega)| < \delta, n \in \mathbb{N} \}$. Clearly $A_{\omega}$ is bounded, so $\overline{A}_{\omega}$ is compact. By uniform continuity over $\overline{A}_{\omega}$, there exists a $\delta(\omega) > 0$ such that whenever $X_n(\omega) \in A_{\omega}$ and $|X_n(\omega) - X(\omega)| < \delta(\omega)$ we must have $|f(X_n(\omega)) - f(X(\omega))| < \epsilon$.
I was thinking to try letting $\delta = \inf \{\delta(\omega): \omega \in \Omega \}$, but this seems problematic since it could be that $\delta = 0$. Instead, I thought maybe to try using the fact that
$$ |f(X_n(\omega)) - f(X(\omega))| \geq \epsilon \implies |X_n(\omega) - X(\omega)| \geq \delta(\omega)$$
but to make an argument that holds for all $\omega \in \Omega$ I would again need to take $\delta = \inf \{\delta(\omega): \omega \in \Omega \}$, which doesn't seem like a good route to go down.
I'm guessing there might be another result in real analysis that I am missing or I need to use but I am unaware. If anyone has any suggestions or solutions I would greatly appreciate the help.
 A: Looking at this element-wise doesn't seem to be the most natural approach given the assumption about convergence in probability.
When $X_n$ and $X$ are close together, you want to exploit the continuity of $f$, while arguing that $X_n$ and $X$ being far apart is a rare event. Here is one way:
\begin{align*}
P(|f(X_n)-f(X)|>\varepsilon)&\le P(|X_n-X|>\delta)+P(|f(X_n)-f(X)|>\varepsilon,|X_n-X|\le\delta).
\end{align*}
On the second event, $|X_n-X|\le \delta$, so if $X$ took values in $[-M,M]$ for some $M>0$, then $f$ restricted to $[-2M,2M]$ would be uniformly continuous, and you could see that this second term would be zero by choosing $\delta$ appropriately.
To make this work without the boundedness assumption of $X$, let $\bar X = X\cdot\mathbf 1[|X|\le M]$ be the truncation of $X$ to level $M$. Then we can estimate the second term like this
\begin{align*}
&P(|f(X_n)-f(X)|>\varepsilon,|X_n-X|\le\delta)\\
&\qquad\le P(|f(X_n)-f(\bar X)|>\varepsilon,|X_n-\bar X|\le\delta) + P(X\ne\bar X).
\end{align*}
By taking $M$ sufficiently large, you can estimate $P(X\ne\bar X)$. What's left to do is to choose an $N$ sufficiently large so that for $n\ge N$, we guarantee the original quantity $P(|f(X_n)-f(X)|>\varepsilon)$ is less than $\varepsilon$. You have to think about what order to choose the parameters $M, N, \delta$.
