Proof. Suppose first that $g$ is uniformly continuous. Given $\epsilon > 0$ there is $\delta > 0$ such that $\norm{g(X_n) - g(X)} < \epsilon$ whenever $\norm{X_n - X} < \delta$ Therefore, $$ \PP \bigl( \norm{g(X_n) - g(X)} \geq \epsilon \bigr) \leq \PP \bigl( \norm{X_n - X} \geq \delta \bigr) \to 0 $$ as $n \to \infty$

Now suppose $g$ is not necessarily uniformly continuous on $\real^k$ But it will be uniformly continuous on any compact set $\{ x \in \real^k \colon \norm{x} \leq m \}$ for $m \geq 0$ Consequently, if $X_n$ and $X$ are bounded (by $m$ , then the proof just given is applicable. Thus we attempt to reduce the general case to the case that $X_n$ and $X$ are bounded.

Let $$ f_m(x) = \begin{cases} x\,, & \norm{x} \leq m \\ mx/\norm{x} \,, & \norm{x} \geq m \end{cases} $$ Clearly, $f_m\colon \real^k \to \real^k$ is continuous; in fact, it can be verified that $f_m$ is uniformly continuous on $\real^k$ (This is geometrically obvious in the one-dimensional case.)

Set $X_n^m = f_m(X_n)$ and $X^m = f_m(X)$ so that $X_n^m$ converge to $X^m$ in probability for each $m \geq 0$

We now show that $g(X_n)$ converge to $g(X)$ in probability by a four-step estimate. Let $\epsilon > 0$ and $\delta > 0$ be given. For any $m \geq 0$ (which we will fix later), $$ \PP\bigl( \norm{g(X_n) - g(X)} \geq \delta \bigr) \leq \PP\bigl( \norm{g(X_n^m) - g(X^m) } \geq \delta \bigr) + \PP\bigl( \norm{X_n} \geq m \bigr) + \PP\bigl( \norm{X} \geq m \bigr)\,. $$

Choose $M$ such that for $m \geq M$ $$ \PP\bigl( \norm{X} \geq m \bigr) \leq \PP \bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4}\,. $$ (This is possible since $\lim_{m \to \infty} \PP \bigl( \norm{X} \geq m \bigr) = \PP \bigl( \bigcap_{m=0}^\infty \{ \norm{X} \geq m \} \bigr) = \PP(\emptyset) = 0$ )

In particular, let $m = M+1$ Since $X_n^m$ converge in probability to $X^m$ and $X_n^m$ $X^m$ are bounded, $g(X_n^m)$ converge in probability to $g(X^m)$ That means for $n$ large enough, $$ \PP\bigl( \norm{g(X_n^m) - g(X^m)} \geq \delta \bigr) < \frac{\epsilon}{4}\,. $$

Finally, since $\norm{X_n} \leq \norm{X_n - X} + \norm{X}$ and $X_n$ converge to $X$ in probability, we have $$ \PP\bigl( \norm{X_n} \geq m = M+1 \bigr) \leq \PP \bigl( \norm{X_n - X} \geq 1 \bigr) + \PP\bigl( \norm{X} \geq M \bigr) < \frac{\epsilon}{4} + \frac{\epsilon}{4} $$ for large enough $n$

Collecting the previous inequalities together, we have $$ \PP\bigl(\norm{g(X_n) - g(X)} \geq \delta \bigr) < \epsilon $$ for large enough $n$