could I possibly have feedback on my attempt to prove the following statement?
If $g$ is a continuous function and $X_{n}\rightarrow X$, where $X_{n}$ is a sequence of random variable, then $g\left( X_{n} \right) \rightarrow g\left(X\right)$
Attempt on Proof. $Start$
We want to prove that for any $\epsilon>0$, $P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)\rightarrow0$ as $n\rightarrow\infty$. For some $\delta>0$, by Law of Total Probability, we have $$ P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)=P\left(\left|g(X_{n})-g(X)\right|>\epsilon,\left|X_{n}-X\right|<\delta\right)+P\left(\left|g(X_{n})-g(X)\right|>\epsilon,\left|X_{n}-X\right|\geq\delta\right). $$
As $\delta\rightarrow0$, the first term on the RHS converget to $0$ by the continuity of $g$. Meanwhile, the latter term also converge to $0$ by convergence of $X_{n}$ to $X$.
Consequently, $P\left(\left|g(X_{n})-g(X)\right|>\epsilon\right)\rightarrow0$, and we have shown that the statement holds. $End$
Thanks in advance for the help.
$\endgroup$ 41 Answer
$\begingroup$BeerR's proof
The proof is correct, although justifying convergence of your first term is not completely trivial.
Pick a sequence $\delta_m \searrow 0$ and $\varepsilon >0$ fixed. Define
$$ A_m := \{\omega: |g(X_n(\omega))-g(X(\omega))| > \varepsilon, |X_n(\omega)-X(\omega)| <\delta_m \text{ for some } n\} $$
Clearly $A_{m'} \subset A_m$ for $m' > m$.
We will show that $A_m \searrow \emptyset$ as $m \to \infty$. Then by continuity of probability measures we will get $\Pr(A_m) \searrow 0$ as $m \to \infty$.
For this fix $\omega$:
By continuity of $g$ at $X(\omega)$, there exists a $\delta$ such that $|g(y) - g(X(\omega))| < \varepsilon$ for all $y$ with $|y-X(\omega)| < \delta$.
In particular, pick $M$ such that $\delta_M < \delta$, then $\omega \notin A_M$. Since $\omega$ was arbitrary it follows that:
$$ \bigcap_{m=1}^{\infty} A_m = \emptyset $$
This shows that the first term is negligible, i.e. for all $n$:
$$ \Pr(\{\omega: |g(X_n(\omega))-g(X(\omega))| > \varepsilon, |X_n(\omega)-X(\omega)| <\delta_m \}) \leq \Pr(A_m)$$
Now, in BeerR's expression, one can first take $n \to \infty$ and then $m \to \infty$ to get the wanted result.
Did's comment
If $g$ is uniformly continuous, then things simplify quite a bit:
For fixed $\varepsilon > 0$ choose $\delta >0$ such that $|g(x)-g(y)| < \varepsilon$ for all $x,y$ with $|x-y| < \delta$.
Then: $ |g(X_n(\omega))-g(X(\omega)| \geq \varepsilon$ implies $|X_n(\omega) - X(\omega)| \geq \delta$.
Therefore:
$$ \Pr(|g(X_n)-g(X)| \geq \varepsilon) \leq \Pr( |X_n - X| \geq \delta) $$
The RHS now goes to $0$ since $X_n$ converges to $X$ in probability ($\delta >0$) .
$\endgroup$ 3
