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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.

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I don't see any problem with this proof. For an alternate proof you can see theorem 20.5 and the discussion below it of Billingsley's Probability and measure. –  Kumara Jan 28 '13 at 7:15
2  
Assuming your hypothesis is the convergence in probability of $X_n$ to $X$, the problem with your argument to deal with the first term on the RHS is that one needs the function $g$ to be uniformly continuous. If one assumes only that $g$ is continuous, the proof is incomplete. –  Did Jan 30 '13 at 11:43

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