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Let $X_n$ be a sequence of random variables converging to $X$ almost surely. $f:{\rm I\!R} \to {\rm I\!R}$ is a continuous function. Then, $f(X_n)$ converges to $f(X)$ almost surely.

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  • $\begingroup$ This should follow from the definition of continuity. $\endgroup$ – Teresa Lisbon Oct 11 '17 at 10:55
  • $\begingroup$ You are welcome. But do you understand how it follows? That is, can you write down a proof of this fact? $\endgroup$ – Teresa Lisbon Oct 12 '17 at 10:20
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I need to show that $P\{ w:f(X_n)(w) \to f(X)(w)\} = 1$

I know that $P\{ w:X_n(w) \to X(w)\} = 1$

Let $w$ be such that $X_n(w)\to X(w)$

Then because $f$ is continuous $f(X_n(w))\to f(X(w))$

so $\{ w:X_n(w) \to X(w)\}$ $\subseteq$ $\{ w:f(X_n)(w) \to f(X)(w)\}$

The smaller set has probability 1 so the bigger set must also have probability 1.

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  • $\begingroup$ @ actoh binna I think I got it right. Please let me know if there is some more arguements that need to be considered. $\endgroup$ – Hili Oct 13 '17 at 5:49
  • $\begingroup$ I am also considering a similar problem where everything is same as in this question only the convergence is in probability. $\endgroup$ – Hili Oct 13 '17 at 5:51

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