I am trying to understand the derivation of a proof of a theorem regarding convergence in probability of continuous functions.
The theorem states: "Suppose $X_n \overset{P}{\rightarrow} a$ and the real function g is continuous in a, then $g(x_n) \overset{P}{\rightarrow} g(a)$".
Proof:
"Since g is cont. $\Rightarrow \exists \delta \gt 0$ s.th. if $|x-a| \lt \delta \Rightarrow |g(x) – g(a)| \lt \varepsilon \Leftrightarrow |g(x) – g(a)| \geq \varepsilon \Rightarrow |x-a| \geq \delta $. Substituting $X_n$ for x: $P(|g(X_n) -g(a)| \geq \varepsilon) \leq P(|X_n -a| \geq \delta)$"
Assuming the above state is true, $\lim_{n\rightarrow \infty} P(|X_n -a| \geq \delta) = 0$, hence $P(|g(X_n) -g(a)| \geq \varepsilon) = 0$ or $g(x_n) \overset{P}{\rightarrow} g(a)$
However, why is $P(|g(X_n) -g(a)| \geq \varepsilon) \leq P(|X_n -a| \geq \delta)$?
Best Answer
\begin{align}P(\lvert g(X_n)-g(a)\rvert\ge \varepsilon)&=P\{\omega\in\Omega\,:\, \lvert g(X_n(\omega))-g(a)\rvert\ge \varepsilon\}=\\&=P(X_n^{-1}[\{x\in\Bbb R\,:\, \lvert g(x)-g(a)\rvert\ge \varepsilon\}])\\ P(\lvert X_n-a\rvert\ge \delta)&=P\{\omega\in\Omega\,:\, \lvert X_n(\omega)-a\rvert\ge \delta\}=\\&=P(X_n^{-1}[\{x\in\Bbb R\,:\, \lvert x-a\rvert\ge \delta\}])\end{align}
The hypothesis already establishes that $$\{x\in\Bbb R\,:\, \lvert x-a\rvert\ge\delta\}\supseteq \{x\in\Bbb R\,:\,\lvert g(x)-g(a)\rvert\ge\varepsilon\},$$ hence the inequality.