I'm not sure about a step in the proof for the following theorem: Let $a\in \mathbb{R},$ $\{X_n\}$ be a sequence of random variables, and $g$ be a real-valued function that is continuous at $a$. Suppose that $X_n \overset{P}{\to} a$ then $g(X_n) \overset{P}{\to} g(a).$

Proof: Fix $\varepsilon >0.$ Then since $g$ is continuous at $a$ there exists a $\delta>0$ such that if $|x-a|<\delta$ then $|g(x)-g(a)|<\varepsilon.$ Thus $$|g(x)-g(a)|\ge \varepsilon \implies |x-a|\ge \delta \tag{1}.$$ Substituting $X_n$ for $x$ in the above implication, we obtain $$\text{Pr}[|g(X_n)-g(a)|\ge \varepsilon]\le\text{Pr}[|X_n-a|\ge\delta]\tag{2}.$$ By hypothesis the last term goes to $0$ as $n \to \infty$, which completes the proof. $\blacksquare$

Where I'm confused is how we went from the implication in $(1)$ to the inequality $(2).$ I understand that the inequalities in the implication (1) are unions of events, I don't see how we get the inequality in (2) from this.


1 Answer 1


Let $X_n \ : \ \Omega \to \mathbb R$.

The implication $$ |g(x)-g(a)|\ge \varepsilon \implies |x-a|\ge \delta \tag{1}. $$ means that the set

$$ A=\big \{ \omega \in \Omega : |g\left (X_n(\omega) \right)-g(a)|\ge \varepsilon \big \} $$

is a subset of

$$ B=\big \{ \omega \in \Omega : |X_n(\omega)-a|\ge \delta \big \} $$ since $\omega \in A \Rightarrow \omega \in B$.

Thus $\mathbb P(A) \leq \mathbb P(B)$ by the monotonicity property of a probability.

  • $\begingroup$ Thank you! I had completely forgotten what material conditional means for sets. $\endgroup$ Jan 13, 2021 at 16:13

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