From Conditional Statement to an Inequality

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.

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.

• Thank you! I had completely forgotten what material conditional means for sets. Jan 13 '21 at 16:13