# $X_n$ converges in probability to $X$ is equivalent to $\inf \{ \epsilon>0: P(|X_n-X|>\epsilon)<\epsilon \} \rightarrow 0$

I am stuck with a problem:

$$X_n$$ converges in probability to $$X$$ is equivalent to $$\inf \{ \epsilon>0: P(|X_n-X|>\epsilon)<\epsilon \} \rightarrow 0$$

See that $$X_n$$ converges to $$X$$ in probability $$\iff$$ $$\exists N$$ such that $$\forall n \geq N$$, $$P[|X_n-X| > \epsilon]<\epsilon \iff \lim_{n \rightarrow \infty} \inf \{ \epsilon: P(|X_n-X|>\epsilon)<\epsilon \}=0$$

I am not sure whether I can write the last step. If I can then it's done.

Can anyone help?

Let $$X_n \to X$$ in probability. Let $$\eta >0$$. Then there exists $$n_0$$ such that $$P(|X_n-X| >\eta) <\eta$$ for $$n \geq n_0$$. Hence $$\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \leq \eta$$ for $$n \geq n_0$$. This proves that $$\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \to 0$$ as $$n \to \infty$$.
Conversely suppose $$\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} \to 0$$ as $$n \to \infty$$. Let $$\eta >0$$. Then $$\inf \{\epsilon >0: P(|X_n-X| >\epsilon) <\epsilon\} <\eta$$ for $$n$$ sufficiently large, say $$n \geq n_0$$. This implies that there exists $$\epsilon \in (0, \eta)$$ such that $$P(|X_n-X| >\epsilon) <\epsilon\}$$. But then $$P(|X_n-X| >\eta) <\eta\}$$ for $$n \geq n_0$$. [Verify this!]. This proves that $$X_n \to X$$ in probability.