Convergence almost surely implies convergence in probability sup

Question

I'm trying to prove that convergence almost surely implies convergence in probability. I know this proof can be found everywhere online, but I don't find one using the fact that:

$$ X_n \rightarrow_{n+\infty}^{p.s.} X \Rightarrow \forall \epsilon > 0, \lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert \geq \epsilon) = 0 $$

My attempt:

Let's fix $\epsilon > 0$. Let $n \in \mathbb{N}$.

$$\lvert X_n - X \rvert \leq \sup_{n} \lvert X_n - X \rvert$$

Hence,

$$ (\sup_n \lvert X_n - X \rvert < \epsilon) \subset (\lvert X_n - X \rvert < \epsilon)$$

Using the fact that the probability mapping is monotonically increasing,

$$\mathbb{P}(\sup_n \lvert X_n - X \rvert < \epsilon) < \mathbb{P}(\lvert X_n - X \rvert < \epsilon)$$

yet,

$$\lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert < \epsilon) = 1$$

So,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert < \epsilon) = 1$$

Eventually,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert \geq \epsilon) = 0$$

So, $(X_n)$ converges in probability towards $X$.

I don't know if it is correct, I feel like I don't manage well the supremum.

Answer 1

Your proof is (almost) correct. Below is the slightly revised proof for mathematical rigor:

Let's fix $\epsilon > 0$. Let $n \in \mathbb{N}$.

$$\lvert X_n - X \rvert \leq \sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert$$

Hence,

$$ (\sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert < \epsilon) \color{blue}{\subseteq} (\lvert X_n - X \rvert < \epsilon)$$

Using the fact that the probability mapping is monotonically increasing,

$$\mathbb{P}(\sup_{\color{blue}{k \geq n}} \lvert X_{\color{blue}{k}} - X \rvert < \epsilon) \color{blue}{\leq} \mathbb{P}(\lvert X_n - X \rvert < \epsilon)$$

yet,

$$\lim_{n+\infty} \mathbb{P}(\sup_{k\geq n} \lvert X_k - X \rvert < \epsilon) = 1$$

So,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert < \epsilon) = 1$$

Eventually,

$$\lim_{n+\infty} \mathbb{P}(\lvert X_n - X \rvert \geq \epsilon) = 0$$

So, $(X_n)$ converges in probability towards $X$.