A basic doubt on probabilistic and almost sure convergence

Suppose we have a sequence of random variable $X_n$ such that $X_n$ converges in probability to $X$. Now, consider the set $\{\omega: X_n(\omega) -> X(\omega)\}$. Assume that the convergence is uniform in the previous set. Does this imply that $X_n$ converges to $X$ in almost sure sense.

I think so, but not able to prove it.

• I'm not sure I understand... You want to prove that $X_n\to X$ a.s. if $\Pr\{\sup_{k\ge n}|X_k-X|>\varepsilon\}\to0$ for every $\varepsilon>0$, right? – Cm7F7Bb Mar 6 '14 at 14:28
• @V.C.: Sorry, I did not ask for that. – aaaaaa Mar 7 '14 at 4:51

It follows from the very definition of (pointwise) convergence that

$$\{\lim_{n \to \infty} X_n = X\} = \bigcap_{n \in \mathbb{N}} \bigcup_{m \in \mathbb{N}} \left\{ \sup_{k \geq m} |X_k-X| < \frac{1}{n} \right\}.$$

Hence, by the continuity of the measure $\mathbb{P}$,

\begin{align*} \mathbb{P}(\lim_{n \to \infty} X_n = X) &= \lim_{n \to \infty} \lim_{m \to \infty} \mathbb{P} \left( \sup_{k \geq m} |X_k-X| < \frac{1}{n} \right) \\ &= 1- \lim_{n \to \infty} \underbrace{\lim_{m \to \infty} \mathbb{P} \left( \sup_{k \geq m} |X_k-X| \geq \frac{1}{n} \right)}_{0} = 1.\end{align*}

Alternative proof: Fix $\varepsilon>0$. It suffices to show that $$\mathbb{P}\left( |X_n-X|>\varepsilon \, \text{infinitley often} \right) = 0.$$

Therefore, the claim follows from the fact that

\begin{align*} \mathbb{P}\left( |X_n-X|>\varepsilon \, \text{infinitely often} \right) &= \mathbb{P} \left( \bigcap_{m \in \mathbb{N}} \bigcup_{k \geq m} \{|X_k-X| > \varepsilon\} \right) \\ &= \mathbb{P} \left( \bigcap_{m \in \mathbb{N}} \left\{ \sup_{k \geq m} |X_k-X|>\varepsilon \right\} \right) \\ &\leq \mathbb{P} \left( \sup_{k \geq m} |X_k-X|>\varepsilon \right) \\ &\to 0 \qquad m \to \infty. \end{align*}

• I could not understand where did we use uniform convergence in the second proof ? – aaaaaa Mar 7 '14 at 1:19
• @aaaaaa In order to see that $$\mathbb{P}\left( \sup_{k \leq m} |X_k-X| > \varepsilon \right) =0$$ as $m \to \infty$; right at the end of the proof. – saz Mar 7 '14 at 7:12