# Non-decreasing sequence of random variable convergence in probability implies it also converges almost surely.

The problem stated as follow:

Suppose $X_1 \leq X_2 \leq \cdots$ and $X_n \xrightarrow[]{p} X$. Show that $X_n \to X$ a.s.

I'm think about may be use the continuity of probability measure, but I don't know if that's correct.

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## 3 Answers

Since $X_n \to X$ in probability, there is a subsequence $\{X_{n_k}\}_{k=1}^\infty$ of $\{X_n\}_{n=1}^\infty$ such that $X_{n_k} \to X$ almost surely. That $X_n \to X$ almost surely now follows from the fact that if a subsequence of a monotone sequence converges, then the original sequence converges to the same limit.

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Is that also true for sequence of random variables? – BigMike Nov 11 '12 at 2:08
@BigMike: The part about monotone sequences? Apply it pointwise at each element of the probability space on which the subsequence converges. – user48944 Nov 11 '12 at 13:54

Alternate proof:

Fix $\epsilon > 0$. Since the sequence $\{ X_n \}_{n=1}^{\infty}$ is monotone and $X_ n \to X$ in probability, then the events $\{ |X_n - X| \geq \epsilon \}$ are monotone decreasing. Therefore,

\begin{align} P(\{ |X_n - X| \geq \epsilon \} \,\, \text{i.o.}) &= P(\cap_{n=1}^{\infty} \cup_{i=n}^{\infty} \{|X_i - X| \geq \epsilon \}) \\ &= P(\cap_{n=1}^{\infty} \{ |X_n - X| \geq \epsilon \}) \\ &= \lim_{n \to \infty} P(|X_n - X| \geq \epsilon) \\ &= 0 . \end{align}

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You don't need to use the subsequence argument. For each $\omega$, $\{X_n(\omega)\}$ is an increasing sequence of real numbers, and so it has a limit $Y(\omega) \in (\infty, +\infty]$. Being a pointwise limit of measurable functions, $Y$ is measurable, i.e. a random variable. It remains to show that $Y = X$ a.s. But since $X_n \to Y$ pointwise, we also have $X_n \to Y$ in probability and limits in probability are unique up to null sets, so indeed $Y=X$ a.s.

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