Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 10 down vote accepted

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.

share|cite|improve this answer
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

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.

share|cite|improve this answer

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,

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.