# almost sure convergence via subsequence arguments

Suppose I have a sequence of random variables $$\{X_n\}_{n \in \mathbf{N}}$$ such that for every subsequence there exists a further subsequence that converges almost surely to $$X$$. Can I prove that $$X_n \to X$$ almost surely?

The confusion I have is that if $$X_n$$ converges in probability to $$X$$, then I have the statement mentioned above. On the other hand, from the real analysis we know one trick to prove convergence of a sequence is via the subsequence argument for deterministic sequences.

However, convergence in probability is weaker than the almost sure convergence.

• No. A sequence of random variables converges in probability if and only if from every subsequence one can extract a further subsequence that converges a.s.; we know a.s. convergence and convergence in probability aren't the same. Commented Jan 25, 2022 at 23:10

I think the answer is no.

Consider $$\Omega = [0,1]$$ with uniform measure, $$\{X_n\} = \{Y_{n,j}: 0 \leq j \leq 2^n - 1\}$$, where $$Y_{n,j}$$ is $$1$$ if $$x \in I_{n,j} = [\frac{j}{2^n}, \frac{j + 1}{2^n}]$$ and $$0$$ otherwise. Let $$X = 0$$.

I claim that this sequence has the subsequence property you claimed. Say we have a sequence $$\{Y_{n_i, j_i}\}$$. Then we can take the subsequence with distinct $$n_i$$, so that $$\sum_{i} \mathbb{P}(Y_{n_i, j_i} \neq X) < \infty.$$ Thus by the Borel-Cantelli lemma, $$Y_{n_i, j_i}$$ converges almost surely to $$X$$. But obviously $$X_n$$ does not converge almost surely to $$X$$.

• Thanks Shengtong. I agree with your answer. One thing that I still do not understand is that in topological spaces if for every subsequence we have a further subsequence that converges to something, then the whole sequence converges to that point. What is the apparent contraction here?
– MMH
Commented Jan 25, 2022 at 21:51
• I don't have good intuition either... Maybe because you cannot install a topology on the random variables such that almost sure convergence is convergence in that topology? Commented Jan 25, 2022 at 21:56
• @Mahdi the topology you are after is that of convergence in probability, which in fact is a metrisable separable topology. (What you claimed is not true in every topological space, you need some nice separabilty properties.) Commented Jan 25, 2022 at 23:11
• @Mahdi I am afraid I don't see any good reference in my books. Donald L. Cohn's Measure Theory has Exercise 5 of section 3.2 "Let $\mu$ be a finite measure on $(X, \mathscr{A}).$ Show that (a) the formula $d(f,g) = \int \dfrac{|f-g|}{1+|f-g|} d\mu$ defines a semimetric on the colection of all real-valued $\mathscr{A}$-measurable functions on $X,$ and (b) $\lim_n d(f_n, f) = 0$ holds if and only if $\{f_n\}$ converges to $f$ in [$\mu$-]measure." Commented Jan 25, 2022 at 23:49
• @Mahdi Robert B. Ash's book "Real Analysis and Probability" in sect. 2.5, exercise 2 asks the reader to "Show that a sequence is Cauchy in measure iff it is convergent in measure." So, both exercises (from different books!) show that convergence in measure (finite measure) is a metrisable complete topology! Commented Jan 25, 2022 at 23:52