I am working with Ash's Probability and Measure Theory, Second Edition, specifically on theorem 6.2.1 (some convergence criteria for a random variable sequence).


We are given a sequence $(X\_i)\_{i \ge 1}$of random variables, and:


1. $\sum\_{i=1}^\infty \text{Var} X\_i$ converges.
2. Wlog, $\forall i.\:E(X\_i)=0$, (so assumption 1 becomes $\sum\_{i=1}^\infty X\_i^2$ converges).
3. $S\_n=\sum\_{i=1}^n X\_i$.
4. $\mu$ is the probability measure.


The proof says at a certain point that


$$\forall \epsilon>0.\:\mu\Bigl[\bigcup\_{k=1}^\infty\{|S\_{m+k}-S\_m|\geq\epsilon\}\Bigr]\to0\text{ as }m\to \infty$$


implies:


$$\forall \epsilon>0.\:\mu\Bigl[\bigcup\_{j,k\geq n}\{|S\_j-S\_k|\geq\epsilon\}\Bigr]\to0\text{ as }n\to \infty$$


I have been butting my head against exactly how to prove this for ages. Does anyone have a proof of this?



Equivalence of unions in probability theory
Let $\epsilon>0$ and $n \ge 1$. Then
$$\bigcap\_{k=1}^\infty\{|S\_{n+k}-S\_n|<\epsilon = \bigcap\_{j \geq n}\{|S\_j-S\_n|<\epsilon\} \subset \bigcap\_{j,k\geq n}\{|S\_j-S\_k|<2\epsilon\} .$$
Hence, taking complements,
$$\bigcup\_{j,k\geq n}\{|S\_j-S\_k|\geq 2\epsilon\} \subset \bigcup\_{k=1}^\infty\{|S\_{n+k}-S\_n|\geq\epsilon\}.$$


