Equivalence of unions in probability theory

Question

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?

Answer 1

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, $$\bigcup_{j,k\geq n}\{|S_j-S_k|\geq 2\epsilon\} \subset \bigcup_{k=1}^\infty\{|S_{n+k}-S_n|\geq\epsilon\}.$$