I want to prove the following lemma, relating to the proof of Kolmogorov's Two Series Theorem:
Obviously, it suffices to show that $\{Y_k\}_{k \in \mathbb{N}}$ is Cauchy for almost every $\omega$. I'm not sure if my approach to this is correct, particularly on the part concerning the set inclusion (I've put this in all caps below). Please help me verify if you can, thanks!
My ideas:
Lemma #1: For every $\delta > 0$, $\lim_{m \rightarrow \infty} P(\cup_{k = m}^\infty|Y_k - Y_m| > \delta) = 0$.
Proof: Let $n \in \mathbb{N}$, $\epsilon > 0$ and choose $N$ such that $m \ge N$ implies $P(\cup_{k = m}^{m+n} |Y_k - Y_m| > \delta) < \epsilon/2$ (possible from assumption). Define $A_n \equiv \cup_{k = m}^{m+n} |Y_k - Y_m| > \delta$. For fixed $m$ we have $A_n \subseteq A_{n+1}$ so that $\forall m \ge N$, $$P\left(\bigcup_{k=m}^\infty |Y_k-Y_m| > \delta\right) = P(\cup_{n=1}^\infty A_n) = \lim_{n \rightarrow \infty}P(A_n) \leq \epsilon/2 < \epsilon \quad \quad \square$$
For the main problem (Lemma 10.3), we try to show that for each $\delta$, $$A_\delta \equiv \bigcap_{N=1}^\infty \bigcup_{m, n \ge N} \{ |Y_m - Y_n| > \delta \}$$ is a nullset, because this will imply Cauchy-ness by taking the union over $\delta \in \mathbb{Q}$ for example.
Fix $N^*$ such that $m \ge N^*$ implies $P(\cup_{k = m}^\infty |Y_k - Y_m| > \delta/2) < \epsilon$. Then (THE SECOND INCLUSION HERE IS THE PART I'M REALLY NOT SURE ABOUT): $$A_\delta \subseteq \bigcup_{n, m \ge N^*} \{|Y_n - Y_m| > \delta\} \subseteq \bigcup_{m=N^*}^\infty \{|Y_{N^*} - Y_m| > \delta/2\}$$
The second inclusion follows from the fact that, if for all $m \ge N^*$, $|Y_m - Y_{N^*}| \leq \delta/2$, then for all $n, m \ge N^*$, $|Y_n - Y_m| \leq |Y_n - Y_{N^*}| + |Y_m - Y_{N^*}| \leq \delta$ (i.e. if we are not in the rightmost set, we cannot be in the previous set in the sequence of inclusions). Thus $P(A_\delta) < \epsilon$. Since $\epsilon$ was arbitrary, we are done. $\quad \square$