Proof of Kolmogorov Convergence Theorem in Durrett Ed5 I am reading the proof of Kolmogorov Convergence Theorem in Durrett Ed5 (Thm 2.5.6). I do not understand the second half of the proof. The proof is here and I have noted parts I am confused about with red arrows.


I am confused about:

*

*Why would $\epsilon^{-2}\sum_{m = M+1}^{\infty}\text{var}(X_{n})\rightarrow 0$ as $M\rightarrow\infty$?


*Why would $w_{M}$ decrease as $M$ increases?


*How did Durrett form the equality: $P(w_{M}>2\epsilon)\leq P(\sup_{m\geq M}|S_{m} - S_{M}|>\epsilon)\rightarrow 0$ as $M\rightarrow\infty$?


*Why would $w_{M}\downarrow 0$ implies $S_{n}(\omega)$ is a Cauchy sequence?
Any idea or help will be greatly apprecaited!

*

*The convergence of the series $\sum_{n=1}^\infty \text{var}(X_n) < \infty$ (assumption in the theorem) is equivalent to the tail sums vanishing ($\sum_{n = M+1}^\infty \text{var}(X_n) \to 0$ as $M \to \infty$).

*$w_M$ is a supremum over a set of numbers $\{|S_m - S_n| : m,n \ge M\}$. As $M$ increases, this set gets smaller and smaller, so the corresponding suprema get smaller and smaller too.

*If $\sup_{m \ge M} |S_m - S_M| \le \epsilon$, then $$w_M = \sup_{n,m \ge M} |S_m - S_n| \le \sup_{n,m \ge M} (|S_m - S_M| + |S_M - S_n|) \le 2 \sup_{m \ge M} |S_n - S_M| \le 2 \epsilon.$$

*Write down the definition of $S_n(\omega)$ being Cauchy, and see that it is implied by $w_M \downarrow 0$.

