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!
 A: (1) Notice that $$\sum_{n=1}^\infty \text{Var}(X_n) < \infty,$$ so it is a convergent sum. We use https://proofwiki.org/wiki/Tail_of_Convergent_Series_tends_to_Zero.
(2) Let $N < M$, consider $w_N = \sup_{n, m \geq N} |X_n - X_m|$ and $w_M = \sup_{n, m \geq M} |X_n - X_m|.$ Notice that $n,m \geq M$ implies $n,m \geq N$, so the set $n,m \geq N$ is larger than $n,m \geq M$. Consequently the supremum gets smaller. See Proving the supremum of a subset is smaller that the supremum of the set
(3) Good question. Edit: I'll be more careful here. Notice
$$ w_M > 2\epsilon \leftrightarrow \sup_{m,n \geq M } |S_m - S_n| > 2\epsilon. $$
Now do the standard "add $0$" trick and the triangle inequality to get
$$ \sup_{m,n \geq M} |S_m - S_n| = \sup_{m,n \geq M} |S_m - S_M + S_M - S_n| \leq \sup_{m,n \geq M} (|S_m - S_M| + |S_n - S_M|).$$
Here use https://proofwiki.org/wiki/Supremum_of_Sum_equals_Sum_of_Suprema to get
$$\sup_{m,n \geq M} (|S_m - S_M| + |S_n - S_M|) = \sup_{m,n \geq M} |S_m - S_M| + \sup_{m,n \geq M} |S_n - S_M| \\ = \sup_{m \geq M} |S_m - S_M| + \sup_{n \geq M} |S_n - S_M| = 2 \sup_{m \geq M} |S_m - S_M|.$$
Notice that after separating the supremums the other variable doesn't matter. In particular they are basically the same thing beyond a name, so relabel both variables to $m$ and add them together.
Using what we've derived, we have
$$ w_M \leq 2 \sup_{m \geq M} |S_m - S_M|.$$
Thus if an element (or event since we're dealing with random variables) is such that $w_M > 2\epsilon$ then $2\epsilon < w_M \leq 2 \sup_{m \geq M} |S_m - S_M|.$ Notice that if we ignore the $w_M$ this is the same as $\epsilon < \sup_{m \geq M} |S_m - S_M|$ after dividing by $2$.  Now use the monotonicity of probability (see https://en.wikipedia.org/wiki/Probability_axioms#Monotonicity) to get
$$ P(w_M > 2\epsilon) \leq P\left( \sup_{m \geq M} |S_m - S_M| > \epsilon\right).$$
(4) Again, definition of Cauchy. If $\sup_{n,m \geq M} |X_n - X_m| \rightarrow 0$ then for all $\epsilon > 0$ there exists an $M$ so that $\sup_{n,m \geq M} |X_n - X_m| < \epsilon.$ Now this says for all $n,m \geq M$ we have $|X_n - X_m| < \epsilon$ by definition of supremum.
A: *

*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 $w_M \downarrow 0$.

A: *

*We are given that $\sum_{n=1}^\infty \mathrm{Var}(X_n) < \infty$.  So this sum converges to some number, $L$.  The definition of convergence to $L$ for a series is $\forall \varepsilon > 0, \exists M$ such that $|S_M - L| < \varepsilon$.  That is, there is a tail of the sequence of partial sums with all its member as close as desired to the convergent.  But,
\begin{align*}
S_M - L &= \sum_{n = 1}^M \mathrm{Var}(X_n) - \sum_{n = 1}^\infty \mathrm{Var}(X_n)  \\
    &= \sum_{n = M+1}^\infty \mathrm{Var}(X_n)  \text{,}
\end{align*}
so the sum you ask about goes to $0$ by the definition of convergence.  The factor of $\varepsilon^{-2}$ doesn't alter this.

*There are four ways to see this:


*

*The sequence $(S_N)_{N \in \Bbb{N}}$ is a Cauchy sequence.  (It's a convergent sequence in $\Bbb{R}$.)

*For any $\varepsilon > 0, \exists M$ such that for $m,n > M$, $|S_m - L| < \varepsilon / 2$ and $|S_n - L | < \varepsilon / 2$.  Apply the triangle inequality to find
$$  |S_m - L - (S_n - L)| = |S_m - S_n| < \varepsilon  \text{.}  $$
Of course, as $\varepsilon$ decreases, the necessary $M$ increases.  Equivalently, as $M$ increases, the minimal compatible choices of $\varepsilon$ decrease to $0$.

*Two partial sums in the part of the tail that are close to $L$ are close to each other.  The further out the tail you go, the closer these must be (or the series fails to converge).

*The supremum of a subset is less than or equal to the supremum of the whole set.



*Triangle inequality.  One $\varepsilon$ from $|S_m - L|$ and one from $|S_M - L|$, hence "$2\varepsilon$".

*Definition.

