\usepackage{amssymb} \newcommand{\dom}1{\operatorname{Dom}_{#1}} \newcommand{\Q}{\mathbb{Q}}
I'm reading the Book Billingsley of Probability and Measure. I am struggling with understanding some parts of the proof of Theorem 25.9:
Theorem 25.9. For every sequence ($F_n$) of distribution functions there exists a subsequence $(F_{n_k})$ and a nondecreasing, right-continuous function $F$ such that $\lim_k F_{n_{k}}(x) = F(x)$ at continuity points $x$ of $F$.
PROOF. An application of the diagonal method [A14] gives a sequence ($n_{k}$) of integers along which the limit $G(r) = \lim_k F_{n_k}(r)$ exists for every rational $r$. Define $F(x) = \inf[G(r): x < r]$. Clearly, $F$ is nondecreasing.
To each $x$ and $\varepsilon$, there is an $r$ for which $x < r$ and $G(r) < F(x) + \varepsilon$. If $x < y < r$, then $F(y) < G(r) < F(x) + \varepsilon$. Hence F is continuous from the right.
If $F$ is continuous at $x$, choose $y < x$ so that $F(x) – \varepsilon < F(y)$; now choose rational $r$ and $s$ so that $y < r < x < s$ and $G(s) < F(x) + \varepsilon$. From $F(x) – \varepsilon < G(r) < G(s) < F(x) + \varepsilon$ and $F_{n}(r) < F_{n}(x) < F_{n}(s)$ it follows that as $k$ goes to infinity $F_{n_{k}}(x) has limits superior and inferior within $\varepsilon$ of $F$
My questions are the following:
(1) On the last sentence of the second paragraph proof. How does he get the conclusion? I don't see connections between it and my definition:
My definition of right continuos at $x_{0}$ is the following:
$$F(x_{0}) = \lim_{x \downarrow x_{0}}F(x)$$ It means that for a monotonically decreasing sequence $\{x_{n}\}_{n \ in \N}$ that goes to $x$ we have that $F(x_{n})$ tends to $F(x)$.
(2) I don't understand how to get the conclusion we have by the end to see that the limit exists. Do you have a way to get it?
Rewritting the Proof
Using ideas from answers!!
An application of the diagonal method [A14] gives a sequence ($n_{k}$) of integers along which the limit $G(r) = \lim_k F_{n_k}(r)$ exists for every rational $r$. Define
$$F(x) = \inf[G(r): x < r]$$
F is nondecreasing Let $x,y \in \dom{F}$ such that $x < y$. Notice that, from the hypothesis, we have the inclusion of sets:
$$[G(r)\,:\,x < r] \subset [G(r)\,:\,y < r]$$
By infimum properties, we have that:
$$\inf[G(r)\,:\,x < r] \leq \inf[G(r)\,:\,y < r]$$
By definition of $F$ we get that:
$$F(x) \leq F(y)$$
As we wanted!
F is right continuous Let $\{x_{n}\}_{n \in \N}$ a decreasing sequence that converges to $x$. We want to verify that $F(x_{n}) \to F(x)$
Fix $x$ and take decreasing sequence $\{x_{n}\}$ so that $x_{n}\to x$. We wish to prove $F(x_{n}) \to F(x) $. In order to do so we must show that for any $\varepsilon$ there is a large enough $n_{0}$ so that $$|F(x)-F(x_{n})|\leq \varepsilon$$ For all $n\geq n_{0}$. We take $r$ as in
Beacuse $x_{n}\to x$ there exist a an $n_{0}$ such that $x_{n}-x < r-x$ for all $n\geq n_{0}$ (We are omitting the absolute value here because we know $x_{n}\geq x$). Then for any such $n$ we have $x\leq x_{n}\leq r$, so $$F(x_{n})\leq F(x)+\varepsilon$$ Because $F$ is non-decreasing we have $$F(x)\leq F(x_{n})$$ so $$|F(x_{n})-F(x)|=F(x_{n})-F(x)\leq \varepsilon$$ For every $n\geq n_{0}$.
2.
Careful there, remember you are dealing with infimums here and not minimums, at most you can take $G_{x}(r_{1})$ such that $G_{x}(r_{1})-F(x)\leq \varepsilon$
I don't exactly see how this follows from what you wrote, you have $x<r_{1}$ and $y<r_{2}$ but no way to know $r_{1}<r_{2}$.
Your approach is not entirely wrong but I suspect that on your way of fixing that last bit of your argument you'll stumble upon the much simpler fact in that if $x<y$ $$\{G(r):y<r\} \subseteq \{G(r): x<r\}$$ so $\inf \{G(r): x<r\} \leq \inf \{G(r):y<r\} $
3.
So far you have that for any $\varepsilon$ there are rationals $r \leq x \leq s$ such that $$F(x)- \varepsilon \leq G(s) \leq G(r)\leq F(x) + \varepsilon $$
Because $F_{n}(r) \leq F_{n}(x) \leq F_{n}(s) $ we have $$ F(x)-\varepsilon \leq G(r) = \lim\limits_{k \to \infty}{F_{n_{k}}(r)}\leq \liminf\limits_{k\to \infty}{F_{n_{k}}(x)} $$ And $$\limsup\limits_{k\to \infty}{F_{n_{k}}(x)} \leq \lim\limits_{k \to \infty}{F_{n_{k}}(s)} = G(s) \leq F(x)+\varepsilon $$
Because $\varepsilon$ was chosen arbitrarily we can conclude that $$F(x)\leq \liminf\limits_{k\to \infty}{F_{n_{k}}(x)} \leq \limsup\limits_{k\to \infty}{F_{n_{k}}(x)} \leq F(x) $$