(PTE) 3.2.2. Vague convergence and uniform tightness
Our next interest is in whether a sequence of distribution functions converges weakly. To be more specific, subsequential convergence of distribution functions are is the topic of this subsection. Helly’s selection theorem shows there always exists a vaguely convergent subsequence. Uniform tightness of a sequence strengthen this result to be weakly convergent.
Helly’s selection theorem
Notice that unlike weak convergence, vague convergence does not guarantee that the limiting function $F$ is a distribution function.
expand proof
(Step 1. Diagonal argument) Let $q_1,q_2,\cdots \in \mathbb{Q}$ be enumeration of rationals. For $k=1,$ $\{F_1(q_1),F_2(q_1),\cdots\}$ is a bounded real sequence so has a limit point. Pick a subsequence $(n_{1k})$ such that $F_{1k}(q_1) \to G(q_1)$ for a limit point $G(q_1).$ For $k=2,$ we do the similar work. $\{F_{n_{11}}(q_2), F_{n_{12}}(q_2),\cdots\}$ is a bounded real sequence so has a limit point $G(q_2).$ Picka further subsequence $(n_{2k})\subset(n_{1k})$ such that $F_{2k}(q_2) \to G(q_2).$ Repeat this process for all consecutive $k\in\mathbb{N}$ and let $m_k = n_{kk}$ so that $F_{mk} \to G(q)$ for all $q \in \mathbb{Q}.$ Then by construction $G$ is nondecreasing. Let $F(x) = \inf\{G(q):~ q\in\mathbb{Q}, q> x\}.$
(Step 2. $F \leftarrow_{v.} F_{m_k}$ is rightcontinuous)
First we prove that $F$ is rightcontinuous. $$ \begin{align} \lim_{x_n\downarrow x} F(x_n) &= \lim_{x_n\downarrow x} \inf\{G(q):~ q\in\mathbb{Q}, q>x_n\} \\ &= \inf\{G(q):~ q\in\mathbb{Q}, q>x_n \text{ for some } n\} \\ &= \inf\{G(q):~ q\in\mathbb{Q}, q>x\} = F(x). \end{align} $$ Next we show that $F_{m_k} \to_{v.} F.$ Given $x$ a continuity point of $F,$ pick $r_1,r_2,s\in\mathbb{Q}$ so that $r_1<r_2<x<s$ and $F(x)\epsilon<F(r_1)\le F(r_2) \le F(x) \le F(s) < F(x)+\epsilon.$ Then $$ F_{m_k}(r_2) \to G(r_2) \ge F(r_1) = \inf\{G(q):~ q\in\mathbb{Q}, q>r_1\}, \\ F_{m_k}(s) \to G(s) \le F(s) = \inf\{G(q):~ q\in\mathbb{Q}, q>s\}. $$ Thus for all large $k$, $$ F(x)\epsilon < F_{m_k}(x) < F(x)+\epsilon. $$ so $F_{m_k} \to_{v.} F.$
Tightness theorem
While Helly’s thereom provides valuable result, vague convergence is not enough. We want additional condition that allows us to ensure weak convergence. Uniform tightness^{1} is the one.
Uniform tightness implies the measure of any set can be approximated from below by some compact set. This condition leads to a stronger result.
expand proof
($\Leftarrow$) We need to show that for $F_{n_k} \to_{v.} F,$ $F$ is a distribution function. It suffices to show $\lim_{x \to \infty} (1F(x)+F(x))=0.$ Given $\epsilon>0,$ there exists $M>0$ such that $\limsup_n (1F_n(M)+F_n(M)) \le \epsilon.$ Let $r<M$ and $s>M$ be continuity points of $F.$ $$ \begin{align} 1F(s)+F(r) &= \lim_k (1F_{n_k}(s)+F_{n_k}(r)) \\ &\le \limsup_n (1F_n(M)+F_n(M)) \le \epsilon. \end{align} $$ Thus $$ \limsup\limits_{x\to\infty} (1F(x)+F(x)) \le \epsilon. $$ ($\Rightarrow$) Suppose $(F_n)$ is not tight and show that $F$ is not a distribution function in that case. There exists $\epsilon>0$ such that $\limsup_n(1F_n(M)+F_n(M))>\epsilon$ for all $M.$ Let $r<0<s$ be continuity points of $F,$ then $$ \begin{align} 1F(s)+F(r) &= \lim_j (1F_{n(k_j)}(s)+F_{n(k_j)}(r)) \\ &\ge \liminf\limits_j (1F_{n(k_j)}(k_j)+F_{n(k_j)}(k_j)) \ge \epsilon. \end{align} $$ As $s\to\infty$ and $r\to\infty,$ $\lim_{x\to\infty} (1F(x)+F(x))>0$ so $F$ is not a distribution function.
There is a criterion for checking whether a sequence is uniformly tight.
Acknowledgement
This post series is based on the textbook Probability: Theory and Examples, 5th edition (Durrett, 2019) and the lecture at Seoul National University, Republic of Korea (instructor: Prof. Johan Lim).

Many textbooks including Durrett and Billingsley use the term tightness to describe this condition. However I want to separate the terms for clarity: uniform tightness for a collection of measures, and tightness for a measure. ↩