Conditions for weak convergence and generalized distribution functions I am having some trouble proving Corollary 6.3.2 in Borovkov's Probability Theory (for reference, this material is on pages 147 to 149 in the book). For convenience, I provide some definitions and related theorems. Skip to the end for a TLDR.

Definition: We say a function $G$ is a generalized distribution function if it satisfies monotonicity and right-continuity. We denote the class of generalized distribution functions to be $\mathcal{G}$ and the class of distribution functions to be $\mathcal{F}$. Of course, $\mathcal{F}\subseteq \mathcal{G}$ and the only difference is that distribution functions have $\lim_{x\to\infty}F(x) = 1$ and $\lim_{x\to-\infty}F(x) = 0$.
Definition: We say that a sequence of generalized distribution functions $\{G_n\}\subseteq\mathcal{G}$ converges weakly to some $G\in \mathcal{G}$ if, for all points of continuity $x\in\mathbb{R}$ of $G$, we have $G_n(x)\to G(x)$.

Note that the above definition is not as "nice" as weak convergence in distribution functions. Recall that, for $\{F_n\}\subseteq \mathcal{F}$ and $F\in\mathcal{F}$ such that $F_n(x)\to F(x)$ for each point of continuity of $F$, an equivalent condition is to say that, for all bounded continuous functions $f$, we have
$$\int f\ \mathrm{d}F_n \to \int f\ \mathrm{d}F.$$
However, this equivalence is not true for the case where $G_n,G\in\mathcal{G}$. Some extra definitions:

Definition: A sequence of probability measures $\{\mathbb{F}_n\}_{n=1}^\infty$ is said to be tight if, for all $\varepsilon>0$, there exists some $N\in\mathbb{N}$ such that
$$\inf_n \mathbb{F}_n([-N,N]) > 1- \varepsilon.$$
Definition: A class of bounded continuous functions $\mathcal{L}$ is said to be distribution determining if, for $F\in \mathcal{F}$ and $G\in \mathcal{G}$,
$$\int f\ \mathrm{d}F = \int f\ \mathrm{d}G \qquad (\forall f\in \mathcal{L})$$
implies that $F=G$.

The book then presents a variation of Helly's selection theorem and a corollary:

Theorem (Helly's Selection Theorem): Let $\{G_n\}_{n=1}^\infty\subseteq \mathcal{G}$ be a sequence of generalized distribution  functions, then there exists some subsequence $\{G_{n_k}\}_{k=1}^\infty$ and $G\in \mathcal{G}$ such that $G_{n_k}$ converges weakly to $G$ for some $G\in \mathcal{G}$. That is, the space $\mathcal{G}$ is sequentially compact with respect to weak convergence.
Corollary: If every convergent subsequence of a sequence $G_n$ converges weakly to the same $G\in\mathcal{G}$ then the entire sequence $G_n$ converges weakly to $G$.

The following theorem is true:

Theorem: Let $\mathcal{L}$ be a distribution determining class and $\{\mathbb{F}_n\}_{n=1}^\infty$ be a sequence of probability measures. Then $\mathbb{F}_n$ converges weakly to some probability measure $\mathbb{F}$ if and only if the sequence $\{\mathbb{F}_n\}_{n=1}^\infty$ is tight and $\lim_n \int f\ \mathrm{d}\mathbb{F}_n$ exists for all $f\in \mathcal{L}$.

I will provide a proof of the converse that basically outlines Borovkov's proof:
Proof. By Helly's selection theorem, there exists a subsequence of distribution functions (that correspond to the probability measures) $F_{n_k}$ that converges weakly to some $F\in \mathcal{G}$. Now, let $\varepsilon>0$. By the tightness of $\mathbb{F}_{n_k}$, can find some $M$ such that, for all $x\geq M$, we have
$$\inf_k F_{n_k}(x) \geq \inf_k \mathbb{F}_{n_k}([-M,M]) > 1- \varepsilon.$$
Let $x$ be a point of continuity of $F$ with $x\geq M$ (this must exist as $F$ only has countably many points of discontinuity). We then have
$$F(x) = \lim_k F_{n_k}(x) \geq \inf_k F_{n_k}(x) >1- \varepsilon.$$
Further, for all $y\geq x$, we would have $F(y) > 1- \varepsilon$. Therefore, we have $\lim_{x\to\infty}F(x) = 1$. A similar argument shows that $\lim_{x\to-\infty}F(x) = 0$. Hence, we have $F\in \mathcal{F}$ is actually a distribution function.
It remains to show that the entire sequence $\mathbb{F}_n$ converges weakly to the $\mathbb{F}$ (given by the distribution function $F$) above. By Corollary to Helly's selection theorem, it suffices to show, for arbitrary $\{F_{n_j}\}_{j=1}^\infty$ subsequence that converges weakly to some $G\in \mathcal{G}$, we have $F=G$. First, notice that the above argument ensures the limit function is actually a distribution function, so we actually have $G\in\mathcal{F}$.
Let $\mathbb{F}$ and $\mathbb{G}$ denote the probability measures induced by distribution functions $F$ and $G$, respectively. Since $F_{n_j}$ converges weakly to $G \in\mathcal{F}$, we have, for all $f\in \mathcal{L}$,
$$\lim_j \int f\ \mathrm{d}\mathbb{F}_{n_j} = \int f\ \mathrm{d}\mathbb{G}.$$
Further, the weak convergence of our original sequence $\{\mathbb{F}_{n_k}\}_{k=1}^\infty$ to $\mathbb{F}$ gives us that, for all $f\in \mathcal{L}$,
$$\lim_k \int f\ \mathrm{d}\mathbb{F}_{n_k} = \int f\ \mathrm{d}\mathbb{F}.$$
Since $\lim_n \int f\ \mathrm{d}\mathbb{F}_n$ exists for all $f\in \mathcal{L}$, all its subsequences must converge to the same limit, which gives us
$$\int f\ \mathrm{d}\mathbb{F} = \lim_k \int f\ \mathrm{d}\mathbb{F}_{n_k} = \lim_j \int f\ \mathrm{d}\mathbb{F}_{n_j} = \int f\ \mathrm{d}\mathbb{G}$$
As the above equality holds for all $f\in \mathcal{L}$, a distribution determining class, we have $\mathbb{F} = \mathbb{F}'$ and thus $F=F'$, as desired.
However, I am having trouble proving the following corollary (I copied and paste straight from the text):

I am having trouble showing that (2) is sufficient. The argument they gave appears to be down the lines of the following (just basing off the proof of the previous theorem)
Consider the sequence $F_n$, by Helly's selection theorem, we have $F_{n_k}$ converges weakly to some $G\in \mathcal{G}$. Further, we have, for each of those subsequences that converge weakly,
$$\lim_{k\to\infty}\int f\ \mathrm{d}F_{n_k} = \int f\ \mathrm{d}F.$$
Now, it would be nice if we also write (I think the author just assumes this, but this does not appear to be true for me)
$$\lim_{k\to\infty}\int f\ \mathrm{d}F_{n_k} = \int f\ \mathrm{d}G$$
If that's true, then we're done because that would imply $G=F$ and every subsequence that converges in $F_n$ converges to the same distribution function. But we cannot do that, because $G\in \mathcal{G}$ and we know that weak convergence (as defined by pointwise convergence to all points of continuity in the limit function) is no longer equivalent to convergence in integrals.
TLDR: I have having trouble proving (2) is sufficient in the above corollary and the hint the author gave is confusing.
 A: Basically I was able to prove my case of interest, which is just Lévy's continuity theorem. It's easy to see that Lévy's theorem falls easily out of this corollary because $\{e^{itx}\}_{t\in\mathbb{R}}$ is a distribution determining class. However, this is a "special" distribution determining class because, as it turns out, if $\varphi_n(t)\to\varphi(t)$, where $\varphi_n$ and $\varphi$ are characteristic functions of $\mathbb{F}_n$ and $\mathbb{F}$, respectively, one could prove that this condition implies $\mathbb{F}_n$ will be a tight sequence of measures. The proof of tightness relies on the fact that $\varphi(t)$ will be a continuous function, in particular it will be continuous at $0$. Of course, this is a very special case of the Corollary and I'm not sure how much generalization does this hold.
Edit: David Williams' Probability with Martingales proves this small bit in the Central Limit Theorem chapter.
A: Let $\mathcal{L} = \{f \in C_b(\mathbb{R}) : f \text{ has period $k$ for some }k \in \mathbb{N}\}$. This family is distribution determining because sub-probability measures on $\mathbb{R}$ are inner regular. The basic idea is that you can approximate the integral of $f \in C_c(\mathbb{R})$ by one of these functions in both $L^1(\eta)$ and $L^1(\mu)$ simultaneously by working on a compact set $K$ which contains the support of $f$ and for which $\eta(K^c),\mu(K^c)< \epsilon$. I can add more details if you need them. Note also that $1 \in \mathcal{L}$.
Call $k_f$ the minimal non-negative integer period of $f$ and let $\mu_n(dx) = \delta_{n!}(dx)$. For each function $f \in \mathcal{L}$ and all $n \geq k_f$, since $k_f$ divides $n!$, we have $f(n!) = \int f(x)\mu_n(dx)= \int f(x) \delta_0(dx) = f(0)$. Clearly $\delta_0$ is a probability measure.
On the other hand, the family $\mu_n$ converges to the zero measure vaguely and in particular is not tight.
A: Edit: The initial argument I suggested was incorrect. I am not deleting the answer because it has accumulated some relevant comments. See the answer by  Chris Janjigian that gives a counterexample to Cor 6.3.2 parts (2) and (3).
