$P_{n}(A)\xrightarrow{n \to \infty} P(A)$ for any $P$-continuity set $A$ iff $F_{n}(x)\xrightarrow{ n \to \infty} F(x)$ for all continuity points

Question

For a while, I have been strunggling to find why Levy's continuity Theorem would imply that:

$P_{n} \xrightarrow{\text{distribution}} P\iff \phi_{n}(t)\xrightarrow{n \to \infty}\phi(t)\; $

since Levy's Continuity Theorem is concerned with weak convergence rather than convergence in distribution. I thus attempted to find an equivalence between convergence in distribution and weak convergence. This leads me to the Portmanteau Lemma which shows many equivalent notions of weak convergence. So I am to attempt the following equivalence in the case of the measure being on $\mathbb R$:

$$ P_{n}\xrightarrow{\text{weakly}}P \iff P_{n}\xrightarrow{\text{distribution}}P$$

For "$\Rightarrow$" consider the equivalent notion of "weak convergence" from the Portmanteau Lemma, namely:

$P_{n}(A)\xrightarrow{n \to \infty} P(A)$ for any $P$-continuity set $A$.

Now let $x$ be a continuity point of $F$ in $\mathbb R$, then the set $(-\infty,x]$ is a continuity set under $P$ and hence: $F_{n}(x)=P_{n}((-\infty,x])\xrightarrow{n \to \infty} P((-\infty,x])=F(x)$ thus $F_{n}\xrightarrow{ \text{distribution}} F$.

I do not know what equivalent definition of weak convergence to prove for "$\Leftarrow$" direction. Continuity sets could take vastly different forms to $(-\infty,x]$ , thus I do not think I could use it. Any help would be greatly appreciated.

Answer 1

I believe from your question that you take as definition of convergence in distribution that $\lim_n F_n(x) = F(x)$ where $x$ is a continuity point of $F$. Hence, I will show that this condition implies $$\liminf_n P_n(G) \geq P(G)$$ for all non-empty open subsets $G$ (you already proved the converse). The latter condition is equivalent with weak convergence, by the Portmanteau lemma.

Fix such a non-empty open subset $G$. For any $x \in G$, the fact that $G$ is open allows us to select $\epsilon_x > 0$ with $]x-2\epsilon_x, x + 2\epsilon_2[\subseteq G$. Then $G= \bigcup_{x \in G} ]x-\epsilon_x, x + \epsilon_x[$. By a theorem of Lindelöf (every open cover has a countable subcover), we can select a sequence $(x_n)_n\subseteq G$ with $G = \bigcup_{n=1}^\infty ]x-\epsilon_{x_n}, x + \epsilon_{x_n}[$ and by our construction also $$]x_n-2\epsilon_{x_n}, x_n + 2 \epsilon_{x_n}[\subseteq G.$$ Since $F$ is monotone, the set $D(F)$ of points of discontinuity is at most countable. Hence, for any $n \geq 1$, we can find $a_n, b_n$ continuity points with $$x_n - 2 \epsilon_{x_n} < a_n < x_n- \epsilon_{x_n}< x_n + \epsilon_{x_n} < b_n < x_n + 2 \epsilon_{x_n}.$$ It follows that $G = \bigcup_{n=1}^\infty]a_n, b_n]$ where $a_n < b_n$ are continuity points for all $n \geq 1$. Put $$\mathcal{P}:= \{]a,b]: a < b, a,b \notin D(F)\}\cup\{\emptyset\}.$$ Then $\mathcal{P}$ is closed under finite intersections. If $C = ]a,b] \in \mathcal{P}$, then \begin{align}\lim_n P_n(C) &= \lim_n \{P_n(]-\infty, b])- P_n(]-\infty, a])\}\\ &=\lim_n {F_n(b)- F_n(a)}\\ &= F(b)-F(a) = P(C).\end{align} If $C_1, \dots, C_m \in \mathcal{P}$, then by inclusion-exclusion principle also $$\lim_n P_n\left(\bigcup_{i=1}^m C_i\right) = P \left(\bigcup_{i=1}^m C_i\right).$$ Hence, for all $m \geq 1$, $$\liminf_n P_n(G) \geq \liminf_n P_n\left(\bigcup_{i=1}^m ]a_i, b_i]\right) = P\left(\bigcup_{i=1}^m ]a_i, b_i]\right).$$ But $\bigcup_{i=1}^m ]a_i, b_i[ \nearrow G$, so by continuity of measure we obtain $$\liminf_n P_n(G) \geq P(G).$$

Feel free to ask for clarification if something is not clear to you!