# Portmanteau Theorem in Weak Convergence ( Lipschitz functions and lim sup / lim inf )

I have recently embarked on an endeavor to understand weak convergence and, consequently, have stumbled across the Portmanteau theorem. As I am not a mathematician, I presume that my questions are rather basic in nature. I would very much appreciate, though, if someone (who might have a little time on their hands) could help me to see what I am missing.

Provided that $\mu_n$ weakly converges to $\mu$, I learned that the following statements are equivalent \begin{align}&\text{a)}\qquad\int_{S}f\,\text{d}\mu_n\to\int_S f\,\text{d}\mu\quad\forall\quad\text{real-valued, bounded, and continuous functions $f:S\to\mathbb{R}$}\\&\text{b)}\qquad\int_{S}f\,\text{d}\mu_n\to\int_S f\,\text{d}\mu\quad\forall\quad\text{real-valued, bounded Lipschitz functions $f:S\to\mathbb{R}$}\\&\text{c)}\qquad\underset{n\to\infty}{\lim\text{sup}}\,\int_S\mathbb{1}_A\,\text{d}\mu_n\leq\int_S\mathbb{1}_A\,\text{d}\mu\quad\forall\quad\text{closed subsets }A\subseteq S\\&\text{d)}\qquad\underset{n\to\infty}{\lim\text{inf}}\,\int_S\mathbb{1}_B\,\text{d}\mu_n\geq\int_S\mathbb{1}_B\,\text{d}\mu\quad\,\forall\quad\text{open subsets }B\subseteq S\\&\text{e)}\qquad\underset{n\to\infty}{\lim}\,\,\,\,\int_S\mathbb{1}_C\,\text{d}\mu_n\geq\int_S\mathbb{1}_C\,\text{d}\mu\quad\,\,\,\forall\quad\mu-\text{continuity sets }C\subseteq S\end{align}

Statement a) is the definition of weak convergence; moreover I understand that the vectorspace of Lipschitz continuous functions is a subset of all continuous function so a) implies b).

I do, however, struggle to see how b) implies c). Building on a couple of proofs I have seen so far, it would seem that it is key to construct a set $A_k$ such that $$A_k=\{x\in S:\text{inf}\{|x-y|:y\in A\}\leq k^{-1}\,\}.$$ Hence, as $k\to\infty$ it follows that $A_k\downarrow A$. Moreover, most proofs go on proposing to define a bounded, Lipschitz function $f_k:S\to\mathbb{R}$ such that $$\mathbb{1}_A\leq f_k\leq\mathbb{1}_{A_k}$$ Unfortunately, I have not been able to find a single proof that features an explicit function, so I am a little bit lost at this point.

Question 1: How would such a function look like and how would one go about proving that this function is, in fact, Lipschitz continuous?

Provided this function exists, it follows that $$\underset{n\to\infty}{\lim\text{sup}} \int_S\mathbb{1}_A\text{d}\mu_n\leq\lim_{n\to\infty}\int_Sf_k\text{d}\mu_n=\int_Sf_k\text{d}\mu\leq\int_S\mathbb{1}_{A_k}\text{d}\mu$$ I understand that the last inequality follows from the increasing property of the Lebesgue integral and the penultimate equality follows from b). I struggle, however, with this part $$\underset{n\to\infty}{\lim\text{sup}} \int_S\mathbb{1}_A\text{d}\mu_n\leq\lim_{n\to\infty}\int_Sf_k\text{d}\mu_n$$ As I am not a mathematician by education, I have looked up the definition of the limit superior on Wikipedia and I have learned that $$\underset{n\to\infty}{\lim\text{sup}} \int_S\mathbb{1}_A\text{d}\mu_n=\lim_{n\to\infty}\sup\left(\int_S\mathbb{1}_A\text{d}\mu_m:m\geq n\right)$$ but I have no idea how that translates to the above inequality.

Question 2: I would be very grateful, if someone could give me some sort of step-by-step explanation of how this first inequality comes about.

Moving on; I do see how $k\to\infty$ then yields the desired result. But...

Question 3: Statement d) is supposed to follow from c) "by taking complements". Since I do not even really see how the fact that all sets $A$ are closed factors into the line of reasoning above, I have no idea how the fact that all sets $B$ are open makes d) equivalent to c) by taking complements.

Thank you so very much.

Best regards,
Jon

1. Define $f_k\colon S\to \mathbb R$ by $$f_k(x)=1-\frac{d(x,A_k)}{k^{-1}+d(x,A_k)},$$ where $d(x,B)=\inf\left\{d(x,y),y\in B\right\}$.
2. This a consequence of the combination of the fact that if $\left(a_n\right)_{n\geqslant 1}$ and $\left(b_n\right)_{n\geqslant 1}$ are two sequence of real numbers such that for all $n$, $a_n\leqslant b_n$, then $\limsup_{n\to +\infty}a_n\leqslant\limsup_{n\to +\infty} b_n$. If the sequence $\left(b_n\right)_{n\geqslant 1}$ converges to $b$, then $\limsup_{n\to +\infty} b_n=b$.
This gives that $\limsup_{n\to +\infty}\mu_n(A)\leqslant \mu(A_k)$ for all $k$. Then notice that the sequence of sets $\left(A_k\right)_{k\geqslant 1}$ is non-increasing and its intersection is $A$.
1. Suppose (c). Let $B$ be an open subset of $S$. Apply (c) to $A=S\setminus B$, which is closed.