Portmanteau Theorem and lipschitz functions In the Portmanteau Theorem, on the equivalence forms of weak convergence for random vectors $X_n\to X$, the fact that $E[f(X_n)]\to E[f(X)]$ for all Lipschitz function is used to prove that $\lim\inf P(X_n\in G)\ge P(X\in G)$ for every _open_ set $G$. The proof says: > For every open set $G$ there exists a sequence of Lipschitz functions with $0\le f_m\uparrow 1_G$. For instance $f_m(x)=(m\cdot d(x,G^c))\wedge 1$. For every fixed $m$, $$\liminf_{n\to\infty}P(X_n\in G)=\liminf_{n\to\infty}E(1_G(X_n))\ge\liminf_{n\to\infty}E(f_m(X_n))=E(f_m(X))$$ and then it uses $m\to\infty$ and the monotone convergence theorem. My question is: why we need $G$ to be open? why it can not be a closed set?

If $G$ is not open, then there does, in general, not exist a sequence of Lipschitz functions $(f_n)_{n \in \mathbb{N}}$ such that $0 \leq f_n \uparrow 1_G$.

Consider for instance $G=[0,1]$. Take a sequence of (Lipschitz) continuous functions $(f_n)_n$ such that $0 \leq f_n(0) \uparrow 1_{[0,1]}(0)=1$. Then we can find for large $n \in \mathbb{N}$ a constant $\delta>0$ such that $$f_n(x) \geq \frac{1}{2} \qquad \text{for $-\delta \leq x \leq 0$}.$$ Hence, $$f_n(x)> 0= 1_{[0,1]}(x) \qquad \text{for all $-\delta \leq x < 0$}.$$ This shows that $f_n(x) \uparrow 1_{[0,1]}(x)$ does not hold true for any $x \in [-\delta,0)$. This means that there cannot exist a sequence of continuous functions $(f_n)_{n \in \mathbb{N}}$ such that $0 \leq f_n \uparrow 1_{[0,1]}$.