I would like an help for showing that the following function is Lipschitz:
consider a function $f_m$ of $x$, $f_m: \mathbb{R} \rightarrow \mathbb{R}$ such that $f_m:=\min\{1, m*d(x, G^c)\}$ where $d(x, G^c)$ is the distance between $x$ and a closed set $G^c \subseteq \mathbb{R}$ and $m>0 \in \mathbb{N}$.
Moreover, why we can say that
For every open set $G$ there exists a sequence of Lipschitz functions $0\leq f_m \rightarrow \mathbb{1}_G$ where $1_G:= \begin{cases} 1 > \text{ if } x \in G\\ 0 \text{ otherwise} \end{cases}$
? (the function $f_m$ above is an example of function converging to $\mathbb{1}_G$ as $m\rightarrow \infty$)
(from van der Vaart, Asymptotic Statistics p. 7).
For $x \in G^c$ we have $f_m(x) = 0$ so it is Lipschitz there.
The open set $G \subset \mathbb{R}$ can be written as a union of countably many open intervals. Let $]a, b[$ be one of those intervals and $x \in ]a,b[$. Wlog $x > \frac{a+b}{2}$ so $x$ is closer to $b$ than to $a$ and we have $d(x, G^c) = b-x$. If $b-x < 1/m$ then $f_m(x) = m(b-x)$, else $f_m(x)=1$, in both cases it is obviously Lipschitz.
With a separation of the cases as above you can also show that $f_m$ converges pointwise to $1_G$.