Approximate measure of a Borel set by a Lipschitz continuous bounded functional

Question

I'm trying to prove this approximation. Could you verify if my attempt is fine?

Let $(X, d)$ be a metric space, $\operatorname{Lip}_{b}(X)$ the space of Lipschitz continuous bounded real-valued maps on $X$, and $\mu$ a finite Borel measure on $X$. Let $U$ be open in $X$.

1. For every $U \subset X$ open and every $\varepsilon>0$ there exists an $f \in \operatorname{Lip}_{b}(X)$ with $0 \leq f \leq \mathbb{1}_{U}$ and $\int ( 1_U - f ) \mathrm d \mu<\varepsilon$.
2. For every $A \in \mathcal{B}(X)$ and every $\varepsilon>0$ there exists an $f \in \operatorname{Lip}_{b}(X)$ with $\int |f-\mathbb{1}_{A} | \mathrm d \mu<\varepsilon$.

I post my proof separately as below answer. This allows me to subsequently remove this question from unanswered list.

Answer 1

1. Consider the map $$ g(x) := \min\{d(x, U^c), 1\} \quad \forall x \in X. $$

Notice that $\min\{a,c\}- \min\{b,c\} \le a-b$ for all $a\ge b$, so $g$ is $1$-Lipschitz. We define $$ g_n := \sqrt[n]{g} \quad \forall n \ge 1. $$

Then $0 \le g_n \nearrow 1_U$. By monotone convergence theorem, there is $N$ such that $$ \int (1_U-g_N) \mathrm d \mu < \varepsilon/2. $$

Notice that the map $[0, \infty) \to \mathbb R,x \mapsto x^{\alpha}$ is unifomly continuous for all $\alpha \in [0,1]$. This implies $g_N$ is uniformly continuous. On the other hand, $\operatorname{Lip}_{b}(X)$ is dense (w.r.t. $\| \cdot \|_\infty$) in the space of uniformly continuous bounded maps. So there is $f \in \operatorname{Lip}_{b}(X)$ such that $$ f\le g_N \quad \text{and} \quad\|f-g_N\|_\infty < \varepsilon/(2 \mu(X)). $$

It follows that $$ \begin{align} \int (1_U-f) \mathrm d \mu &= \int (1_U-g_N) \mathrm d \mu + \int (g_N-f) \mathrm d \mu \\ &\le \varepsilon/2 + \varepsilon/2 = \varepsilon. \end{align} $$

2. Every finite Borel measure on a metric space is outer regular, so there is $U$ open in $X$ such that $\mu(U\setminus A) < \varepsilon/2$. By 1., there is $f \in \operatorname{Lip}_{b}(X)$ such that $\int |1_U - f| \mathrm d \mu<\varepsilon/2$. The claim the follows.