Lemma 1: Let $\mu, \mu_1, \mu_2,\ldots \in \mathcal{M}$ and $g \in \mathcal C_b(X)$. If $\mu_i \to \mu$ weakly, then $\mu_i(A) \to \mu(A)$ for all Borel set $A \subseteq X$ with $\mu(\partial A) = 0$.
Lemma 2: If $X$ is separable and $\mu \in \mathcal{M}$. Then for each $\delta>0$ there are countably many open (or closed) balls $B_{1}, B_{2}, \ldots$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$, the radius of $B_{i}$ is less than $\delta$, and $\mu\left(\partial B_{i}\right)=0$ for all $i$.
Fix $\varepsilon>0$. We want to show that $\exists N, \forall i \geq N: d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$, i.e., $\mu_{i}(B) \leq \mu\left(B_{\varepsilon}\right)+\varepsilon$ and $\mu(B) \leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon$ for all Borel subset $B$.
Fix $\delta \in (0, \varepsilon/4)$. By Lemma 2, there are countably many open balls $B_{1}, B_{2}, \ldots$ with radius less than $\delta/2$ such that $\bigcup_{i=1}^{\infty} B_{i}=X$ and $\mu\left(\partial B_{i}\right)=0$ for all $i$. Fix $k$ such that
$$
\mu\left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu(X)-\delta.
$$
Let $\mathcal A$ be the finite collection of subsets built by combining the balls $B_1, \ldots, B_k$, i.e.,
$$
\mathcal{A}:=\left\{\bigcup_{j \in I} B_{j} \,\middle\vert\, J \subset \{1, \ldots, k\}\right\}.
$$
We will use this collection to approximate any Borel set. For each $A \in \mathcal{A}, \partial A \subset \partial B_{1} \cup \cdots \cup \partial B_{k}$, so $\mu(\partial A) \leq$ $\mu\left(\partial B_{1}\right)+\cdots+\mu\left(\partial B_{k}\right)=0$. By Lemma 1, $\mu_{i}(A) \rightarrow \mu(A)$ for all $A \in \mathcal{A}$. Fix $N$ such that
$$
\left|\mu_{i}(A)-\mu(A)\right|<\delta \quad \forall i \geq N, \forall A \in \mathcal{A}.
$$
In particular,
$$
\mu_i \left(\bigcup_{j=1}^{k} B_{j}\right) \ge \mu \left(\bigcup_{j=1}^{k} B_{j}\right) -\delta \ge \mu(X) - 2 \delta \quad \forall i \ge N.
$$
Now we fix a Borel set $B$ and approximate it by
$$
A := \bigcup \{B_j \mid j = 1,\ldots,k \text{ such that } B_j \cap B \neq \emptyset\}.
$$
Then
- $A \subset B_{\delta} := \{x \mid d(x, B)<\delta\}$ because $\operatorname{diam} B_{j}<\delta$,
- $B=\left[B \cap \bigcup_{j=1}^{k} B_{j}\right] \cup\left[B \cap\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right] \subset \left [ A \cup\left(\bigcup_{j=1}^{k} B_{j}\right)^{c} \right ]$,
- $\left|\mu_{i}(A)-\mu(A)\right|<\delta$ for all $i \geq N$, and
- $\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \delta$ and $\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \leq \mu_i(X)-\mu(X)+ 2 \delta \le \mu_i(X) + 3\delta$ for all $i \geq N$.
It follows that for every $i \geq N$ :
\begin{aligned}
\mu(B) & \leq \mu(A)+\mu\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\
& \leq \mu(A)+\delta \\
& \leq \mu_{i}(A)+2 \delta \\
& \leq \mu_{i}\left(B_{\delta}\right)+2 \delta \\
&\leq \mu_{i}\left(B_{\varepsilon}\right)+\varepsilon \\
\mu_{i}(B) & \leq \mu_{i}(A)+\mu_{i}\left(\left(\bigcup_{j=1}^{k} B_{j}\right)^{c}\right) \\
&\leq \mu_{i}(A)+ 3 \delta \\
&\leq \mu(A)+4 \delta \\
& \leq \mu\left(B_{\delta}\right)+4 \delta \\
&\leq \mu\left(B_{\varepsilon}\right)+\varepsilon.
\end{aligned}
This is true for every $B \in \mathcal{B}$, so $d_{P}\left(\mu_{i}, \mu\right) \leq \varepsilon$ for all $i \geq N$.
- 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
$$
\begin{align}
\int (1_U-f) \mathrm d \mu &= \int (1_U-g_N) \mathrm d \mu + \int (g_N-f) \mathrm d \mu \\
&\le \varepsilon/2 + \varepsilon/2 = \varepsilon.
\end{align}
$$
- 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.
Best Answer
Let $f$: $X$ $\rightarrow$ $\mathbb{R}$ be uniformly continuous and bounded. Let $$ f_n(x) := \inf_{y\in X}\{f(y)+n\text{d}(x,y)\}. $$
Let $\alpha >0$ such that $|f(x)| \le \alpha$ for all $x\in X$, so $|f(x)-f(y)| \le 2 \alpha$ for all $x,y\in X$.
It's clear that $-\alpha \le f_n \le f$ is bounded and $n$-Lipschitz continuous. Let's prove that $f_n \to f$ uniformly. Fix $\varepsilon>0$, there is $\delta>0$ such that $d(x,y)<\delta$ implies $|f(x)-f(y)|<\varepsilon$. Then for all $x\in X$, we have
$$ \begin{align} 0\le f(x) - f_n(x) &= f(x)- \inf \{ f(y)+n\text{d}(x,y) \mid y\in X \} \\ &= \sup \{ f(x)-f(y)-n\text{d}(x,y) \mid y\in X \} \\ &= \sup \{ f(x)-f(y)-n\text{d}(x,y) \mid y\in X \text{ s.t. } d(x,y) \le 2\alpha/n \}. \end{align} $$
If $n$ such that $2\alpha/n <\delta$ or equivalently $n>2\alpha/\delta$. Then $$ f(x) - f_n(x) \le \sup \{ \varepsilon-n\text{d}(x,y) \mid y\in X \text{ s.t. } d(x,y) \le 2\alpha/n \} \le \varepsilon \quad \forall x\in X. $$
It follows that $\|f-f_n\|_\infty \le \varepsilon$.