Are the bounded Lipschitz functions dense in $L^1(\mu)$?

Question

I am currently reading a paper (L. Ambrosio and B. Kirchheim. Currents in metric spaces) and I stumbled uppon a fact which I don't know how to prove. I have the following setting:

Let $X$ be a complete metric space, $\mu$ a finite Borel measure and let $\text{Lip}_b(X)$ denote the bounded Lipschitz functions $X \rightarrow \mathbb{R}$. Then $\text{Lip}_b(X)$ is supposed to be dense in $L^1(X,\mu)$.

I assume I need to do something with some density of $\text{Lip}_b(X)$ in $C(X, \mathbb{R})$, but since we don't have any compactness assumptions, we can't apply Stone-Weierstrass and I don't know how we got that fact.

Answer 1

1. Bounded continuous functions are dense in $L^1(X,\mu)$: Every finite Borel measure on a metric space is regular, so it suffices to show that characteristic functions of closed subsets can be approximated by bounded continuous functions in $L^1$. For that purpose let $C\subset X$ be closed and $f_n=(1-nd(\cdot,C))_+$. The function $f_n$ is continuous and satisfies $0\leq f_n\leq 1$ and $f_n(x)\to 1_C(x)$ for all $x\in X$. Thus $f_n\to 1_C$ in $L^1$ by dominated convergence.
2. Every nonnegative bounded continuous function on $X$ is the increasing limit of nonnegative Lipschitz functions: For lower semicontinuous $f\colon X\to [0,\infty)$ let $$f_n(x)=\inf_y (f(y)+n d(x,y)).$$ This function is $n$-Lipschitz, satisfies $f_n\leq f$ and $f_n(x)\to f(x)$ for all $x\in X$. In particular, $f_n\to f$ by monotone convergence.

Edit: I don't think you need completeness of $(X,d)$.