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

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.

• I don't see how to get started on this either. In fact it's not even clear to me that the (bounded) continuous functions are dense in $L^1(\mu)$. The typical setup where that sort of thing is clear is a locally compact Hausdorff space... Mar 13 '19 at 14:21

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 $$(X,d)$$.
• I was about to post an answer saying this is true if $\mu$ is regular. How obvious is it that every finite Borel measure on a metric space is regular? Mar 13 '19 at 14:58
• I don't know about obvious. You look at the family of all Borel sets $A$ for which $\mu(A)=\sup\{\mu(C)\mid C\subset A\text{ closed}\}=\inf\{\mu(U)\mid U\supset A\text{ open}\}$. This is a $\sigma$-algebra. If $X$ is a metric space, this family also contains all closed subsets of $X$, because they are countable intersections of open sets. Mar 13 '19 at 15:09
• Thanks. Obvious or not, certainly trivial. (And maybe obvious, given that the standard way to show every Borel set has some property is to show that the class of all sets with that property is a $\sigma$_algebra...) Mar 14 '19 at 14:12