I know that for every probability measure $\mu$ on doubling metric space $X$, the set of lipschitz functions (and therefore continuous functions) is dense in $L_1(\mu)$. Is it true for $L_{\infty}$ as well? i.e. for every $\epsilon >0$ and a bounded function $g$ there is a continuous $f$ s.t. for all $x \in X$ we have $|f(x) - g(x)| <\epsilon $ (I assume $f:X \to \mathbb{R}$)
If it is not true, is it correct with additional constraints on $X$?
False even in very nice spaces like $[0,1]$. It is known that pointwise limits of continuous functions are continuous at points of a dense set. Since a bounded (measurable) function need not be continuous at any point we cannot write it as a pointwise limit of continuous functions.