Closed subset with measure arbitrarily close to original set

Question

Let $(\mathbb{R}^n, \mathcal{B}(\mathbb{R}^n))$ be the measurable space formed by the real coordinate space of $n$ dimensions and the associated Borel algebra. We let $\mu$ be a measure for the space.

Let $S$ be an arbitrary measurable set. For any $\epsilon > 0$, is it true that there exists a $C \subseteq S$ such that

1) $C$ is closed

2) $\mu(S\setminus C) < \epsilon$

That is, does any set contains a closed subset with measure arbitrarily close to that of the original set?

It would seem to me that this is the case, and that we can find a union of intervals and singletons contained in $C$ such that (2) is satisfied. However, I have no idea how to prove this rigorously.

Answer 1

This is not true in general. For a counterexample, take $\mu$ be the counting measure. However, if you add one more restriction on $\mu$ such that $\mu(K)<\infty$ for ever compact set $K$, then $\mu$ will have that property. In fact, with this additional restriction, $\mu$ is regular. See Rudin's Real and Complex Analysis, theorem 2.18, page 48.

Answer 2

Let the set $S$ be measurable with $\mu(S)<\infty$. Let $\mu_*$ denote the inner measure with respect to $\mu$. We have the following definition,

$$\mu_*(S):=\sup\{\mu(C): C\subset S, C\mbox{ compact}\}$$

Further, since the set $S$ is measurable w.r.t. $\mu$ we have $\mu(S)=\mu_*(S)$. Let $\epsilon>0$, if $\mu_*(S)=0$ then we are done so assume that $\mu_*(S)>0$. Then,

$$\sup\{\mu(C): C\subset S, C\mbox{ compact}\} >0\implies \exists C\subset S: \mu(S)-\epsilon<\mu(C)\leq\mu(S)\\ \implies \mu(C)<\mu(S)+\epsilon$$

Since $C$ is compact, it is also closed.