Finding a set $U$ such that $\mu(U) < \mu(E) + \epsilon$ Suppose we are working on some measure space $(X, M, \mu)$, and suppose $E \in M$. Given $\epsilon > 0$, I have been wondering if we can always find sets of the following form:

*

*$U \supset E$ such that $\mu(U) < \mu(E) + \epsilon$.

*$F \subset U$ such that $\mu(F) > \mu(U) - \epsilon$.

The motivation for this questions comes from me seeing this construction quite a bit in measure theoretic proofs. For example, when proving that every Radon measure is inner regular on all of its $\sigma$-finite sets. However, I have deliberately omitted any specific theorem as I would like to know if (why) such sets can be found in general.
So far I have seen that this can indeed be done if $E$ is $\sigma$-finite, $U$ is open, and $F$ is compact. But why? Does it hold in the general case?
 A: The reason why you see such arguments being used, for instance, in the proof of inner regularity of the Radon measure is because a Radon measure is already defined to be  outer regular on all Borel sets and inner regular on all open sets. This means precisely that given a Borel set $E$ $$\mu(E)=\text{inf} \{ \mu(U): U \text{ is open and } E \subset U \}$$
Notice this immediately implies that given any $\epsilon>0$ there exist $U$ open such that $$\mu(E)+\epsilon>\mu(U)$$ Because if such a set $U$ doesn't exist then the infimum would be $\mu(E)$$+\epsilon$ instead of just $\mu(E)$.
Inner regularity on open sets precisely means that for each open set $U$ and $\epsilon>0$ there exist a compact set $K$ such that
$$\mu(K)+\epsilon>\mu(U)$$
Essentially the intuition should be: Inner regularity means you can arbitrarily approximate by compact sets from within, and outer regularity means you can arbitrarily approximate by open sets from "outside".
Given all that, Notice that a Radon measure is only inner regular on open sets, not arbitrary $\sigma$-finite Borel sets, and this is what the theorem you cited is trying to prove. But if you go back and carefully read your theorem, the reason why you can use the arguments you cited is precisely because a Radon measure is defined to be outer regular on Borel sets and inner regular on all open sets.
