Let $(X, \Sigma, \mu)$ be a measure space. Define $$ \mu^*(S) = \inf\{\mu(U)|U \in \Sigma, S \subseteq U\}: \mathcal{P}(X) \to [0, \infty] $$
Theorem: $\forall S \subseteq X, \exists U \in \Sigma|\mu(U) = \mu^*(S)$
How to prove this theorem? Intuitively there seem to be some counterexamples. (Approximating a sphere with rectangles, etc.)
Let $(X, \Sigma, \mu)$ be a measure space. Define $$ \mu^*(S) = \inf\{\mu(U)|U \in \Sigma, S \subseteq U\}: \mathcal{P}(X) \to [0, \infty] $$
Proof. For $n>1$, let $U_n \in \Sigma$ such that $S \subseteq U_n$ and $\mu^*(S)\le \mu(U_n) \le \mu^*(S)+1/n$. Consider $\cap_{k \in N}U_k$. We have $S \subseteq \cap_{k \in N}U_k$. On the one hand, $\mu^*(S) \le \mu(\cap_{k \in N}U_k)$. On the other hand, we have $\mu(\cap_{k \in N}U_k) \le \mu(U_n)\le \mu^*(S)+1/n$ for each $n \in N$. Hence $\mu(\cap_{k \in N}U_k)=\mu^*(S)$.
Now under $U$ we can take $\cap_{k \in N}U_k \in \Sigma$.