# Outer measure (induced by a premeasure) on a set can be arbitrarily close to a subset

This is a homework problem from Folland (1.18):

Let $\mathcal{A} \subset \mathcal{P}(X)$ be an algebra, $\mathcal{A}_\sigma$ the collection of countable unions of sets in $\mathcal{A}$, and $\mathcal{A}_{\sigma\delta}$ the collection of countable intersections of sets in $\mathcal{A}_\sigma$. Let $\mu_0$ be a premeasure on $\mathcal{A}$ and $\mu^*$ the induced outer measure.

(a) For any $E\subset X$ and $\epsilon > 0$ there exists $A \in \mathcal{A}_\sigma$ with $E\subset A$ and $\mu^*(A)\le \mu^*(E)+\epsilon$.

What I have so far:

Let $\mathcal{U}(E) = \{ A \in \mathcal{A}_\sigma\ \mid\ E \subset A\}$. We show that $\mathcal{U}(E)$ is non-empty for all $E$: $X \in \mathcal{A}$ for all algebras $\mathcal{A}$, and since $E \subset X$, we simply observe that $X \in \mathcal{A}_\sigma$.

For any $A \in \mathcal{U}(E)$, we have $A \in \mathcal{A}_\sigma$ and hence $A = \bigcup_1^\infty A_j$; furthermore, since $\mu^*$ is an outer measure, the following property holds: $$\mu^*(A) = \mu^*\left(\bigcup_1^\infty A_j\right) \le \sum_1^\infty \mu^*(A_j) = \sum_1^\infty \mu_0(A_j),$$ with the latter equality justified by the fact that $\mu^*$ is induced by the premeasure $\mu_0$ and $A_j \in \mathcal{A}$.

By definition of $\mu^*$, we have $$\mu^*(E) = \inf \left\{ \sum_1^\infty \mu_0(A_j)\ \mid\ A_j \in \mathcal{A},\ E\subset \bigcup_1^\infty A_j \right\}.$$

Now I'm stuck. I feel like I have to combine these notions, but I can't seem to justify existence of an $A$ that gets me within $\epsilon$ of this infimum.

The thing you are missing is the definition of $\inf$, you dummy! For any $\inf$ of a sequence of real numbers, we can either always get within $\epsilon$ of the $\inf$, or otherwise $\inf$ is the smallest such element of the sequence (which would also satisfy the condition).