From Royden's Real Analysis 4th edition, section 17.5, problem 36:
Let $\mu$ be a finite premeasure on an algebra $\mathcal S$, and $\mu^*$ the induced outer measure. Show that a subset $E$ of $X$ is $\mu^*$-measurable if and only if for each $\varepsilon>0$ there is a set $A\in\mathcal S_\delta$, $A\subseteq E$, such that $\mu^*(E\smallsetminus A)<\varepsilon$.
Primarily my question is: is this even true? This seems like it's stated backward. I'm a noob so maybe I have this wrong, but since $\mathcal S_\delta$ is the set of all intersections of sets in $\mathcal S$, then I would have thought that we were looking for a set $E\subseteq A$. When I try to do the proof that seems like the direction that pops out naturally. Also, in section 2.4 where there is a corresponding proof for the Lebesgue measure, we claim "There is a $G_\delta$ set $G$ containing $E$ for which ..."
But when I look up the errata, this is not mentioned in it.
If the answer is "the book stated it correctly" then I'd appreciate a hint. I've tried using the sequence of sets such that $\mu^*(E)-\varepsilon < \sum \mu^*(E_k)$ where $\{E_k\}$ covers $E$. Intersecting these seems like a dead-end. I could union them, prove the theorem for $E\subseteq B$, and then try to use $B$ to construct $A$. But any way that I can see of doing that, I lose the property that $A\in \mathcal S_\delta$.
This might be a typo. What can be shown is the following: If $\mu^*(E)< \infty$ then $E$ is $\mu^*$-measurable if and only if for all $\epsilon>0$ there exists $B$ in $S_\sigma$ with $E\subset B$ and $\mu^*(B/E)< \epsilon$.