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Let $A$ be a set, let $B$ be a Borel set such that $B \subseteq A$.

Because $B$ is a Borel set, can I automatically say that I can represent it as a countable union of closed sets. Thus I can demonstrate that $m(A\setminus B)=0$.

Alternatively, let $C$ is a set and $D$ is a Borel set such that $C \subseteq D$. Then, since $D$ is Borel we can write it as a countable intersection of open sets. So $m(D \setminus C)=0$.

I'm trying to find a relationship between Borel sets and the Inner/Outer approximation theorems.

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Sadly, we can do nothing so simple. Countable unions (intersections) of closed (open) sets are called $F_\sigma$ $(G_\delta)$ sets. Countable unions (intersections) of $G_\delta$ $(F_\sigma)$ sets are called $G_{\delta\sigma}$ $(F_{\sigma\delta})$ sets. Similarly, we have $G_{\delta\sigma\delta}$ and $F_{\sigma\delta\sigma}$ sets, $G_{\delta\sigma\delta\sigma}$ and $F_{\sigma\delta\sigma\delta}$ sets, and so on. All of these are Borel sets (along with basic open and closed sets)--and in fact comprise the entirety of the collection of Borel sets. For a more explicit description of this transfinitely recursive construction of the Borel heirarchy--in $\omega_1$ steps, not countably many (Thanks, Trevor, for pointing that out!)--see here.

Taking $A$ to be the overlying set, $B'$ to be any Borel subset of $A$ of positive measure, and $B=A\smallsetminus B'$, we have that $B$ is a Borel set and $A\smallsetminus B=B'$, and furthermore $m(A\smallsetminus B)=m(B')>0$. On the other hand, take $D$ to be any Borel set of positive measure, and let $C=\emptyset$, so that $D\smallsetminus C=D$, and so $m(D\smallsetminus C)=m(D)>0$.

It is worth noting that we can approximate any measurable set from without (within) by some $G_\delta$ $(F_\sigma)$ set, in exactly the way you described. We just can't do it with every $G_\delta$ $(F_\sigma)$ superset (subset). Also, we can't necessarily do this for arbitrary sets, which may not be measurable.

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It is perhaps worth mentioning that the "and so on" in the first paragraph means "for $\omega_1$ many steps" and not "for countably many steps". –  Trevor Wilson Oct 17 '12 at 3:52
    
Interesting, Trevor! I'd never heard or suspected that. Can you link me to a source? –  Cameron Buie Oct 17 '12 at 4:05
    
Wait...I fumbled my positioning of $B$ and $A$ in the set differences. –  emka Oct 17 '12 at 4:39
    
Ah! In that case, you run into a similar problem, taking $A$ to be the overlying set, and $B$ to be a set not of full measure. I'll correct my answer to reflect that. –  Cameron Buie Oct 17 '12 at 5:06
    
@CameronBuie See en.wikipedia.org/wiki/Borel_hierarchy. The reason is that each of the finite stages you mentioned properly contains the previous ones, so we can pick for each $n$ a set that doesn't appear until stage $n$. Then we can take a single set coding these countably many sets, and this set will not appear at any finite stage. –  Trevor Wilson Oct 17 '12 at 15:55

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