You showed that $l^*(E) = \inf l^*(\mathcal{O})$ (the infimum runs over all open sets $\mathcal{O}$ containing $E$).
Suppose $E$ is of finite exterior measure -then for any $n\in \mathbb{N}$ we have an open set $\mathcal{O_n}$ such that
$$l^*(\mathcal{O_n}) \leq l^*(E)+n^{-1}<\infty$$
And since $l^*(\mathcal{O_n}) = l^*(E)+l^*(\mathcal{O_n} \setminus E)$ we have $l^*(\mathcal{O_n} \setminus E) \leq n^{-1}$ (The equality is infact a restatement of the measurability of $E$, see Equivalent Definition of Measurable set)
Taking $\mathcal{O} = \bigcap_{n=1}^\infty \mathcal{O_n}$ we have come up with a $G_\delta$ set (countable intersection of open sets) which is obviously a borel set, with the property: $$l^*(\mathcal{O}\setminus E) \leq l^* (\mathcal{O_n}\setminus E) \leq n^{-1}$$ for all $n\in\mathbb{N}$ which is the first corollary.
If $E$ is of infinte exterior measure then we denote $E_n = E \cap B(0,n)$ ($E$'s intersection with the ball of radius $n$ centered at the origin). Each $E_n$ is bounded and so has finite exterior measure (Since it obviously is encompassed within $B(0,n)$) And we may extract $G_n$ sets such that $$l^*(G_n \setminus E_n)=0$$ Putting $G=\bigcup_{n=1}^\infty G_n$ we have $$l^*(G\setminus E) = l^*\left(\bigcup_{n=1}^\infty G_n \setminus \bigcup_{n=1}^\infty E_n \right)\leq \sum_{n=1}^\infty l^*(G_n \setminus E_n ) =0$$
This shows the first corollary.
For the second: For every measurable $E^C$ we obviously have a borel set $G\supset E^C$ such that $l^*(G\setminus E^C) =0$ (Take the set from the first claim) that shows that $G\setminus E^C$ is a lebesgue measurable set as it is a null set.
Then obviously $E^C= G \setminus (G\setminus E^C)= G \cap (G\cap E)^C$. Taking completements on both sides we have:$$ E = G^C \cup (G\setminus E^C)$$
$G^C$ is again a borel set (As a completement of borel set).
As a final comment on this excercise: One notices we used 2 main properties of the Lebesgue measure:
- It is outer regular meaning that for every measurable subset $E$ , for every $\varepsilon>0$ we have an open set $\mathcal{O}\supset E$ with the property $l^*(\mathcal{O} \setminus E) \leq \varepsilon$ which we had to prove.
- It is defined on borel sets, bounded sets have finite exterior-measure.
It turns out that the second implies the first and the first property is what we actually needed. In this proof I made a slight detour (as i wasn't sure which Lebesgue measurability criterion you are using).
The following turns my comments above into an answer (modulo exercises :P).
Throughout, I'll assume that $X$ is a non-measurable set which has a positive-measure measurable subset.
The process in the OP is too loosely-defined to be guaranteed to terminate - maybe we just happen to keep grabbing "inefficient" subsets of our starting set. Instead, it's better to try to carve out a "maximal(ish) measurable subset" (and think about what's left over).
Note my careful use of the suffix "ish" in the above. There is not going to be a literally maximal measurable subset of $X$: if $A\subseteq X$ is measurable and $a\in X\setminus A$, then $A\cup\{a\}$ is also a measurable subset of $X$. Similarly, we can't do something like taking the union of all the positive-measure subsets of $X$: that will just yield $X$ itself (again thinking about how singletons don't affect measurability), which isn't measurable. This second obstacle brings up a related issue, namely that measurability is fragile: if I take a union of a bunch of measurable sets, the result may not be measurable. Only countable unions are safe: the union of countably many measurable sets is always guaranteed to be measurable.
So we want to find a measurable subset of $X$ which in some sense is as big as possible, but $(i)$ that sense has to be somewhat nuanced and $(ii)$ we have to only take a small number (= countably many) of unions of measurable pieces. What can we do?
Well, besides literal maximality, we do have at hand a way of measuring when something is as big as possible: measure! Say that $A\subseteq X$ is a quasimaximal-measurable subset of $X$ (not an actual term) iff $A$ is measurable and, for every measurable $B$ with $A\subseteq B\subseteq X$, we have $m(A)=m(B)$ (or equivalently, if $B$ is any measurable subset of $X$ we have $m(A)\ge m(B)$). So maybe $A$ isn't literally as big as possible, but we can't do any better measure-wise.
Using basic facts about measurability, you should be able to prove the following statements:
If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ is non-null.
If $A$ is a quasimaximal-measurable subset of $X$, then $X\setminus A$ does not have a positive-measure measurable subset.
At this point you're basically done ... except that you need to prove that quasimaximal-measurable subsets exist at all! HINT: consider the number $$\alpha=\sup\{m(A): A\subseteq X\mbox{ is measurable}\}$$ (for simplicity let's assume $\alpha<\infty$). Can you see how to build a measurable subset of $X$ with measure exactly $\alpha$ (think about countable unions!), and why such a subset must be quasimaximal-measurable?
Best Answer
Yes. A standard construction goes by building the set by transfinite recursion to ensure that neither $A$ nor its complement contains a perfect set. Note that if this is the case, then any measurable subset of $A$ has measure $0$, because (by regularity of Lebesgue measure) if $E$ is measurable, $\lambda(E)$ is the supremum of $\lambda(K)$ for $K$ compact contained in $E$. But any compact set that has no perfect subset is in fact countable, so it has measure $0$. The same happens with $A^c$, but then neither $A$ nor $A^c$ can be measurable: If $A$ is measurable, then it has measure $0$, so $A^c$ has full measure and therefore contains a perfect set.
The transfinite recursion goes as follows: There are $\mathfrak c=|\mathbb R|$ many perfect subsets of $\mathbb R$. Identifying cardinals with ordinals, list them in a well-ordering of type $\mathfrak c$. We build $A$ and its complement by recursion. At stage $\alpha$ we have a set $A_{<\alpha}$ and a disjoint set $B_{<\alpha}$. (When $\alpha=0$, both are empty.) Consider the $\alpha$-th perfect set. Pick two elements of the perfect set not in $A_{<\alpha}\cup B_{<\alpha}$. Let $A_{\alpha}=A_{<\alpha+1}$ be the union of $A_{<\alpha}$ and one of these points, and let $B_{\alpha}$ be the union of $B_{<\alpha}$ and the other point. That the points exist follows from the fact that there are $\mathfrak c$ reals in any perfect set, but at stage $\alpha$ at most $2|\alpha|<\mathfrak c$ of them have been used. (When $\lambda$ is limit, $A_{<\lambda}$ is just the union of the previous $A_\alpha$, same for $B_{<\lambda}$.) At the end, one of the sets so built is $A$. This works, because by construction, given any perfect set $P$, at least one of its points is in $A$ (so $P$ is not contained in $A^c$), and at least one of its points is not (so $P$ is not contained in $A$).
The details of the construction can be varied a bit, but any argument must use choice in a nontrivial way, since it is consistent with the usual axioms of set theory (without the axiom of choice) that all sets of reals are measurable. That said, I haven't seen a presentation of the result I'm describing that avoids the use of a well-ordering of $\mathbb R$. The sets so obtained are called Bernstein sets, and were first exhibited by Felix Berenstein (of Bernstein-Schroeder fame) in $1908$.