Why isn't every set of finite outer measure Lebesgue measurable if every set can be approximated by an open set? We proved that every subset E of $R^n$ can be approximated arbitrarily by an open set in terms of outer measure. That is $\forall \epsilon$ > 0, there exists an open set $O_\epsilon$ s.t. E $\subset$ $O_\epsilon$, and m*(E) $\le$ m*($O_\epsilon$) $\le$ m*(E) + $\epsilon$, where m* is the outer measure. 
Lebesgue measurable sets were defined as sets E in $R^n$ s.t.  $\forall \epsilon$ > 0,there exists an open set $O_\epsilon$ s.t. m*(O$_\epsilon$\E) < $\epsilon$. 
Wouldn't the first result imply that all sets of finite out measure are Lebesgue measurable? Since if E has finite outer measure, then m*(O$_\epsilon$\E) = m*(O$_\epsilon$) - m*(E) $\le \epsilon$. Is it the strict inequality all that is missing here, or is there something else? 
 A: How do you know that $$m^*(O_\epsilon\setminus E)=m^*(O_\epsilon)-m^*(E)?$$ This equality doesn't hold in general . . . unless $E$ is measurable.

Along similar lines, outer measure isn't even finitely additive in general! Consider a Vitali decomposition $\{V_i: i\in\mathbb{N}\}$ of $[0, 1)$ - that is, the $V_i$s partition $[0, 1)$ and have the same nonzero outer measure $\mu$. Since $\mu>0$, there is some $k\in\mathbb{N}$ such that $k\mu>1$. But then if outer measure were finitely additive, we'd have $m^*(U)>1$ where $U=V_1\cup V_2\cup . . . \cup V_k$; and this clearly contradicts the fact that $U\subset [0, 1)$.
(How do we get a Vitali decomposition in the first place? If you haven't seen this before, it's a neat argument! Consider the equivalence relation $\sim$ on $[0, 1)$ given by $$a\sim b\iff a-b\in\mathbb{Q}.$$ Let $V\subseteq[0, 1)$ be a transversal of this equivalence relation - that is, $V$ contains exactly one element of each equivalence class. Note that in order to get $V$ we need the axiom of choice. Indeed, without AC it is consistent that every set is measurable!
Now, if you imagine wrapping $[0, 1)$ into a circle, the rational translates of $V$ partition $[0, 1)$! Formally, for $q\in\mathbb{Q}\cap [0, 1)$ let $V+q=\{v+q: v\in V\}$, and let $V_q=\{v+q: v+q\in [0, 1)\}\cup\{v+q-1: v+q\in [1, 2)\}$ be the "wrapped around" version of $V+q$. Then you can show that:


*

*The $V_q$s partition $[0, 1)$,

*$m^*(V_p)=m^*(V_q)$ for $p, q\in\mathbb{Q}\cap [0, 1)$ (this is basically translation-invariance of $m^*$), and

*$m^*(V_p)>0$ for every $p\in\mathbb{Q}\cap [0, 1)$ (this uses the facts that $\mathbb{Q}$ is countable, the countable union of measure-zero sets is measure-zero, and $[0, 1)$ is not measure zero).
This is, I believe, the historically first example of a non-measurable set. (There are others.)
