equivalence of measurability This is an exercise in real analysis:

Let $E\subset {\Bbb R}^d$. For every $\varepsilon>0$, one can find a Lebesgue measurable set $E_{\varepsilon}$ such that $m^*(E_{\varepsilon}\Delta E)\leq\varepsilon$. Show that $E$ is Lebesgue measurable.

The definition of Lebesgue measurable used here is 

Let $E\subset{\Bbb R}^d$, $E$ is Lebesgue measurable if for any $\varepsilon>0$, there exists an open set $U\supset E$ such that $m^*(U\setminus E)\leq\varepsilon$, where $m^*$ is Lebesgue outer measure. 

Directly using the definition above might be difficult. I'm trying to use the fact that the Lebesgue measurable sets form a $\sigma$-algebra. But I don't see the way to write $E$ as a union of Lebesgue measurable set. Any help?
 A: Here's an alternative approach, working more or less directly from the definitions. Let an arbitrary $\varepsilon>0$ be given; I want to produce an open $U\supseteq E$ with $m^*(U-E)\leq\varepsilon$. Apply the given hypothesis with $\varepsilon/2$ in place of $\varepsilon$ to get a measurable set $E'$ such that $m^*(E\triangle E')\leq\varepsilon/2$. Since $E'$ is measurable, there is an open $V\supseteq E'$ with $m^*(V-E')\leq\varepsilon/2$. Also, by definition of $m^*$, there is an open $W\supseteq E\triangle E'$ with $m(W)\leq\varepsilon/2$.  Let $U=V\cup W$.  Then $U$, being the union of two open sets, is open. Also, $U\supseteq E$; indeed, for any $x\in E$, if $x\in E'$ then $x\in V$, and otherwise $x\in E\triangle E'\subseteq W$, so in either case $x\in U$.  It remains to prove that $m^*(U-E)\leq\varepsilon$, and for this purpose it suffices to show that $U-E\subseteq(V-E')\cup W$, because both $V-E'$ and $W$ have (outer) measure $\leq\varepsilon/2$ and outer measure is subadditive and monotone.  So consider any $x\in U-E$; I want to show that $x$ is in (at least one of) $V-E'$ or $W$.  By definition of $U$, $x$ is in $V$ or in $W$, and if it's in $W$ then we're done, so from now on suppose $x\in V$.  If $x\notin E'$ then $x\in V-E'$ so we're again done.  There remains only the case that $x\in E'$.  But $x\notin E$, so $x\in E\triangle E'\subseteq W$, and we're again done.
A: For any $\varepsilon>0$, there exists a family of L-measurable sets $\{E_{\varepsilon,n}\}_{n=1}^{\infty}$ such that
$$
m^*(E_{\varepsilon,n}\Delta E)\leq \varepsilon/2^n
$$
Let 
$$
E_{\varepsilon}=\cup_{n=1}^{\infty}E_{\varepsilon,n}.
$$
Then $E_{\varepsilon}$ is L-measurable and
$$
E\setminus E_{\varepsilon}\subset E\setminus E_{\varepsilon,n}
$$
for all $n$. By monotonicity of Lebesgue outer measure,
$$
m^*(E\setminus E_{\varepsilon})\leq m^*( E\setminus E_{\varepsilon,n})\leq \varepsilon/2^n
$$
Thus $m^*(E\setminus E_{\varepsilon})=0$ which implies that $E\setminus E_{\varepsilon}$ is L-measurable. It follows that for any $\varepsilon>0$, we have a L-measurable set
$$
E'_{\varepsilon}=(E\setminus E_{\varepsilon})\cup E_{\varepsilon}\supset E
$$
with $m^*(E'_{\varepsilon}\setminus E)=m^*(E_{\varepsilon}\setminus E)\leq\sum_{n=1}^{\infty} \varepsilon/2^n=\varepsilon$. 
Now consider $E'_{1/n}$ and let $E'=\cap_{n=1}^{\infty}E'_{1/n}$. Then $E'$ is L-measurable and $E'\supset E$. We also have 
$$
m^*(E'\setminus E)\leq m^*(E'_{1/n}\setminus E)\leq 1/n
$$
for all $n$. Thus $m^*(E'\setminus E)=0$ and hence $E'\setminus E$ is L-measurable. It follows that $E=E'\setminus(E'\setminus E)$ is L-measurable.  Q.E.D
