If $E$ is Lebesgue-measurable, then Borel sets exist with $B_1 \subset E \subset B_2$ Let $\lambda$ be the Lebesgue measure on $\mathbb{R^n}$.
How to prove that for all Lebesgue-measurable sets $E \subset \mathbb{R^n}$ Borel sets $B_1,B_2 \subset \mathbb{R^n}$ exist with $B_1 \subset E \subset B_2$ and $\lambda(B_2$ \ $B_1)=0$ ?
To prove that $B_1$ and $B_2$ exist, I tried it with the definition:
Let $\lambda ^*(E)$ be the outer measure of $E$ with $\inf \left\lbrace \lambda(O):O \subset E\mbox{ open}\right\rbrace$.
Then an open set $E \subset O_n$ exists with $\lambda(O_n)<\lambda^*(E)+\frac{1}{n}$. Here I don't know how to continue.
Is this step right? Or how can it be shown?
 A: Suppose first that $E$ is of finite measure. Since $E$ is measurable, for each $\varepsilon >  0$ there exists by definition $E \subseteq G$ open with $|G \setminus E| < \varepsilon$. Similarly we have $F \subseteq E$ open with $|E \setminus F| < \varepsilon$ (just observe that $E^\complement$ is measurable and use the former). Thus, for each $n \geq 1$ we have $F_n \subseteq E \subseteq G_n$ with $F_n$ closed, $G_n$ open and $|E\setminus F_n|,|G_n \setminus E| < \frac{1}{n}$. Now set $F = \bigcup_n F_n$ and $G = \bigcap_n G_n$ which are $F_\sigma$ and $G_\delta$ respectively (in particular, Borel sets), and note that 
$$
|E \setminus F| \leq |E \setminus F_n| < \frac{1}{n} \to 0
$$
and 
$$
|G \setminus E| \leq |G_n \setminus E| < \frac{1}{n} \to 0.
$$
So we have $|G \setminus F| = |G| - |F| = 0$, because $|G| = |E| = |F|$. 
Finally, suppose that $E$ has infinite measure an let $E_n = E \cap \overline{B}(0,n)$. By the previous result, we have Borel sets $F_n \subseteq E_n \subseteq G_n$ with $|F_n| = |E| = |G_n|$ and so if $F = \bigcup_nF_n$, $G = \bigcup_n G_n$, we get $F \subseteq E \subseteq G$. To conclude, recall that since $F$ and $G$ are increasing union of measurable sets,
$$
|F| = \lim_n |F_n| = \lim_n|E_n| = |E| = +\infty
$$
and similarly for $G$.
