# 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?

• You are on the right track. If the $O_n$ are Borel sets then so is their intersection. If $O$ denotes this intersection then $E\subseteq O$ and $\lambda(O)\leq\lambda(O_n)$ for every $n$. – drhab Oct 25 '18 at 14:10

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 $$G$$.