2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 $n$. $\endgroup$ – drhab Oct 25 '18 at 14:10
0
$\begingroup$

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 $G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.