# A compact in a measurable set

The problem is as follows:

Given a Lebesgue-measurable set $M\subset\mathbb{R^n}$ find a compact $C_\epsilon$ and an open $A_\epsilon$, $C_\epsilon\subset M\subset A_\epsilon$, such that $\lambda(A_\epsilon)-\lambda(C_\epsilon)<\epsilon$.

$\lambda$ is the Lebesgue-measure, in particular, if $M$ is not bounded, $k\in\mathbb{N}$, $$\lambda(M)=\lim_{k\to+\infty}\lambda(M\cap[-k,k]^n).$$

Now, if $M$ is bounded I can find $A_\epsilon$ and $C_\epsilon$ with the required properties.

if $M$ is not bounded but $\lambda(M)<\infty$, how can I find the open and the compact?

Furthermore, if $\lambda(M)=\infty$, I think we can not found a compact such that the difference between its measure and the measure of $M$ is small, is it correct?

If $M$ is unbounded and $\lambda(M)<\infty$, start by taking a $k>0$ such $\lambda(M)-\lambda\bigl(M\cap[-k,k]\bigr)<\frac\varepsilon2$.

If $\lambda(M)=\infty$, then you are right: there is no such compact set.

• thanks for your answer. I thought about two proofs.... – Jack J. May 23 '18 at 19:42
• @JackJ. They look fine. – José Carlos Santos May 23 '18 at 21:37

1) For all $k\in\mathbb{N}$, let a compact $C_k\subset M\cap [-k,k]^n$ such that $\lambda(M\cap [-k,k]^n\setminus C_k)<\frac{\epsilon}{2^{k+1}}$.
Since, $\lim_{k\to\infty}\lambda(M\cap[-k,k]^n)=\lambda(M)$, $\exists\bar{k}\in\mathbb{N}$ such that for $k>\bar{k}$ $$\lambda(M)-\lambda(M\cap[-k,k]^n)<\frac{\epsilon}{2}.$$
Now, for $k>\bar{k}$ we consider $$\begin{split} \lambda(M)-\lambda(C_k) &= \lambda(M)-\lambda(M\cap[-k,k]^n)+\lambda(M\cap[-k,k]^n)-\lambda(C_k) \\ & = \lambda(M)-\lambda(M\cap[-k,k]^n)+\lambda(M\cap[-k,k]^n\setminus C_k) \\ &< \frac{\epsilon}{2}+\sum_{k=1}^{\infty}\frac{\epsilon}{2^{k+1}}=\epsilon \end{split}$$
2) Since, $\lim_{k\to\infty}\lambda(M\cap[-k,k]^n)=\lambda(M)$, $\exists\bar{k}\in\mathbb{N}$ such that for $k>\bar{k}$ $$\lambda(M)-\lambda(M\cap[-k,k]^n)<\frac{\epsilon}{2}.$$ At this point just take $M\cap[-k,k]^n$, for same $k>\bar{k}$ and to observe that $M\cap[-k,k]^n$ is bounded.