Lebesgue measureable --> outer measure = 0

Question

Let $A \subset \mathbb R^n$.

Why does follow from "$A$ is Lebesgue measurable" that "$\exists A_1 \in F_{\sigma}, A_2 \in P(\mathbb R^n)$ such that $\lambda_n^* (A_2) = 0$ and $A=A_1 \cup A_2$"?

I guess that it has to do with the regularity of the Lebesgue measure.

Answer 1

First assume that $A$ is bounded. By regularity of Lebesgue measure there exist compact sets $K_n \subset A$ such that $m(A)< m(K_n)+\frac 1 n$. Take $A_1=\cup_n K_n$ and $A_2=A\setminus \cup_n K_n$. For the general case apply the result to $A \cap [-N,N]$ for each $N$. I will leave the rest to you.