Lebesgue Measure – Lebesgue Measurable if and only if Outer Measure = 0

Question

Let $A \subset \mathbb R^n$. Then the following two statements hold true:

• $A$ is Lebesgue-measurable.
• It exists $B \in F_{\sigma}$ such that $\lambda_n^* ((A \setminus B) \cup (B \setminus A)) = 0$

Why?

(Where $\lambda_n^*$ is the outer measure.)

Answer 1

If exists an $F_{\sigma}$ set $B$ satisfying the condition then you have that $m(A \setminus B)=m(B \setminus A)=0$

So $A \setminus B,B\setminus A$ are measurable.

Also $ (A \cap B)=B \setminus (B\setminus A) $

So $(A \cap B)$ is measurable.

From this you can conclude that $A$ is measurable.

The other part is just the regularity of the Lebesgue measure which you can find it in many textbooks.