Approximate a measurable set with an elementary set of equal measure?

Question

Let $A \subset [0,1]$ be a measurable set. we know that A is measurable iff for every $\varepsilon>0$ there is an elementary set B such that $\mu(A \Delta B) <\varepsilon$ Where $A\Delta B$ is their symmetric difference.

For a given $\varepsilon$. Can we choose B such that $\mu(B) =\mu(A) $?

Answer 1

You can do this. We have $\mu (A\Delta B) \leq \mu (A\Delta C)+\mu (B\Delta C)$. So first make the first term less than $\epsilon /2$ and then increase or decrease the length of one of the intervals in $C$ to make $\mu (A)=\mu (B)$. Use the fact that $|\mu(A)-\mu (C)| <\epsilon /2$.