## Outer measure of Symmetric Difference Zero implies Measurability

Just came across this neat beginner’s Lebesgue Theory question. As students of analysis know, just to show a set is measurable is no easy feat. The usual way is to use the Caratheodory definition, where  a set E is said to be measurable if for any set A, $m^*(A)=m^*(A\cap E)+m^*(A\cap E^c)$. This can be quite tedious.

Question: Suppose E is a Lebesgue measurable set and let F be any subset of $\mathbb{R}$ such that $m^*(E\Delta F)=0$ (Symmetric Difference is Zero). Show that F is measurable.

The short way to do this is to note that $m^*(E\Delta F)=0$ implies $m^*(E\setminus F)=0$, and $m^*(F\setminus E)=0$. This in turn (using a lemma that any set with outer measure zero is measurable) implies the measurability of $E\setminus F$ and $\setminus F\setminus E$.

Next comes the critical observation: $\boxed{E\cap F=E\setminus (E\setminus F)=E\cap (E\setminus F)^c}$. Using the fact that the collection of measurable sets is a $\sigma$-algebra, we can conclude $E\cap F$ is measurable.

Thus $F=(E\cap F)\cup (F\setminus E)$ is the union of two measurable sets and thus is measurable.

Interesting indeed! 