Outer measure of Symmetric Difference Zero implies Measurability

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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!

 

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