Let $E_1$ be a measurable set and $E_2$ be any bounded set, how to prove $m^*(E_1\cup E_2)+m^*(E_1\cap E_2) = m(E_1)+m^*(E_2)$

measure-theoryreal-analysis

Let $E_1$ be a measurable set and $E_2$ be any bounded set. How to prove
$$m^*(E_1\cup E_2)+m^*(E_1\cap E_2) = m(E_1)+m^*(E_2)?$$
the fact that the outer measure does not have finite additivity caused me a lot of trouble.

Best Answer

Note that $m^* E_1$ is infinite iff $m^*(E_1 \cup E_2) $ is infinite, so the equation holds vacuously in this case.

So we can presume that $E_1$ has finite measure.

For any set $A$ we have $m^*A = m^*(A\cap E_1) + m^*(A \setminus E_1)$.

With $A=E_2$ we get $m^* E_2 = m^*(E_1 \cap E_2) + m^*(E_2 \setminus E_1)$.

With $A=E_1 \cup E_2$ we get $m^*(E_1 \cup E_2) = m E_1 + m^*(E_2 \setminus E_1)$.

Adding gives the desired result.

Boundedness is used to ensure that subtraction is valid.

Related Question