If $A \In B$ are both measurable, then $B \RM A$ is measurable, and $m(B \RM A) = m(B) - m(A)$.

We need to show that for any subset $E \In \R^n$… wait a second, do we really need to go by definitions again? After all these preparations, we should be able to exploit our sweat. Hint

• $B \RM A = B \cap A^c$.
• $B = A \sqcup (B \RM A)$, a decomposition into measurable subsets.

Let $\{E_j\}_{j=1}^\infty$ be a countable collection of disjoint subsets. Then, their union $E$ is measurable, and we have $$ m^*(E) = \sum_{j=1}^\infty m^*(E_j} $$

Proof: Tao's proof is really clever, let's first try to go through the proof, then discuss how we can come up with it ourselves.

First, we start by showing $E$ is measurable from the definition: we want to show for any subset $A$, we have $m^*(A) = m^*(A \cap E) + m^(A \RM E)$ Suffice to show $\geq$ direction. Let $F_N = \sum_{j=1}^N E_j$. We have two expressions

• $m^*(A \cap E) \leq \sum_{j=1}^\infty m(A \cap E_j) = \sup_{N > 1} \sum_{j=1}^N m^*(A \cap E_j) = \sup_{N > 1} m^*(A \cap F_N)$
• $m^*(A \RM E) \leq m^*(A \RM F_N)$ for all $N$

Hence, $m^*(A \RM E) + m^*(A \cap E) \leq \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM E)) = \sup_{N > 1} (m^*(A \cap F_N)) + m^*(A \RM F_N)) = \sup_{N > 1} m^*(A) = m^*(A)$