Today we continue going over Tao's sequence of Lemma 7.4.2 - 7.4.11

I will prove the easy case (1-dim), you will do the general case in HW.

Let $E = (0,+\infty)$ be the open half space in $\R$. For any subset $A \In \R$, we need to prove that $$ m^*(A) = m^*(A \cap E^c) + m^*(A \cap E). $$ Let $A_+ = A \cap E$ and $A_- = A \cap E^c$, note that if $0 \in A$, then $0 \in A_-$. First, we show that $$ m^*(A) \leq m^*(A_-) + m^*(A_+)$$. This is because any open cover of $A_-$ and open cover of $A_+$, union together form an open cover of $A$. Next, we need to show that, for any $\epsilon>0$, we have $$ m^*(A) + 2\epsilon \geq m^*(A_-) + m^*(A_+).$$ The plan is the following. Given an open covering of $A$ by intervals $\{B_j\}_{j=1}^\infty$, such that $m^*(A) + \epsilon > \sum_j |B_j|$. We define $$ B_j^+ = B_j \cap (0,\infty) , \quad B_j^- = B_j \cap (-\infty, \epsilon / 2^j) $$ then $\{B_j^+\}$ forms a cover of open interval of $A_+$, similarly $\{B_j^+\}$ forms a cover of open interval of $A_-$. $|B_j^+| + |B_j^-| \leq |B_j| + \epsilon / 2^j$. Thus, we have $$ m^*(A) + 2\epsilon \geq \sum_j |B_j| + \epsilon \geq \sum_j |B_j^+| + \sum_j |B_j^-| \geq m^*(A_+) + m^*(A_-). $$

That finishes the proof for $n=1$ case. How to generalize to higher dimension? In the above discussion, we used the trick of $1 = \sum_j 1/2^j$, which is same trick as we prove outer-measure has countable sub-additivity. We can prove the general $n$ case in two steps, first treat the case of $A$ being an open box, then for general subset $A \In \R^n$, and given an open box cover $\{B_j\}$ of $A$ with $\sum_j |B_j| < m^*(A) +\epsilon$, we have $$ m^*(A) + \epsilon > \sum_j |B_j| = \sum_j m^*(B_j) = \sum_j m^*(B_j \cap E) + m^*(B_j \RM E) \geq m^*(\cup_j(B_j \cap E)) + m^*(\cup(B_j \RM E)) \gep m^*(A \cap E) + m^*(A \RM E) $$