I was doing exercise 18.4.1. from Terence Tao's analysis book that is : if $A$ is a subset of $\Bbb R$, show that $m^*(A) = m^*(A \cap (0,\infty)) +m^*(A\backslash(0, \infty))$. This basically asks us to show that $(0,\infty)$ is Lebesgue measurable.
By sub-additivity of the outer measure, we already have that $m^*(A) \leq m^*(A \cap (0,\infty)) +m^*(A\backslash(0, \infty))$.
It remains to show that $m^*(A) \geq m^*(A \cap (0,\infty)) +m^*(A\backslash(0, \infty))$. I tried to sort some things out of this. Since $m^*(A)=\inf\{ \sum_{j \in J} \text{vol } B_j \lvert \{B_j\} \text{ is countable covering by boxes of } A\}$, then I can show that for any such covering $\{B_j\}$ of $A$, $\sum_{j \in J} \text{vol } B_j \geq m^*(A \cap (0,\infty)) +m^*(A\backslash(0, \infty))$ and it would imply the claim. The right side of this new inequality if made of a sum of two infimums and $\inf(A+B)=\inf(A)+ \inf(B)$. It is becoming very hard to handle.
Is there a way to show this inequality in a cleaner way ? (And is there any advice for trying to show measurability by hand in such questions ?)