I am trying to do exercise 7.4.3. in Tao's Analysis II: Prove that the half space $E:=\{(x_1,\cdots, x_n)\in \mathbb{R}^n| x_n>0\}$ is measurable. i.e. $m^{\ast}(A)=m^{\ast}(A\cap E)+m^{\ast}(A\setminus E)$ for any subset $A\subseteq \mathbb{R}^n.$
The hint he gave is to first prove that if $A$ is an open box $(a_1,b_1)\times \cdots \times (a_n,b_n)$ in $\mathbb{R}^n$, and $E$ is the half-plane $E:=\{(x_1,\cdots, x_n)\in \mathbb{R}^n| x_n>0\}$, then $m^{\ast}(A)=m^{\ast}(A\cap E)+m^{\ast}(A\setminus E)$.
I was able to prove the hint but I am not sure how it helps to prove half spaces are measurable.
I have to show that for any set $A$, we have $m^{\ast}(A)=m^{\ast}(A\cap E)+m^{\ast}(A\setminus E)$. We have showed that it is true if $A$ is an open box but I don't know how to extend that to arbitrary sets. Any help would be appreciated.Thanks.
Recall we define the outer measure by $$m^{*}(A)=\text{inf} \sum_{n \geq 1} m(I_n)$$
where the inf is taken over all families of open boxes $(I_n)_{n \geq 1}$ that cover $A$(or at least I hope that's how Tao defines it, as I've not read his book. This seems pretty standard though)
Let $(I_n)$ be such a family and note, $A\subset \bigcup I_n$ imples $A\setminus E \subset \bigcup(I_n \setminus E)$ and similarly $A\cap E \subset \bigcup(I_n \cap E)$.
Recall also the outer measure is increasing and subadditive. In our case, $$m^{*}(A\cap E)\leq m^{*}(\bigcup (I_n\cap E))\leq \sum m^{*}(I_n\cap E))$$ and of course
$$m^{*}(A\setminus E)\leq m^{*}(\bigcup (I_n\setminus E))\leq \sum m^{*}(I_n\setminus E))$$
Because we're working with non-negative terms we're free to regroup and rearrange the sum as we please. So we find $$m^{*}(A\cap E)+m^{*}(A\setminus E)\leq \sum (m^{*}(I_n \cap E)+m^{*}(I_n \setminus E))$$
But the point is now $I_n$ is a box for every $n$ so as you've proved $$(m^{*}(I_n \cap E)+m^{*}(I_n \setminus E))=m^{*}(I_n)=m(I_n)$$ which together with the above inequality yields $$m^{*}(A \cap E)+m^{*}(A \setminus E)\leq \sum m(I_n)$$
You can now take inf over all sequences of boxes $(I_n)$ that cover $A$ to find $$m^{*}(A \cap E)+m^{*}(A \setminus E)\leq m^{*}(A)$$
Finally, $$m^{*}(A) \leq m^{*}(A \cap E)+m^{*}(A \setminus E)$$ follows by subadditivity of the outer measure, so equality must actually hold, proving $E$ is measurable.