Lebesgue measure - upper half plane

Lebesgue measure - upper half plane

Let $A := \prod_{i=1}^n (a_i,b_i)$ and $E := \{(x_1,\cdots,x_n) \mid x_i >
0 \} = \prod_{i=1}^n (0,\infty)$. Then we have $$ m^*(A) = m^*(A \cap E) +
m^*(A \cap E^c) $$ where $m^*$ is the outer Lebesgue-measure.
I first proved this for $n = 1$. How can I expand this to the general case
? Induction ?



Maybe this proof is not clean but here it is.
Induction on $n$. $n=1$ is proven. Assume it holds for $n-1$. Define
$\tilde A := \prod_{i=1}^{n-1} (a_i,b_i)$ and $\tilde E :=
\prod_{i=1}^{n-1} (0,\infty)$. Let further $I:= (0,\infty)$. By the
induction hypothesis we know $$ m^*( \prod_{i=1}^{n-1} (a_i,b_i) \cap I) =
m^*(\tilde A \cap \tilde E) = m^*(\tilde A)-m^*(\tilde A \cap \tilde E^c)
\leq m^*(\tilde A) $$ Similar $$ m^*( \prod_{i=1}^{n-1} (a_i,b_i) \cap
I^c) \leq m^*(A) $$ This gives \begin{align*} m^*(A \cap E) + m^*(A \cap
E^c) \\ = m^* \left ( \prod_{i=1}^n (a_i,b_i) \cap I \right ) + m^* \left
( \prod_{i=1}^n (a_i,b_i) \cap I^c \right ) \\ \leq m^* \left (
\prod_{i=1}^{n-1} (a_i,b_i) \cap I \right ) m^*((a_n,b_n) \cap I) +m^*
\left ( \prod_{i=1}^{n-1} (a_i,b_i) \cap I^c \right) m^*((a_n,b_n) \cap
I^c ) \\ \leq m^*(\tilde A)m^*((a_n,b_n)\cap I) + m^*(\tilde
A)(m^*((a_n,b_n)\cap I^c)) \\ = m^*(\tilde A)m^*((a_n,b_n)) = m^*(A)
\end{align*} For the 2nd to 3rd line I use that $m^*(A \times B) \leq
m^*(A)m^*(B)$ and further that $$ m^*\left ( \left ( \prod_i A_i \right )
\cap \left( \prod_i B_i \right ) \right ) = m^* \left ( \prod_i A_i \cap
B_i \right) $$