# Why open and closed boxes are measurable

Let $B := \prod_{i=1}^n (a_i,b_i) := \{(x_1,\cdots,x_n) \in \mathbb R^n \mid \forall i: x_i \in (a_i,b_i) \}$. Can someone help me to show that $B$ is a measurable set, i.e. that if $A \subseteq \mathbb R^n$ then $m^*(A) \geq m^*(A \cap B) + m^*(A \cap B^c)$ where $$m^*(E) = \inf \left \{ \sum_{j=0}^\infty vol(B_j) : E \subseteq \bigcup_{i=0}^\infty B_i \text{ where } (B_i)_{i=0}^\infty \text{ at most countable } \right \}$$ and the $B_i$ have to be open boxes. Further is $vol(B) := \prod_{i=1}^n (b_i-a_i)$ for each open box $B$.

-
First see if you can do the case $n=1$. –  GEdgar Dec 29 '12 at 1:50
Yes, I can :D How can I proceed after that ? Can I show that the Cartesian product is measurable ? I.e. $B = (a_1,b_1) \times \cdots \times (a_n,b_n)$ where each $(a_i,b_i)$ is measurable. –  André Dec 29 '12 at 2:12
It would be nice to show that $(a,\infty)^n$ is measurable for all $a \in \mathbb R$.Then an open box $B$ is an intersection of finitly many boxes of the form $(a,\infty)^n$ but I already have proven that finite intersection preserves measurability. –  André Dec 29 '12 at 2:40
It would be nice to show that $(a,\infty)^n$ is measurable for all $a \in \mathbb R$ and $(-\infty,a)^n$, too .Then an open box $B$ is an intersection of finitly many boxes of the form $(a,\infty)^n$ or $(-\infty,a)^n$ but I already have proven that finite intersection preserves measurability. (Edit of the above comment) –  André Dec 29 '12 at 2:46
For the case $n=1$ –  leo Dec 29 '12 at 17:53