Let A R^{n} and let |A|_{e} be the Lebesgue outer measure.
Suppose A, B R^{n} and d(A,B)>0. How do I prove that |A B|_{e}=|A|_{e}+|B|_{e}?
Having thought about this problem for days, I still can't solve it. Any suggestion?
Hi, I would try this:
If $\displaystyle 0<d(A,B)=\inf\{\delta (x,y)\,;\,x\in A\,,\,y\in B\}$ where $\displaystyle \delta (x,y)=\sqrt{\sum_{i=1}^n |x_i-y_i|^2} $
then there exists open set $\displaystyle G\subset\mathbb{R}^n$ such that :
1. $\displaystyle A\subset G$
2. $\displaystyle B\cap G=\emptyset $ .
Since every open set in $\displaystyle (\mathbb{R}^n, \delta) $ is Lebesgue measurable, there holds
$\displaystyle \forall S\subset\mathbb{R}^n \,:\,|S|_e=|S\cap G|_e + |S - G|_e $.
Rewriting this for $\displaystyle S=A\cup B$ and using properties 1,2 of set $\displaystyle G$ seems to be a good way to solve your problem.