# Lebesgue outer measure.

• Oct 27th 2013, 08:45 PM
Rita
Lebesgue outer measure.
Let A http://upload.wikimedia.org/math/4/2...9b30f02246.png Rn and let |A|e be the Lebesgue outer measure.
Suppose A, B http://upload.wikimedia.org/math/4/2...9b30f02246.png Rn and d(A,B)>0. How do I prove that |A http://upload.wikimedia.org/math/d/3...2ec3ce0dac.png B|e=|A|e+|B|e?
• Nov 7th 2013, 01:40 PM
alteraus
Re: Lebesgue outer measure.
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.