I am trying to argue that exterior measure has nice properties that Jordan outer measure doesn't have. One of them is finite additivity, but I can't find a simple way to show Jordan outer measure is not finitely additive on positively separated sets in $\mathbb{R^n}$? Can someone give me a simple proof or a counter example?



Is Jordan outer measure finitely additive on positively separated sets in $\mathbb{R^n}$?
It is finitely additive for separated sets. A sufficiently small cube can not 
cut both sets. 


