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?
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It is finitely additive for separated sets. A sufficiently small cube can not cut both sets.
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$\begingroup$ Sorry I didn't get what you mean. Do you mean positively separated? And what do you mean by "cutting both sets"? $\endgroup$ Commented May 5, 2012 at 13:03
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1$\begingroup$ One way to compute the Jordan measure is to divide $\mathbb R^n$ into a mesh of small cubes of side $\delta$, and count how many of the cubes cut the set (that is, have non-empty intersection with the set). Multiply by the volume $\delta^n$, take the limit as $\delta \to 0$. If two sets are positively separated, then for small enough $\delta$ this is additive, so in the limit this is additive. $\endgroup$ Commented May 5, 2012 at 15:34
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1$\begingroup$ Cutting a set = the intersection with this set is non empty. You asked for "positively separated sets". I understodd that the infimum of the distances between points one of each set is greater than a certain $d > 0$. I called this separated sets in my answer. So my answer is the same as Edgar. $\endgroup$– juanCommented May 5, 2012 at 15:50