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Let $m^*$ the Lebesgue exterior measure. We have by certain observation that :

1) if $E=E_1\cup E_2$ and $d(E_1,E_2)>0$ then $$m^*(E)=m^*(E_1)+m^*(E_2)$$

2) If a set $E$ is the countable union of almost disjoint cubes $E=\bigcup_{i=1}^\infty Q_i$, then $$m^*(E)=\sum_{i=1}^\infty |Q_i|$$

And after I have a remark: Despite observation, one cannot conclude in general that if $E_1\cup E_2$ is a disjoint union of subset of $\mathbb R^d$, then $$m^*(E_1\cup E_2)=m^*(E_1)+m^*(E_2).$$

I definitly don't understand, in one hand we have that $d(E_1,E_2)>0$ (i.e. a union od dijoint set) imply $m^*(E_1\cup E_2)=m^*(E_1)+m^*(E_2)$ and on the other hand it doesn't. So what did I didn't understand ?

Rick
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1 Answers1

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The important thing to notice is, that $d(E_1,E_2)>0$ is much stronger than being disjoint, the set do have some from each other, so the covering cubes for $E_1$ and $E_2$ can be choosen to be disjoint (so that no cube of the covering of $E_1$ does have a non empty intersection with a cube of the covering of $E_2$).

If your sets are only disjoint you can't make something like that, just think of $\mathbb{Q}$ and it's complement in $\mathbb{R}$ (for sure here your equation holds too, but the sets for which doesn't hold will surely be non measurable and hence some kind of weird)

Dominic Michaelis
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