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Let $X,Y\subseteq\Bbb R$ where $\inf\{|x-y|:x\in X\mbox{ and }y\in Y\}>0$. I want to show that $\mu^*(X\cup Y)=\mu^*(X)+\mu^*(Y)$ where $\mu^*$ is denoted to be the outer measure.

Here are my thoughts. Because we are given the inf of $|x-y|$ is greater then zero then $A$ and $B$ should be disjoint. If there disjoint, it means there intersection is empty. Here we can define the union as stated in the problem except we could also subtract there intersection on the right hand side. Thats usually how the probability of a union of two sets is defined, but since that intersection is empty it just goes away and we have what we want. I am a little uncomfortable with this argument. Mainly because outer measure is not additive, so I feel like I am just assuming it is. The help would be appreciated!

Jake Barrows
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    it won't be enough if you only require that $A$ and $B$ are disjoint. If $B$ is a Bernstein set contained in $[0,1]$ and $A$ is its complement in $[0,1]$, then each of them has outer measure $1$, yet their union has measure $1$ (and $1+1\not=1$). For Bernstein sets you may see http://en.wikipedia.org/wiki/Bernstein_set and https://dantopology.wordpress.com/tag/bernstein-set/ and http://math.stackexchange.com/questions/169714/whats-application-of-bernstein-set and http://mathoverflow.net/questions/71575/vitali-sets-vs-bernstein-sets – Mirko Apr 01 '15 at 02:06

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If union of closed intervals $I_k$ contains $A\cup B$ s.t. $$ \sum v(I_k) \leq \mu^\ast (A\cup B) + \epsilon $$

and $${\rm diam}\ (I_k) < \frac{1}{2} d(A,B) $$ where $d(A,B) := \inf \{ |x-y|| x\in A,\ y\in B\} $, then we define $$ F:=\{ I_k\}_{I_k\cap A\neq \emptyset },\ G:=\{ I_k\}_{I_k\cap B\neq \emptyset} $$

Then $$\mu^\ast (A)+\mu^\ast (B) \leq \sum_{I\in F}v(I) + \sum_{I\in G} v(I) \leq \mu^\ast (A\cup B) + \epsilon $$

HK Lee
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