Hausdorff measure with union and intersection Is it true that 
$$\mathcal{H}^{n}(\partial(E \cup F)) + \mathcal{H}^{n}(\partial(E \cap F)) \leq \mathcal{H}^{n}(\partial E) + \mathcal{H}^{n}(\partial F),$$
where $E$ and $F$ are two sets and $\mathcal{H}^n$ is the $n$-dimensional Hausdorff measure?
I think so, but I can't convince myself and I don't know much about measure theory!
Thanks
 A: 
It's more a question of topology than of measure theory. Measure theory only plays into it for the equality $\mu(A) + \mu(B) = \mu(A\cup B) + \mu(A\cap B)$ and the inequality $C\subset D \implies \mu(C) \leqslant \mu(D)$. -- Daniel Fischer

For completeness, here's the topology part. I write $\operatorname{int} E$ and $\operatorname{ext}E$ for the interior and exterior. 


*

*$\partial (E\cup F) \subset (\partial E \cup \partial F)\setminus (\operatorname{int}E   \cup   \operatorname{int}F )=:A$, saying that a boundary point of the union has to be on the boundary of one of two sets and cannot be interior for the other. 

*$\partial (E\cap F) \subset (\partial E \cup \partial F)\setminus (\operatorname{ext}E   \cup   \operatorname{ext}F )=:B$, saying that a boundary point of the intersection has to be on the boundary of one of two sets and cannot be exterior for the other. 

*$A\cup B=\partial E\cup \partial F$ since a point of $\partial E\cup \partial F$ can't get subtracted in both $A$ and $B$. 

*$A\cap B=\partial E\cap \partial F$, which is clear.
Thus, 
$$\mu(\partial (E\cup F))+\mu(\partial (E\cap F)) \le \mu(\partial E) +\mu(\partial F)$$
for any measure $\mu$.
