If $d(A, B) > 0$, then it's true that $m^*(A\cup B) = m^*(A) + m^*(B)$. If there are disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$, doesn't it still hold that $m^*(A\cup B) = m^*(A) + m^*(B)$, even in the case $d(A, B) = 0$?
My thinking:
Since subadditivity gives the other direction, we just need to show $m^*(A\cup B) \geq m^*(A) + m^*(B)$. For some $\epsilon>0$, we have a covering of $A\cup B$ by disjoint closed cubes $\bigcup_{j=1}^{\infty} Q_j$, so that $\sum_{i=1}^{\infty} Q_j < m^*(A\cup B) + \epsilon$. If the cubes can be separated into those lying inside $U$ or $V$, we're done. In the case where some cubes intersect both $U$ and $V$, then for such $Q_j$, we can split the interior of $Q_j \cap U$ and $Q_j \cap V$ into coutably many disjoint open cubes. But then we'd have to worry about the boundaries of these sets, which I $think$ we can cover with cubes whose volumes totals no more than $\frac{\epsilon}{4^j}$. Now we are left with disjoint covers of $A$ and $B$ that are very close in volume to $\bigcup_{j=1}^{\infty} Q_j$.
Is this right? If so, is there a slicker proof?
Best Answer
HINT: For every $A$ subset of $\mathbb{R}^n$ we have $$m^{\star}(A)= \inf_{U \textrm{ open } \supset A} m(U)$$
$\bf{Added:}$ In fact, the above statement implies that for every $A$ subset of $\mathbb{R}^n$ we have $$m^{\star}(A)= \inf_{U \textrm{ measurable } \supset A} m(U)$$ Moreover, the equality is achieved for some $U \supset A$, $U$ countable intersection of open sets containing $A$. Therefore, if $A \subset U$, $B\subset V$, and $U$, $V$ are disjoint and ${ \it measurable}$ subsets then $m^*(A\cup B) = m^*(A) + m^*(B)$. Indeed: take $W\supset A \cup B$ measurable so that $m^*(A\cup B) = \mu(W)$. However, $\mu(W) \ge \mu( W \cap (U\cup V) ) = \mu(W \cap U) + \mu ( W \cap V) \ge m^*(A) + m^*(B)$.