Lebesgue outer measure of disjoint sets in $\mathbb{R}^n$ 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?
 A: 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, $\mu(W)  \ge \mu( W \cap (U\cup V) ) = \mu(W \cap U) + \mu ( W \cap V) \ge m^*(A) +  m^*(B)$.
A: Your thinking is OK and it can be developed into a complete proof. 
Suppose there are disjoint open sets $U, V$ such that $A \subset U$ and $B \subset V$. For each $x\in A$, let $C_x$ be an open cube such that $x\in C_x$ and $\overline{C_x}\subseteq U$. We have that $\{C_x \,|\, x\in A\}$ is open cover of $A$. Since $\mathbb{R}^n$ is separable, there is a countable sub-cover $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$. Note that 
$$A\subseteq \bigcup_{n\in \mathbb{N}} C_{x_n}  \subseteq  \bigcup_{n\in \mathbb{N}}\overline{ C_{x_n}} \subseteq U$$
In a similar way, we can find a countable family $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$ of open cubes such that 
$$B\subseteq \bigcup_{n\in \mathbb{N}} D_{y_n}  \subseteq  \bigcup_{n\in \mathbb{N}}\overline{ D_{y_n}} \subseteq V$$
Since $U$ e $V$  are disjoint, we can easily use $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$ and $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$ (or, if you prefer, $\{\overline {C_{x_n}} \,|\, x_n\in A, n\in \mathbb{N} \}$ and $\{\overline{D_{y_n}} \,|\, y_n\in B, n\in \mathbb{N} \}$) to prove that 
$$m^*(A\cup B) = m^*(A) + m^*(B)$$
In fact, let us show the details. 
From subadditivity, we know that $m^*(A\cup B) \leq m^*(A) + m^*(B)$. we just need to show $m^*(A\cup B) \geq m^*(A) + m^*(B)$. 
If $m^*(A\cup B) =+\infty$ then it is clear that $m^*(A\cup B) \geq m^*(A) + m^*(B)$. 
Suppose $m^*(A\cup B) <+\infty$. For any $\epsilon>0$, there is a countable collection of cubes $\{Q_j\}_{j\in\mathbb{N}}$, such that $A\cup B\subseteq \bigcup_{j=0}^{\infty} Q_j$ and 
$$\sum_{i=0}^{\infty} m(Q_j) < m^*(A\cup B) + \epsilon \tag{1}$$ 
Since the family of cubes in $\mathbb{R}^n$ is a semi-ring, then from $\{C_{x_n} \,|\, x_n\in A, n\in \mathbb{N} \}$, we can have $\{E_k \,|\, k\in \mathbb{N} \}$ a collection of disjoint cubes, such that $A\subseteq \bigcup_{n\in \mathbb{N}} C_{x_n}=\sum_{k\in \mathbb{N}} E_k$.  And from $\{D_{y_n} \,|\, y_n\in B, n\in \mathbb{N} \}$, we can have $\{F_l \,|\, l\in \mathbb{N} \}$ a collection of disjoint cubes, such that $B\subseteq \bigcup_{n\in \mathbb{N}} D_{y_n}=\sum_{l\in \mathbb{N}} F_l$. Note that, for all $k,l  \in  \mathbb{N}$, $E_k\subseteq U$ e $F_l\subseteq V$, so $E_k \cap F_l=\emptyset$. 
Now, we have 
$A\subseteq \sum_{j,k=0}^\infty(Q_j\cap E_k)$
and 
$$m^*(A)\leq  \sum_{j,k=0}^\infty m(Q_j\cap E_k) \tag{2}$$
(remember that the inersection of two cubes is a cube). 
And we also have 
$B\subseteq \sum_{j,l=0}^\infty(Q_j\cap F_l)$
and 
$$m^*(B)\leq  \sum_{j,l=0}^\infty m(Q_j\cap F_l) \tag{3} $$
Combining $(1)$, $(2)$ and $(3)$, we have 
$$m^*(A)+m^*(B)\leq  \sum_{j,k=0}^\infty m(Q_j\cap E_k)+ \sum_{j,l=0}^\infty m(Q_j\cap F_l)\leq\sum_{i=0}^{\infty} m(Q_j) < m^*(A\cup B) + \epsilon$$ Since this is valid for an arbitrary $\epsilon>0$, we have 
$$m^*(A)+m^*(B)\leq  m^*(A\cup B)$$
Remark: IF you already know that, for all $A$ subset of $\mathbb{R}^n$, we have 
$$m^{\star}(A)= \inf_{U \textrm{ open } \supset A} m(U)$$
THEN there are much shorter and elegant proofs (see, for instance, orangeskid's answer). 
A: WLOG $m^*(A), m^*(B) < \infty$, otherwise both sides are infinite.
Choose open $W_n\supseteq A \cup B$ decreasing with $m(W_n) \searrow m^*(A \cup B)$. Then $W_n \cap U \supseteq A$ and $W_n \cap V \supseteq B$.
Thus
$$
m^*(A \cup B) = \lim_n m(W_n) \geq \limsup_n [m(W_n \cap U) + m(W_n \cap V)]\geq m^*(A) + m^*(B).
$$
Of course the other inequality is given immediately by subadditivity.
