Example of two unbounded sets separated by a positive value that are not finitely additive It is true that given bounded $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{R}$ with some $\gamma>0$ such that $\forall a \in A$ and $\forall b\in B$ $|a-b|≥\gamma$, then $m^{\star}(A \bigcup B) = m^{\star}(A) + m^{\star}(B) $ where $m^{\star}$ denotes Lebesgue outer measure.
I'm curious if there is a quick example of unbounded sets $A \subseteq \mathbb{R}$ and $B \subseteq \mathbb{R}$ where this fact fails.
 A: Here's a proof for the fact that the Lebesgue outer measure, $m^*$, on $\Bbb{R}^n$ is a metric outer measure.
Let $A,B\subset \Bbb{R}^n$ be non-empty sets with $d(A,B) := \inf\{d(a,b)| \, a \in A, b \in B\} > 0$ ($d$ being the standard metric on $\Bbb{R}^n$). Now, put $\delta := \frac{d(A,B)}{4}$ (you could probably get away with a $2$ in the denominator, but this certainly doesn't hurt). Now, recall that one possible definition of the Lebesgue outer measure of a set $S\subset\Bbb{R}^n$ is as the infimum over the total volume of covers of $S$ by open rectangles. But actually, it doesn't matter whether we use open rectangles or closed rectangles, or even half-open rectangles in the definition. All of these will turn out to give the same number.
So, given $\epsilon>0$, we can find a countable cover $\{H_j\}_{j=1}^{\infty}$ of the set $A\cup B$ using $n$-dimensional left-closed, right-open rectangles (i.e products of $n$ intervals of the form $[\alpha,\beta)$) such that
\begin{align}
\sum_{j=1}^{\infty} \text{vol}(H_j) \leq m^*(A\cup B) + \epsilon
\end{align}
Now, a-priori, the rectangles $H_j$ might be "too large", but by chopping them up enough times along the coordinate axes, we can write each $H_j$ as a finite disjoint union of left-closed right-open rectangles, each of which has diameter $\leq\delta$ (for example, think of the interval $[0,5)$, and suppose $\delta = 2$; then we can chop it up as $[0,2)\cup [2, 4)\cup[4,5)$). Since a finite union of countable sets is still countable, we lose no generality by supposing each of the $H_j$ itself has diameter $\leq \delta$.
Now, I leave it to you to prove (using the fact each $H_j$ has diameter $\leq \delta$ and the definition of $\delta$) that if $H_j\cap A \neq \emptyset$ then $H_j \cap B = \emptyset$ (and similarly if the roles of $A,B$ are reversed). Thus, we have "disjoint subsequences" of rectangles $\{H_{A_k}\}_{k=1}^{\infty}$ and $\{H_{B_k}\}_{k=1}^{\infty}$ which cover $A$ and $B$ respectively. Thus,
\begin{align}
m^*(A) + m^*(B) &\leq \sum_{k=1}^{\infty} \text{vol}(H_{A_k}) + \sum_{k=1}^{\infty} \text{vol}(H_{B_k}) \\
&\leq \sum_{j=1}^{\infty} \text{vol}(H_j) \\
&\leq m^*(A\cup B) + \epsilon.
\end{align}
Since $\epsilon>0$ was arbitrary, we actually get the inequality $m^*(A)+m^*(B)\leq m^*(A\cup B)$; the other inequality is trivially true by subadditivity of outer measures. This completes the proof. Notice how the proof does not rely on $A$ or $B$ being bounded; it is true even without that assumption.
