Is this result true? I cannot find counterexample Suppose $Q$ is a rectangle in $\mathbb{R}^n$. Suppose $\{Q_1,Q_2,Q_3,\ldots\}$ is a countable collection of boxes such that 
$$ Q \subset \bigcup_{i=1}^{\infty} Q_i $$
Does it follow that $ v(Q) \leq \sum_{i=1}^{\infty} v(Q_i)$? Here $v$ is the Euclidean volume on $\mathbb{R}^{n}$.
 A: Yes, it is true. It follows from the general fact that $v$, as a measure, is monotonic and countably subadditive. We use monotonicity to go from
$$
Q\subset\bigcup_{i\in\mathbb{N}}Q_{i}
\quad\text{to}\quad
v(Q)\leq v\left(\bigcup_{i\in\mathbb{N}}Q_{i}\right)
$$
and we use countable subadditivity to get
$$
v\left(\bigcup_{i\in\mathbb{N}}Q_{i}\right)\leq\sum_{i\in\mathbb{N}}v(Q_{i})
$$
Now all you need to know is that $v$ is a (pre-)measure (which is a bit of a pain to show but is done in measure theory textbooks, e.g. Stein and Shakarchi's "Real Analysis" page 11).
Here I have paraphrased Stein and Shakarchi's argument: from $\{Q_{i}\}_{i\in\mathbb{N}}$ create a collection $\{O_{i}\}_{i\in\mathbb{N}}$ of open rectangles with $Q_{i}\subset O_{i}$ and $v(O_{i})\leq(1+\epsilon)v(Q_{i})$ (swell the rectangles a little bit). $\{O_{i}\}_{i\in\mathbb{N}}$ is now an open cover of $Q$, and so by compactness there exists some finite subcover $\{O_{i}\}_{i=1}^{N}$ (after renumbering). We now have
$$
v\left(\bigcup_{i=1}^{N}O_{i}\right)\leq\sum_{i=1}^{N}v(O_{i})\leq(1+\epsilon)\sum_{i=1}^{N}v(Q_{i})\leq(1+\epsilon)\sum_{i\in\mathbb{N}}v(Q_{i})
$$
We now only need that the leftmost term bounds $v(Q)$ from above. This is given as a lemma in Stein and Shakarchi and is pretty intuitive. You can make a formal argument for it by taking the closed rectangles $\{\overline{O_{i}}\}_{i=1}^{N}$ and making from them a finite collection of disjoint (except for the boundaries) rectangles that together cover $Q$. Then you can use the definition of $v$ to show that
$$
v(Q)\leq\sum_{i=1}^{N}v(\overline{O}_{i})
$$
This is the part that had caused me to remember this proof as being annoying. Also, note that depending on your definition of volume you might only be allowed to evaluate $v$ on closed rectangles, but that doesn't cause any real problems here.
