Consider a rectangle $L = [a_1, b_1 ] \times \dots \times [a_n, b_n]\subseteq \mathbb{R}^n$ such that $v(L) = r$, can we cover $L$ by countably many rectangles $Q_i$ such that $\sum_{i=1}^{\infty}v(Q_i) < r$?
Note that $v(L) = (b_1 - a_1)\dots (b_n-a_n)$. Now intuitively I think the answer to this is no, however I'm not really sure how I could prove this, in the sense that if I suppose that I can find such a covering I don't see any way to reach a contradiction.
On
I'll suppose that $v(A)$ coincides with the $n$-dimensional Lebesgue measure of $A$. Assuming we can find such a cover, then $$ v(L) \le v\left(\bigcup_{i=1}^\infty Q_i\right) $$ since $v$ is monotonic and $L\subset \cup_{i=1}^\infty Q_i$. However, $v$ is also subadditive, i.e. $v\left(\bigcup_{i=1}^\infty A_i\right) \le \sum_{i=1}^\infty v(A_i)$ (and equality holds when $A_i$ are disjoint), hence $$ r=v(L) \le v\left(\bigcup_{i=1}^\infty Q_i\right) \le \sum_{i=1}^\infty v(Q_i) <r. $$ This is a contradiction.
Edit: If your definition of the function $v$ only involves rectangular set, then it is possible to extend $v$ so that its domain is a pretty large class of subsets of $\Bbb R^n$ called the measurable sets. You can look up Caratheodory extension theorem if you want to know precisely what I meant.
No, of course this follows from the theory of Lebesgue measure.
If you don't want to appeal to that, then consider this. One can expand the $Q_i$ slightly so they become open but still the sum of their measures is less than $v(L)$. As $L$ is compact, Heine-Borel shows that a finite number of the $Q_i$ cover $L$. Now it is quite elementary that $\sum_{i=1}^n v(Q_i)\ge v(L)$ for a finite covering of rectangles.