Volume of countable collection of rectangles

Show that if $Q_1,Q_2,\ldots\in\mathbb{R}^n$ is a countable collection of rectangles covering $Q$, then $v(Q)\le\sum v(Q_i)$, where $v(x)$ denotes the volume of $x$.

If the collection $Q_i$ is finite, this is easy. Just take the rectangle $Q'$ that covers all of $Q,Q_1,Q_2,\ldots,Q_n$, and partition it using the endpoints of $Q,Q_1,Q_2,\ldots,Q_n$ in each axis. Then each subrectangle of $Q$ determined by this partition is also a subrectangle of one of the $Q_i$'s, and the result follows.

When the collection $Q_i$ is countable but infinite, I can no longer define a "partition" like that, because it could contain infinitely many points in each axis. How can I modify the argument?

Fix $\epsilon>0$. Replace each $Q_i$ with a rectangle $P_i$ s.t. $Q_i$ is in the interior of $P_i$ and s.t. $v(P_i)< v(Q_i)+\epsilon\times 2^{-i}$, and thus $\sum v(P_i)<\sum v(Q_i)+\epsilon$. Then, by compactness, $Q$ is in a finite collection of $P_i$'s (as the interiors of $P_i$'s cover $Q$). Use what you proved in the finite case and take $\epsilon\to0$.
edit: I suppose that $Q$ is a rectangle (implicit in the question) and that we can't use Lebesgue measure (otherwise it would be pointless)
• So, are you taking the $P_i$'s to be open? (Because we need open covers in order to invoke compactness, right?) Commented Aug 30, 2013 at 19:03
• Where do you get that $Q$ is compact? Commented Aug 30, 2013 at 19:06
• $P_i$'s are closed. But the interiors of $P_i$'s cover $Q$, so it doesn't matter. Commented Aug 30, 2013 at 19:07
• @user2566092 $Q$ is a rectangle Commented Aug 30, 2013 at 19:08