# Volume of a countable collection of sets Calculation

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 $$x$$.

I have trouble understand the following solution from Volume of countable collection of rectangles

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.

For inifinitely countable: 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 $$\epsilon\to0$$.

1) what does it mean to partition using endpoints in each axis?

2) why is it necessary to construct Pi?

If someone can expand the solution a bit, that would be very helpful.

Prove A subset B $$\cup$$ C implies v(A) <= v(B) + v(C).
Q subset $$\cup$${ Q$$_n$$ : n in finite index set }
implies v(Q) <= $$\sum_n$$ v(Q$$_n$$).