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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.

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Alternative proof for the finite case.

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

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  • $\begingroup$ thats a good way thank you , but it doesnt answer my confusion on the original solution if you can explain that? thanks $\endgroup$ Commented Mar 19, 2020 at 2:52

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