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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 $\mathbb{R}^{n}$.

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    $\begingroup$ What do you mean by $\;v\;$ ? Borel measure, other measure...something else? $\endgroup$
    – DonAntonio
    Commented Apr 26, 2014 at 0:00
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    $\begingroup$ Assuming OP means $v$ to be volume, this is a consequence of monotonicity/sub-additivity of measure. Also, it should be $v(Q) \leq \sum v(Q_i)$. $\endgroup$ Commented Apr 26, 2014 at 0:10

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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.

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  • $\begingroup$ Thanks for your reply! I would like to ask if there is a more elementary way to do this since I haven't learned what is a measure. $\endgroup$
    – user145801
    Commented Apr 26, 2014 at 0:17
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    $\begingroup$ Volume is countably subadditive, and monotone as well. An undergraduate text like Bartle's "Real Analysis" treats this in the case of Jordan outer content, and Jordan outer content agrees with volume on cubes. $\endgroup$ Commented Apr 26, 2014 at 0:22
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    $\begingroup$ And here kicks in my comment again: the OP says he hasn't learned what a measure is, so: what in the world is that $\;v\;$ in the OP ?! $\endgroup$
    – DonAntonio
    Commented Apr 26, 2014 at 0:27
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    $\begingroup$ I mean $v$ is volume $\endgroup$
    – user145801
    Commented Apr 26, 2014 at 0:40
  • $\begingroup$ Great. I made an edit to clarify that in the original question. $\endgroup$ Commented Apr 26, 2014 at 0:44

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