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Abdelmalek Abdesselam
Abdelmalek Abdesselam
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This is a bit of a long comment and perhaps a useful complement to the other answers.

Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for $n=1$ using Carathéodory's Theorem. After removing the wrapping paper and other cosmetic trivialities,some easy simplifications it boils down to showing: if the finite semiopen interval $(a,b]$ is the disjoint union of $(a_i,b_i]$, $i\in\mathbb{N}$, then $$ b-a\le\sum_{i=1}^{\infty}(b_i-a_i)\ . $$ Now this is close to what the OP asks. Suppose the set contains a nonflattened cube, i.e., $b-a>0$. Show that the sum on the right or outer measure of the set is $>0$, e.g., $\ge b-a$.

After one more round of trivialitiessimplification (give yourself an $\varepsilon$ of room by shortening $(a,b]$ into a compact interval $[a,b]$, enlarging the $(a_i,b_i]$ into open intervals $(a_i,b_i)$ and finally extracting a finite cover), we arrive at the following purely combinatorial lemma.

Combinatorics Lemma: Suppose $[a,b]\subset\cup_{i=1}^{p}(a_i,b_i)$ then $$ b-a\le\sum_{i=1}^{p}(b_i-a_i)\ . $$

The lemma is not really difficult, but it is not trivial. I think the best way to see that is to imagine a game of Tetris where the covering intervals fall from the sky in the order in which they are numbered from $1$ to $p$ and they pile up (a bit also like in Xavier Viennot's theory of heaps of pieces), possibly in very complicated structures. The lemma can be proved by an easy induction on $p$: look at the interval $(a_i,b_i)$ which covers the endpoint $b$ and argue according to the relative position of $a_i$ and $a$. If $a_i<a$ that interval alone does all the work. If $a_i\ge a$, then the remaining $p-1$ intervals cover $[a,a_i]$, etc. etc.

Now the Lemma has an obvious $n$-dimensional generalization with finitely many open boxes covering a closeclosed box. I suppose one could prove it by double induction on $n$ and $p$. Trying to go around that is perhaps why we typically use some Fubini integration argument instead.

This is a bit of a long comment and perhaps a useful complement to the other answers.

Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for $n=1$ using Carathéodory's Theorem. After removing the wrapping paper and other cosmetic trivialities, it boils down to showing: if the finite semiopen interval $(a,b]$ is the disjoint union of $(a_i,b_i]$, $i\in\mathbb{N}$, then $$ b-a\le\sum_{i=1}^{\infty}(b_i-a_i)\ . $$ Now this is close to what the OP asks. Suppose the set contains a nonflattened cube, i.e., $b-a>0$. Show that the sum on the right or outer measure of the set is $>0$, e.g., $\ge b-a$.

After one more round of trivialities (give yourself an $\varepsilon$ of room by shortening $(a,b]$ into a compact interval $[a,b]$, enlarging the $(a_i,b_i]$ into open intervals $(a_i,b_i)$ and finally extracting a finite cover), we arrive at the following purely combinatorial lemma.

Combinatorics Lemma: Suppose $[a,b]\subset\cup_{i=1}^{p}(a_i,b_i)$ then $$ b-a\le\sum_{i=1}^{p}(b_i-a_i)\ . $$

The lemma is not really difficult, but it is not trivial. I think the best way to see that is to imagine a game of Tetris where the covering intervals fall from the sky in the order in which they are numbered from $1$ to $p$ and they pile up (a bit also like in Xavier Viennot's theory of heaps of pieces), possibly in very complicated structures. The lemma can be proved by an easy induction on $p$: look at the interval $(a_i,b_i)$ which covers the endpoint $b$ and argue according to the relative position of $a_i$ and $a$. If $a_i<a$ that interval alone does all the work. If $a_i\ge a$, then the remaining $p-1$ intervals cover $[a,a_i]$, etc. etc.

Now the Lemma has an obvious $n$-dimensional generalization with finitely many open boxes covering a close box. I suppose one could prove it by double induction on $n$ and $p$. Trying to go around that is perhaps why we typically use some Fubini integration argument instead.

This is a bit of a long comment and perhaps a useful complement to the other answers.

Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for $n=1$ using Carathéodory's Theorem. After some easy simplifications it boils down to showing: if the finite semiopen interval $(a,b]$ is the disjoint union of $(a_i,b_i]$, $i\in\mathbb{N}$, then $$ b-a\le\sum_{i=1}^{\infty}(b_i-a_i)\ . $$ Now this is close to what the OP asks. Suppose the set contains a nonflattened cube, i.e., $b-a>0$. Show that the sum on the right or outer measure of the set is $>0$, e.g., $\ge b-a$.

After one more round of simplification (give yourself an $\varepsilon$ of room by shortening $(a,b]$ into a compact interval $[a,b]$, enlarging the $(a_i,b_i]$ into open intervals $(a_i,b_i)$ and finally extracting a finite cover), we arrive at the following purely combinatorial lemma.

Combinatorics Lemma: Suppose $[a,b]\subset\cup_{i=1}^{p}(a_i,b_i)$ then $$ b-a\le\sum_{i=1}^{p}(b_i-a_i)\ . $$

The lemma is not really difficult, but it is not trivial. I think the best way to see that is to imagine a game of Tetris where the covering intervals fall from the sky in the order in which they are numbered from $1$ to $p$ and they pile up (a bit also like in Xavier Viennot's theory of heaps of pieces), possibly in very complicated structures. The lemma can be proved by an easy induction on $p$: look at the interval $(a_i,b_i)$ which covers the endpoint $b$ and argue according to the relative position of $a_i$ and $a$. If $a_i<a$ that interval alone does all the work. If $a_i\ge a$, then the remaining $p-1$ intervals cover $[a,a_i]$, etc. etc.

Now the Lemma has an obvious $n$-dimensional generalization with finitely many open boxes covering a closed box. I suppose one could prove it by double induction on $n$ and $p$. Trying to go around that is perhaps why we typically use some Fubini integration argument instead.

Source Link
Abdelmalek Abdesselam
Abdelmalek Abdesselam
  • 23k
  • 1
  • 67
  • 127

This is a bit of a long comment and perhaps a useful complement to the other answers.

Having quickly read the question, I almost proceeded to downvote. But on second thought, I think it is not as trivial as it seems and, in fact, lies at the very heart of the standard construction of Lebesgue measure even for $n=1$ using Carathéodory's Theorem. After removing the wrapping paper and other cosmetic trivialities, it boils down to showing: if the finite semiopen interval $(a,b]$ is the disjoint union of $(a_i,b_i]$, $i\in\mathbb{N}$, then $$ b-a\le\sum_{i=1}^{\infty}(b_i-a_i)\ . $$ Now this is close to what the OP asks. Suppose the set contains a nonflattened cube, i.e., $b-a>0$. Show that the sum on the right or outer measure of the set is $>0$, e.g., $\ge b-a$.

After one more round of trivialities (give yourself an $\varepsilon$ of room by shortening $(a,b]$ into a compact interval $[a,b]$, enlarging the $(a_i,b_i]$ into open intervals $(a_i,b_i)$ and finally extracting a finite cover), we arrive at the following purely combinatorial lemma.

Combinatorics Lemma: Suppose $[a,b]\subset\cup_{i=1}^{p}(a_i,b_i)$ then $$ b-a\le\sum_{i=1}^{p}(b_i-a_i)\ . $$

The lemma is not really difficult, but it is not trivial. I think the best way to see that is to imagine a game of Tetris where the covering intervals fall from the sky in the order in which they are numbered from $1$ to $p$ and they pile up (a bit also like in Xavier Viennot's theory of heaps of pieces), possibly in very complicated structures. The lemma can be proved by an easy induction on $p$: look at the interval $(a_i,b_i)$ which covers the endpoint $b$ and argue according to the relative position of $a_i$ and $a$. If $a_i<a$ that interval alone does all the work. If $a_i\ge a$, then the remaining $p-1$ intervals cover $[a,a_i]$, etc. etc.

Now the Lemma has an obvious $n$-dimensional generalization with finitely many open boxes covering a close box. I suppose one could prove it by double induction on $n$ and $p$. Trying to go around that is perhaps why we typically use some Fubini integration argument instead.