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In the lecture on measure theory I attended last semester, we had a sort of complicated technical proof for the existence of the Lebesgue-premeasure. However, I can't see why this easier argument does not work:

Let $\lambda((a, b]) = b - a$ finitely additive on the ring $R = \{ \mathrm{finite \, disjoint \, unions \, of \,} (a, b] \}$. Having a disjoint union of elements $A_n \in R$ that lies again in $R$, so $\bigsqcup_{n \geq 1} A_n \in R$ we may assume w.l.o.g. that $\bigsqcup_{n \geq 1} A_n = (a, b]$. Now one can easily see that $$\sum_{n \geq 1} \lambda(A_n) \leq \lambda((a, b])$$ hence the infinite sum is absolutely convergent and may be reordered. Reorder it such that $A_i = (x_i, x_{i + 1}]$ with $x_i < x_{i + 1}$ for all $i$. We then get a telescopic sum $$\sum_{i \geq 1} \lambda((x_i, x_{i + 1}]) = \sum_{i \geq 1} (x_{i + 1} - x_{i}) = \lim_{i \rightarrow \infty} x_i - x_1 = b - a$$ Hence we have equality.

The question now is: Does this proof work? Or is there something missing?

Steven
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  • "Now one can easily see that $\sum_{n \geq 1} \lambda(A_n) \leq \lambda((a, b])$"

    I don't see it easily. Would you elaborate on it?

    "Reorder it such that $A_i = (x_i, x_{i + 1}]$ with $x_i < x_{i + 1}$ for all $i$".

    Would you explain why this reordering is possible?

     – JHW Aug 23 '15 at 18:57
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    @JHW He's right about the first thing being easy. The point is that if $c_n\ge 0$ and you want to show that $\sum_{n=1}^\infty c_n\le a$ it's enough to show that $\sum_{n=1}^N c_n\le a$. But he's not going to be able to explain how to do the reordering, because it can't be done. – David C. Ullrich Aug 23 '15 at 19:01
  • Okay, I guess David is right about the reordering. Now, about the first thing: $\lambda ((a, b]) = \lambda \left( \bigcup_{i = 1}^n A_n \right) - \lambda \left( (a, b] - \bigcup_{i = 1}^n A_n \right) \geq \lambda \left( \bigcup_{i = 1}^n A_n \right) = \sum_{i = 1}^n \lambda(A_n)$ and take the limit. – Steven Aug 23 '15 at 19:04
  • @DavidC.Ullrich About the first one, I knew that. The point is how you prove that the finite sum is smaller than or equal to $b - a$. – JHW Aug 23 '15 at 19:13
  • @JHW Finitely many intervals can be ordered the way he says. So we have $a\le a_1<b_1\le a_2<b_2\dots <b_n\le b$. Now $b-a$ is equal to the sum of the $b_j-a_j$ plus some other terms,,, – David C. Ullrich Aug 23 '15 at 19:16
  • @DavidC.Ullrich Finitely many intervals can be ordered the way he says. How do you prove this? – JHW Aug 23 '15 at 22:24
  • @JHW You're serious? These are disjoint intervals. So given two of them, $I$ and $J$, wlog $x<y$ for all $x\in I$ and $y\in J$. Say $I<J$ when that happens. We have a finite totally ordered set. Show by induction that any totally ordered set with $n$ elements is order-isomorphic to ${1,2,\dots, n}$. – David C. Ullrich Aug 23 '15 at 23:12
  • @DavidC.Ullrich So given two of them, $I$ and $J$, wlog $x \lt y$ for all $x\in I$ and $y\in J$. How do you prove this? I'm serious. I know you can prove this. My point is that the OP's claim that the proof is easy is misleading. – JHW Aug 23 '15 at 23:39
  • @JHW If I'd known you were asking questions you knew the answer to, just so you could use my replies to illustrate your points regarding the OP's claims, I would have dropped this long ago. If you want to talk about math some time stop by... – David C. Ullrich Aug 24 '15 at 00:01

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I'm going to talk about $[a,b)$ instead of $(a,b]$. The reason being that that's the sort of half-open interval I tend to work with; hence really a lot of things in the first version of this post were simply wrong. The easiest way to fix it is just to change $(a,b]$ to $[a,b)$ everywhere...

You can't necessarily reorder that way! I don't mean you're not allowed to or things don't work if you do, I mean you literally cannot; in general there is no such reordering. The way those intervals fit together can be much more complicated than that.

Simple example: Say $I_n=[1-1/n,1-1/(n+1))$, so $[0,1)=\bigcup_{n=1}^\infty I_n$. Those intervals can be ordered the way you want, in fact they already are ordered that way.

But now say $J_n=[2-1/n,2-1/(n+1))$. So $[0,2)=\bigcup_{n=1}^ \infty I_n \cup\bigcup_{n=1}^\infty J_n$. You can't "reorder" the collection of all the $I_n$ and the $J_n$ the way you want.

That's enough to show your proof doesn't work. But it can be much worse. Say $I_n$ is as above. Now for each $n$ you can find countably many disjoint half-open intervals $I_{n,m}$ so $I_n=\bigcup_{m=1}^\infty I_{n,m}$. Now we have $$[0,1)=\bigcup_{n=1}^\infty\bigcup_{m=1}^\infty I_{n,m},$$and that collection of intervals is much worse than what you had in mind when you said "reorder"...

It really is trickier than you think. Folland points out that the way the intervals are arranged from left to right can be isomorphic to any countable ordinal, if you know what that means. His point is that it can be complicated. One can also say "oh, so it's no worse than a countable ordinal" and then give a cute proof by transfinite induction. But never mind that for now.

David C. Ullrich
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  • I THINK that even using countable ordinals, a proof based only in transfinite induction will not work. At some point the topology of $\mathbb{R}$ must play a key role. Otherwise, the proof based only in transfinite induction would also show that $\lambda((a,b]\cap \mathbb{Q})=b-a$ defined on ${(a,b] \cap \mathbb{Q} ,|, a,b \in\mathbb{Q}}$ is a premeasure (countably additive). And we know that such $\lambda$ is not a premesure. – Ramiro Aug 24 '15 at 17:41
  • I can't imagine why you think I said otherwise. I mean this is like I say one can prove some inequality by induction and you say no, we also need the fact that $(a+b)^2=a^2+2ab+b^2$. – David C. Ullrich Aug 24 '15 at 18:24
  • You wrote: "Folland points out that the way the intervals are arranged from left to right can be isomorphic to any countable ordinal" and " One can also say 'oh, so it's no worse than a countable ordinal' and then give a cute proof by transfinite induction".

    It sounded like you were saying that using a countable ordinal to represent the order the intervals one might produce a "cute" proof just using transfinite induction.

     – Ramiro Aug 24 '15 at 19:11
  • I have just pointed out that an idea like ordering the intervals, even using countable ordinals, and then adding their measures, even if it is done through transfite induction, would not be enough to prove the result. At some point the topology of $\mathbb{R}$ must play a key role in the proof. – Ramiro Aug 24 '15 at 19:25
  • And my point is that well yes of course, your point is obvious. My point was that the proof I have in mind does make the proof much simpler and more straightforward. – David C. Ullrich Aug 24 '15 at 21:40
  • Could you explain what was the proof (using countable ordinals and transfinite induction) you had in mind? How it would be "much simpler and more straightforward" than the classical proof using compacity of $[a,b]$ in $\mathbb{R}$? – Ramiro Aug 24 '15 at 22:23
  • @Ramiro Ok, see http://math.stackexchange.com/questions/1408533/lebesgue-premeasure-vis-transfinite-induction – David C. Ullrich Aug 24 '15 at 23:36
  • @Ramiro Actually wasn't sure until I wrote it out, but the place where the topology of $\Bbb R$ comes in is in the proof that the collection of intervals is well-ordered. – David C. Ullrich Aug 25 '15 at 00:54
  • David, Thanks a lot. I am not sure we can say it is "much simpler and more straightforward" than the classical proof using compacity of $[a,b]$, BUT, it is surely a VERY NICE and INTERESTING proof. And yes, you are right, the place where the topology of $\mathbb{R}$ comes in the proof is that the collection of intervals is well-ordered by the order induced by the order $<$ in $\mathbb{R}$. And when we look in the details in the proof, we see that this fact is, in some way, a "compacity"-like argument, which is per si an interesting fact. Very nice proof. Thanks. – Ramiro Aug 25 '15 at 02:18