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Let $x_0,x_1,\ldots$ be an infinite sequence of real numbers with $x_n \in [0,1],\,\forall n\in\mathbb{N}$. Let $\mathcal{F}_0,\mathcal{F}_1,\ldots,$ be an infinite sequence of disjoint Lebesgue measurable subsets of $[0,1]$ with $\bigcup_{n\in\mathbb{N}}\mathcal{F}_n = [0,1]$. Consider the real number $\delta = \sum_{n\in\mathbb{N}}\int_{\mathcal{F}_n} (x-x_n)^2 \mathrm{d}x$.

So, what I did is the following: Since $\sum_{n\in\mathbb{N}}\int_{\mathcal{F}_n}\mathrm{d}x = 1$, there is an index, say $m$, with $\int_{\mathcal{F}_m}\mathrm{d}x >0$. Now, $\delta \geq \int_{\mathcal{F}_m} (x-x_m)^2 \mathrm{d}x$, and since the integrand is non-zero almost everywhere (except at $x_m$) we have $\delta > 0$.

I guess my derivations are fine. What "worries" me is that I can make $\delta$ arbitrarily small but never $0$. Somehow, and I don't know why, I "feel" that I should be able to take $\{x_0,x_1,\ldots\}$ to be all the rational numbers in $[0,1]$, and get $\delta = 0$ with a "nice" choice of $\mathcal{F}_n$. Am I wrong in my calculations, and if I am right, why is this happening at an intuitive level? Is there any "system" (or a different kind of measure) where I would get $\delta = 0$ (with respect to that measure).

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Consider ${\cal F}_0$. We can assume it's not empty by shifting indices. If it has measure zero, fine, we'll skip it and find the first ${\cal F}_n$ with non-zero measure. (We know one of them is non-zero because they add to $1$.) Then it contains an interval, and on that interval, the function is non-zero everywhere except at one point. Thus that piece has to integrate to something non-zero. (I'm not sure if there are more complicated measurable sets that don't have any intervals of positive measure in them, but if you "push all the pieces together until they are adjacent" you can still conclude the same thing.)

The intuitive idea is that although you can define a point-set that is dense in the interval, the first few had to "stake out a territory" (the ${\cal F}_n$), since they can't all just take the region closest to themselves viz. they are dense.

You could get $\delta=0$ if your measure assigned a non-zero value to a finite number of points, and zero everywhere else. Then you could overlap each point individually, and place $x_n$ on that point, so that the only possible nonzero contributions were "stamped out" individually. If there is a denumerable number of "valuable" points with a single accumulation point, say $\mu(\{1/n\})=2^{-1-n}$, then you should still be able to make this procedure work, using a single cover ${\cal F}_0=\{0\}$ and then ${\cal F}_n=[2^{1-n},2^{-n})$, with $x_n$ in the obvious place. I'm not sure how far you can take this. I think the boundary line is when you try to make the point-set dense, so that you can't partition the points into intervals of one each.

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Hi Mario, thank you very much for the reply. I guess the first paragraph is a reiteration of my argument. For your intuitive explanation (the second paragraph), I thought something very similar; at least one $\mathcal{F}_n$ has to "stake out a territory," like you said. But I was wondering why it is happening the way it is, or is there a Lebesgue-like measure that somehow avoids this issue. I thought about the discussion you have in paragraph 3, yes in that case I can get zero, but the measure in that case does not feel as natural as the Lebesgue measure. – Anon Dec 18 '12 at 6:20
Forgive me, by the way, if what I have written above sounds critical about your reply, that is not my intention at all. I should admit that I am not exactly sure what kind of an answer I would like to get myself. E.g. maybe "This is something not cool about the Lebesgue measure, you can do this with x measure," or "No, you have the same issue with any kind of measure that assigns an interval its length, and the reason is nothing but sets being dense or not (as you have mentioned)," or some "magic wand"ish answer that clears everything. – Anon Dec 18 '12 at 6:21
My interpretation of your question was whether there existed some measure such that you could get $\delta=0$ as an answer. It seems to me that my argument/your argument (I realized after the fact that it wasn't very different, but it was the obvious solution to me) pretty clearly indicates that the Lebesgue measure is not going to cut it. There are certainly some curiosities here. ... – Mario Carneiro Dec 18 '12 at 6:33
... I keep thinking about the fact that an accumulation point doesn't mess anything up. Note that the acc. pt. needs to be covered by a one-element set. If there was space on the other side, the interval could extend in that direction, but if there were other points accumulating on the other side, it couldn't. It relies on the neighboring intervals to cover the neighborhood of the acc. pt. If there was an "adjacent" acc. pt. there would be a hole in between. ... – Mario Carneiro Dec 18 '12 at 6:40
... So with ${\cal F}_n=\{x_n\}$ ranging over the rationals, you miss all the irrationals that way (holes in the set), and of course you've got no measure. So it seems that $\delta>0$ iff there is a subinterval of $[0,1]$ such that every open sub-subinterval has non-zero measure. – Mario Carneiro Dec 18 '12 at 6:40

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