A measure theory question 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).
 A: 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.
