Easy proof for existence of Lebesgue-premeasure 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? 
 A: 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.
