Talk:Lebesgue measure argument for uncountability of the reals

Isn't this article more complicated than necessary? The measure of a singleton is zero. By countable additivity, the measure of the set containing all members of a sequence is therefore zero. QED. Michael Hardy 20:55, 16 October 2006 (UTC)

I shortened it according to the suggestion of User:Michael Hardy. Leocat 17 Oct 2006


 * Yes, it is more complicated than necessary. When I first saw it, it wasn't clear to me what the point of the original contributor was, since I was wondering if there was something "constructive" about this proof. User:Trovatore has said there is nothing especially constructive about it.--CSTAR 21:16, 16 October 2006 (UTC)


 * It is already less complicated. This is an important proof, because it shows that continuum is a higher order infinity than aleph zero. Leocat 17 Oct 2006


 * But there are simpler arguments showing the same thing, so that doesn't mean this one is important. It's only the conclusion or this argument, supportable by other, different arguments, that is important for that reason. Michael Hardy 19:37, 17 October 2006 (UTC)


 * The Lebesgue measure argument is an excellent supporting argument to Cantor's diagonal argument and should be presented to any interested reader. Leocat 23:25, 17 Oct 2006 (GMT)
 * It's not worth a whole article. Proofs get articles only in rather unusual cases. It's worth a mention in some other article. The title Lebesgue measure argument is unacceptable even as a redirect; there are lots of Lebesgue measure arguments. --Trovatore 05:14, 18 October 2006 (UTC)