# Talk:Complete measure

In measure theory, a complete measure is a measure in which every subset of every null set is measurable (having measure 0).

Why do they say "every" null set ... i thought there could be only one, unique null set --> no elements, empty set !! ... and then "every subset of every null set" !! ... a null set will have ony one subset always (itself, and nothing else).

What a piece of cake. "m(A) = 0" is for every set!!! Not only for the empty one! —Preceding unsigned comment added by 149.156.124.9 (talk) 19:32, 2 February 2010 (UTC)

So, what is the meaning of "every subset of every null set"

A null set is not to be confused with the empty set... a null set is a measurable set (a set element of a sigma algebra which itself has a measure attached to it) that has measure zero. A countable set of points in the reals has Lebesgue measure zero for example. In fact, there's more subtleties about a null set than it being of measure zero, see the article about it.

I have no idea how to fix this, but this article doesn't come up under a search for "complete measure space" and it ought to. 134.50.3.40 (talk) 10:59, 10 March 2008 (UTC)

Can someone improve the examples section by giving an example of a Borel set and a non-measurable subset of it, explaining why the Borel measure is not complete. —Preceding unsigned comment added by 82.31.209.115 (talk) 20:20, 7 July 2008 (UTC)

I don't know of a constructive proof, but I know of an existence proof, based on the Cantor function (the only use for that function that I know of...)

First, you have to start knowing that every positive measure set contains a nonmeasurable subset. Let K be the Cantor set, and let f be the Cantor function. Obviously, f(Kc) is the Cantor set, which has measure zero, so f(K) has measure one. We need a strictly monotonic function, so consider $g(x)=f(x)+x$ , which is obviously strictly monotonic, hence one-to-one, hence a homeomorphism. Now, $g(K)$ has measure one. Let $E\subset g(K)$ be non-measurable, and let $F=g^{-1}(E)$ . Because $G$ is injective, we have that $F\subset K$ , and so $F$ is a null set. However, if it were measurable, then $g(F)$ would be measurable (here I use the fact that the preimage of a Borel set by a continuous function is measurable, since continuous functions are measurable; $g(F)={g^{-1}}^{-1}(F)$ is the preimage of $F$ through the continuous function $h=g^{-1}$ .)

Therefore, $F$ is a null, but non-Borel measurable set. Loisel (talk) 04:16, 8 July 2008 (UTC)

Actually, I just regurgitated that mostly from memory, but in my half-drunken state, right now I can't determine if that's correct. I've edited the article, but if I got it wrong, just delete the whole thing. Loisel (talk) 04:27, 8 July 2008 (UTC)