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)

Different definition of the completion of a measure space
Studying Measure Theory in University, I came across the following definition for the completion of a measure space: let $$(X,\mathcal{E},\mu)$$ be a measure space; then the set
 * $$\overline{\mathcal{E}}=\left\lbrace A\subseteq X:\exist B,C\in\mathcal{E}:A\triangle B\subseteq C\wedge\mu(C)=0\right\rbrace$$

is a $$\sigma$$-algebra and, extending $$\mu$$ to $$\overline{\mu}$$ defined on $$\overline{\mathcal{E}}$$ by letting $$\overline{\mu}(A)=\mu(B)$$ for any set $$B\in\mathcal{E}$$ such that $$A\triangle B$$ is contained in a null set, the measure space $$(X,\overline{\mathcal{E}},\overline{\mu})$$ is complete. $$A\triangle B$$ is defined as $$(A\smallsetminus B)\cup(B\smallsetminus A)$$. I've tried to prove this. So far, I've proved $$\varnothing,X\in\overline{\mathcal{E}}$$, or in fact that $$\mathcal{E}\subseteq\overline{\mathcal{E}}$$, since for any $$A\in\mathcal{E}$$ we have $$A\triangle A=\varnothing$$ and obviously $$\mu(\varnothing)=0$$, and that, supposing $$\overline{\mu}$$ is well-defined, it extends $$\mu$$, as one possible $$B$$ set for which $$\overline{\mu}(A)=\mu(B)$$ is $$A$$ itself, as used in the short proof above. My doubts are:

1) Give $$A,B\in\overline{\mathcal{E}}$$, are $$A\cup B$$ and $$A\smallsetminus B$$ elements of $$\overline{\mathcal{E}}$$? In other words, is $$\overline{\mathcal{E}}$$ a $$\sigma$$-algebra?

2) Is it true that for any two such sets $$B,B'$$ as enter the definition of $$\overline{\mu}$$, we have that $$\mu(B)=\mu(B')$$? In other words, is $$\overline{\mu}$$ well-defined?

3) Is this the same as what there is in this article under the name of «completion of a measure space»?

MGorrone (talk) 11:18, 15 March 2014 (UTC)

OK http://math.stackexchange.com/questions/713084/completion-of-a-measure-space/713140#713140 now has an answer to this. I suggest this be placed in the article. MGorrone (talk) 14:24, 15 March 2014 (UTC)