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(K^c) 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

${\overline {\mathcal {E}}}=\left\lbrace A\subseteq X:\exists B,C\in {\mathcal {E}}:A\triangle B\subseteq C\wedge \mu (C)=0\right\rbrace$

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)