Talk:Signed measure

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[edit] (possibly infinite) signed measures

From the book "Measure theory", by Donald L. Cohn, ISBN 0817630031, page 121 (emphasis added):

Let (X,A) be a measurable space, and let μ be a function on A with values in [-∞,+∞] (...) If μ is countably additive and satisfies μ(∅)=0, then it is a signed measure.

So it can, after all, take values in the whole extended real number line. It is later proven (not assumed) that it can't take both +∞ and -∞ as values, for it must be defined in all the sets of A (among which, of course, is X).

I see absolutely no reason to write the article as if μ couldn't, by definition, take -∞ as a value. Why are you reverting my changes? --Fibonacci 03:41, 6 December 2005 (UTC)

I did not revert your change. I did some changes. I left your definition with a measure taking −∞ as value. But at some point I wrote that from now on we will only work with measures not having −∞ as value. This is a trick used a lot in math books to not complicate the exposition, when the more general version of the statement is easily obtainable from the slightly more particular case. Would you like to carefully read the diff of what I wrote and also the current article version and then comment here? Oleg Alexandrov (talk) 03:46, 6 December 2005 (UTC)
Okay then...
"One allows a signed measure can either take +∞ as value but not −∞, or viceversa"
Nope, one does not allow it to take one or the other value. One allows it (in principle) to take any value of the extended real line, even +∞ and -∞ if you want, and then it is shown that it cannot possibly take both. I'm reverting that.
"We will also assume that a signed measure does not take −∞ as a value, for simplicity reasons."
I'm thinking of a better way of stating that, if you want to use the math book trick (strangely enough, Cohn doesn't use it, except when he needs to prove a theorem, and even in those cases, he says "similarly for the -∞ case" or something along those lines at the end of the proof). I'm changing that too - though I won't delete it.
\int_{\{f(x)< 0 \}} \! |f(x)| \, d\nu (x) < \infty.
Different ways of saying the same thing. Still, I think that \int_X f^-(x) \, d\nu(x)<\infty is clearer, and sorry for writing TeX in the middle of a text line.
"both μ+ and μ- are nonnegative measures, with the latter taking only finite values"
Under your assumption above, this is true. Again, I'm thinking of a way to rephrase that.
--Fibonacci 03:59, 6 December 2005 (UTC)
Update: I edited it. Please read it and comment here. --Fibonacci 04:08, 6 December 2005 (UTC)
You say that \int_X f^-(x) \, d\nu(x)<\infty is clearer. I don't agree, because you did not define this extra notation. Once you do so, we will see. Oleg Alexandrov (talk) 04:23, 6 December 2005 (UTC)
See Lebesgue integration#Integration. On the other hand, you did not define what {f(x) < 0} means - {x ∈ X | f(x) < 0} would be a different thing, but besides being too cumbersome, you'd first have to define integrals over a subset of X.
But if you still think your notation is clearer, please reply here. --Fibonacci 04:30, 6 December 2005 (UTC)
Yes, yours is clearer, if you make a pointer to notation in the text. :) Gosh, you are so quick. Oleg Alexandrov (talk) 04:32, 6 December 2005 (UTC)