Talk:Fubini's theorem

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Might this Fubini be the one referred to in "Fubini's Law"? Probably not, but I can't find anything on who the other Fubini might be.

http://www.steptwo.com.au/columntwo/archives/000872.html

Contents

[edit] Not needed section

Section 3.1.1 (`Proof') appears unnecessary - it lends little to the article, and I cannot see what it is trying to show. If you agree, please remove it! HyDeckar 13:24, 6 November 2007 (UTC)

[edit] Missing proof

May someone proof why Fubini's theorem is correct?

Also, I don't get what the applications example within this article (the one with Gaussian integral) is supposed to show, as it doesn't contain two integrals but one. --Abdull 20:12, 13 July 2005 (UTC)


An outline proof is now there - you can probably see why noone put it there before now. The trick with the Gaussian integral is explained on the page Gaussian integral --HyDeckar 13:23, 6 November 2007 (UTC)

[edit] Sigma-finiteness

Shouldn't one of the conditions of Fubini's theorem, as well as Tonelli's, state that A and B are σ-finite? I have the book Real and Complex Analysis by Walter Rudin that shows that this condition is necessary. --Propower 01:36, 19 October 2006 (UTC)

Fubini's theorem, in the texts I have read, is stated with the restriction that the measure spaces are complete (in e.g. http://books.google.com/books?id=WkApkbp4WLUC), but not sigma finite (sigma finiteness comes into Tonelli's theorem though). I think perhaps both these sets of hypotheses are sufficient, but completeness is a "nicer" condition than sigma finiteness, I'm not sure though. 121.209.52.168 13:06, 4 October 2007 (UTC)
I agree with 121.209.52.168, the weaker statement is that the spaces are complete, and sigma finiteness comes in with Tonelli (where we also need completeness). I have changed the statement to reflect this, and hope to put up a proof soon. HyDeckar 12:12, 6 November 2007 (UTC)

[edit] Quicker way of doing the integral...

Notice

\frac{d}{dx}\left[\frac{-x}{x^2+y^2}\right]=\frac{x^2-y^2}{(x^2+y^2)^2}

I stole this idea from the solutions to a sheet of problems I was working through, trying to do the integral with trig substitutions is a pain! Cdyson37 (T) 17:53, 19 December 2006 (UTC)

One can just work it out normally...
\int \frac{x^2 - y^2}{(x^2 + y^2)^2} \, dy
= \int \frac{x^2 + y^2 - 2y^2}{(x^2 + y^2)^2} \, dy
= \int \frac{1}{x^2 + y^2} \, dy + \int \frac{-2y^2}{(x^2 + y^2)^2} \, dy
= \int \frac{1}{x^2 + y^2} \, dy + \int y \left(\frac{d}{dy} \frac{1}{x^2 + y^2}\right) \, dy
= \int \frac{1}{x^2 + y^2} \, dy + \left(\frac{y}{x^2 + y^2} - \int \frac{1}{x^2 + y^2} \, dy\right) by parts
= \frac{y}{x^2+y^2} + C.
-- 129.78.64.102 08:36, 2 August 2007 (UTC)
okay, I went ahead and changed it... -- 129.78.64.102 08:44, 2 August 2007 (UTC)