Talk:Fubini's theorem

message: As it is presented now, Fubini-Tonelli's Theorem is not complete since it doesn't consider the case of non negative and non integrable measurable functions, for which the equlity for the three integrals are actually stated in Tonelli's theorem. I propose the following variant in which firstly the integrals with the absolute value of the integrand function are stated to be equal, then the Fubini's theorem follows when at least one among the three integrals with the absolute value are shown to be finite.       — Preceding unsigned comment added by 79.43.153.251 (talk) 12:20, 9 October 2018 (UTC)
thread_title:  Enunciation of Fubini-Tonelli's Theorem not complete 

message: It is now stated only for dx, dy, meaning lebesgue measure. For some people it will be enough, but not for all  — Preceding unsigned comment added by 83.4.207.202 (talk) 16:16, 8 September 2013 (UTC)
thread_title:  Generalization for arbitrary measure 

message: There to be a problem with the statement of the corollary. It looks as if there is a condition missing on g and h. Should it say "if f(x,y) = g(x) h(y)"?
:The math code for the condition didn't render; I fixed it. Arthena(talk) 13:04, 7 February 2010 (UTC)
replies: text: I further corrected it; since we are integrating over pairs, we need to use the function whose domain is a set of pairs.  —Preceding unsigned comment added by 76.235.223.38 (talk) 20:17, 19 December 2010 (UTC)
thread_title:  Problem with Corollary 

message: 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)
thread_title:  Not needed section 

message: 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)
replies: text: 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)
thread_title:  Missing proof 

message: Although Fubini's theorem can naturally result from intuition with double integrals, this article fails to give a strict proof (i.e. one that would prove that the theorem holds under the definition of a limit of a Riemann sum).--Jasper Deng (talk) 05:11, 3 May 2013 (UTC)
thread_title: Proof still missing

message: Why is this listed under see also?  I went to read the article, and it did not directly mention Fubini's theorem or anything similar. Thenub314 (talk) 15:09, 8 October 2008 (UTC)
thread_title:  Clairaut's theorem 

message: Am I just very confused, or should f have to be measureable?
Assume card([0,1])=ω1.  This follows from AC+CH.
Let g be a bijection from [0,1] to ω1.
:::-1  if g(x)<g(y)
Let f(x,y) = .... 0  if g(x)=g(y)
:::+1  if g(x)>g(y)
A and B are [0,1] with Lebesgue measure.
::0  if x=y
|f(x,y)| = 
::1  else
Since |f(x,y)| = 1 almost everywhere, the integral of |f(x,y)| is 1.
If we integrate over x first, we get g(x)<g(y) for almost all y, giving us -1, then we integrate over y and get -1.
If we integrate over y first, we get g(y)<g(x) for almost all x, giving us +1, then we integrate over x and get +1.
Lebesgue measure is finite and so σ-finite on [0,1], and both iterated integrals of |f(x,y)| are 1, so this applies to the alternate theorem statement too.
JumpDiscont (talk) 03:44, 28 September 2009 (UTC)
thread_title:  Theorem Statements 

message: is there a generalization for arbitrary product spaces, or at least for countable product spaces?  —Preceding unsigned comment added by 91.18.108.36 (talk) 17:52, 8 January 2010 (UTC)
replies: text: My elementary multivariable calculus book gives a "Fubini's theorem for triple integrals" that states that the Riemann integral of a continuous function on (a subset of) R3 is the same as the triple iterated integral. I would expect this to generalize to Rn too.--Jasper Deng (talk) 04:56, 8 July 2014 (UTC)
thread_title:  Generalization 

message: The "Alternate" statement has 3 major problems:
1.  It is "alternative" not "alternate" (I know lots of Americans are fond of this malaprop, but in the case of analysis I'd be careful as "alternate" means something quite specific, as in "alternate signs" or "alternate series".)
2.  The result as stated is not an alternative, but rather the converse.  In fact, this is what goes by the name of Tonelli in the literature.  See  
3. The result as stated is wrong.  Same problem with absolute values has been pointed out.
--cerniagigante (talk) 19:44, 18 December 2012 (UTC)
thread_title:  Three problems with the "Alternate" statement 

message: I think this article can be improved by stating which types of \sigma-finite measure satisfy the conditions needed to exchange the order of integration. Obviously Lebesgue measures work, but do Borel measures? Mostly this information is useful for completeness (math joke). For some background see (https://en.wikipedia.org/wiki/Infinite-dimensional_Lebesgue_measure). Mouse7mouse9 07:32, 10 March 2015 (UTC)
thread_title:  Measure specifics 

message: Can someone give me a reference to where this is proved for the maximal product measure? Every proof I can find assumes either that both X and Y are σ-finite or that both are complete. Rickhev1 (talk) 18:57, 9 April 2015 (UTC)
thread_title:  Fubini's theorem for integrable functions 

message: It might be helpful for some readers to explicitly state a version for Riemann integrals so that it is easier to understand for readers not being familiar with Lebesgue integration of measure theory.--Kmhkmh (talk) 23:53, 8 July 2015 (UTC)
replies: text: I agree, and I will tag this article with . D.Lazard (talk) 16:09, 24 August 2020 (UTC)
thread_title:  Less abstract versions