Talk:Fubini's theorem

Enunciation of Fubini-Tonelli's Theorem not complete
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)

Generalization for arbitrary measure
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)

Problem with Corollary
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)
 * 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)

Which Fubini
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

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)

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)

Proof still missing
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)

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)


 * I think completeness is required too for these theorems. With Fubini we assume the function is integrable, so its support is sigma-finite and we can use that subspace therefore sigma-finiteness is unnecessary. Scineram (talk) 10:00, 3 October 2008 (UTC)


 * Completeness of the measure space is not required. In addition to Rudin, one could see for example N. V. Krylov "Introduction to Diffusion Processes" pp23-24.)  With these proof sigma-finiteness is assumed, but the function &fnof;  does not have to be assumed to be integrable on the the product. Instead the theorem says if one of iterated integrals for |&fnof;| is finite, then fnof; is integrable and the integral in the product space is the same as both iterated integrals.  Comparing the two proofs it seems that you can either assume completeness and integrability on the product, or sigma-finiteness and one of that an iterated integral makes sense for |&fnof;|.  Interestingly looking at the book that is linked above, the author sites both the papers by Fubini and Tonelli.  Looking at the dates it seems that the statement that we make "Tonelli's theorem (named after Leonida Tonelli) is a predecessor of Fubini's theorem"  is false, since it came 2 years later.  Instead, Tonelli dispensed with assuming completeness and introduced sigma finiteness, and was considered a generalization of Fubini's theorem. Thenub314 (talk) 12:01, 3 October 2008 (UTC)


 * Also it seems it would be better if we included the second form of Fubini's theorem, and maybe included the counter examples from Rudin rather then giving an outline to the proof. What to people think? Thenub314 (talk) 13:27, 3 October 2008 (UTC)


 * I am trying to find a proof of the main statement as of 09/25/2013, which requires completeness of the factors but no sigma-finiteness. If anyone can point me to one, I would appreciate it. I have a feeling that whether the main statement is true or not depends on what you mean by "mu cross nu." diBenedetto's proof (p147, Real Analysis, ISBN 978-0817642310) seems to be based on the definition of product measure as the one from the Caratheodory extension theorem. I have seen other people use this convention, but it seems to me that the majority of people use the definition that appears in the "Product Measure" Wikipedia article, i.e. any measure whose value on the rectangles is the product of the values of the factors. I think that the main statement of Fubini's theorem is probably not true for an arbitrary product measure. I have an idea for a counterexample, but I haven't worked out the details. I will sketch it below; feel free to remove what follows if it doesn't conform to style conventions.


 * Let μ' be the Lebesgue measure on B([0,1]) and let ν' be the counting measure on B([0,1]). Let A and B be the respective completions of B([0,1]) under μ' and ν', and let μ and ν be the respective unique extensions of μ' and ν' to A and B. For a set T, let I_T be its indicator function. Let S:={(x,x): x∈[0,1]}. By an exercise in P.R. Halmos's Measure Theory (p145, ISBN 0-387-90088-8), ∫∫I_S dν' dμ' = 1 but ∫∫I_S dμ' dν' = 0; also, the functions T→∫∫I_T dμ' dν' and T→∫∫I_T dν' dμ' are both product measures on the sigma-algebra generated by B([0,1])xB([0,1]). I'm pretty sure that these things remain true after completion, i.e. that ∫∫I_S dν dμ = 1, ∫∫I_S dμ dν = 0, and T→∫∫I_T dμ' dν' and T→∫∫I_T dν' dμ' are both product measures on the sigma-algebra generated by AxB. I'm also pretty sure that S is in the sigma-algebra generated by B([0,1])xB([0,1]), so that I_S is measurable. If that's all true, then A, B, and I_S form a counterexample to the current theorem statement on the Wikipedia article if you are using the second definition of product measure that I gave. However, if we use the Caratheodory version of the product measure, then I think the hypothesis is no longer satisfied because the integral of I_S is then infinite. (I put this comment in the sigma-finiteness section since the product measure is unique if the factors are sigma-finite, at least if you use some version of AC.) 98.222.72.97 (talk) 00:36, 26 September 2013 (UTC)

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)

Clairaut's theorem
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)

Fubini's Theorem is about reversing the order of integration; Clairaut's Theorem is about reversing the order of differentiation. Ah! I see the problem: the link was to the wrong Clairaut's Theorem; I'll fix that. —Toby Bartels (talk) 12:56, 13 June 2015 (UTC)

Theorem Statements
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)

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)

Theorem Statements
The theorem, as currently written, is false.

Here is the current version;

Suppose A and B are complete measure spaces. If
 * $$\int_{A\times B} |f(x,y)|\,d(x,y)<\infty,$$

where the integral is taken with respect to a product measure on the space over $A &times; B$, then


 * $$\int_A\left(\int_B f(x,y)\,dy\right)\,dx=\int_B\left(\int_A f(x,y)\,dx\right)\,dy=\int_{A\times B} f(x,y)\,d(x,y),$$

Here is a counter example;

Let A and B both be the Lebesgue measurable subsets of [0,1].

Let P be a nonmeasurable subset of the unit square [0,1] X [0,1].

Let


 * $$ f(x,y) = \left\{\begin{matrix} 1 & \mbox{if} \quad (x,y) \in P \\ -1 & \mbox{otherwise} \end{matrix}\right. $$

This f(x,y) satisfies the hypothesis of the theorem statement as currently written, but not the conclusion.

The problem is that this f(x,y) is not measurable, but the absolute value of f(x,y) is measurable.

Therefore, I will edit this current version of the theorem to include the additional hypothesis that f(x,y) itself is A X B measurable.

Generalization
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)
 * 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)

Three problems with the "Alternate" statement
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)

Measure specifics
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) — Preceding unsigned comment added by Mouse7mouse9 (talk • contribs)

Fubini's theorem for integrable functions
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)

Less abstract versions
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)
 * I agree, and I will tag this article with technical. D.Lazard (talk) 16:09, 24 August 2020 (UTC)

Product of two integrals
Maybe I'm just making a mistake, but it seems like the proof in this section adds in the parameter y wrong: It claims:
 * $$\int_{0}^{1} x\,v(xy) \,\mathrm{d}y = \biggl[V(xy)\biggr]_{y = 0}^{y = 1} = V(x) $$

but, when I work it out, I get
 * $$\int_{0}^{1} x\,v(xy) \,\mathrm{d}y = x\,\biggl[V(xy)\biggr]_{y = 0}^{y = 1} = x\,V(x) $$

while
 * $$\int_{0}^{1} v(xy) \,\mathrm{d}y = \biggl[V(xy)\biggr]_{y = 0}^{y = 1} = V(x) $$

This section has no sources, and I haven't found an alternative source for what the product of two integrals is, so I thought I shouldn't trust what's written and should note this possible mistake here. If this is wrong, then the identity at the top of the section is probably wrong, too. DubleH (talk) 16:55, 24 January 2024 (UTC)
 * Reformbenediktiner's version is correct. We want to simplify
 * $$I=\int_0^1 xv(xy)\,\mathrm dy.$$
 * Let $$xy=z$$. Then $$\mathrm dy/\mathrm dz=1/x$$ and
 * $$I=\int_0^x \frac{z}{y}v(z)\frac{1}{x}\, \mathrm dz=\int_0^x v(z)\, \mathrm dz=V(x).$$
 * So I'm removing the "dubious" tags. A1E6 (talk) 09:48, 6 April 2024 (UTC)