# Fubini's theorem without completeness or $\sigma$-finiteness conditions

The usual Fubini's theorem(see the Wikipedia article for example) assumes completeness or $\sigma$-finiteness on measures. However, I think I came up with a proof of the Fubini's theorem without those assumptions. Am I mistaken?

I restate the theorem to avoid confusion: If a function is integrable on a product measure space, its integral can be calculated by iterated integrals.

The idea of my proof is to use a fact that if a function is integrable on a product measure space, the function must be zero outside a $\sigma$-finite subset of the product measure space.

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I added the link to the wikipedia article. – David Roberts Apr 19 '12 at 4:15
Makoto above gave an idea for proving Fubini's theorem in the non $\sigma$-finite case, using the fact that if $f$ is integrable then it has $\sigma$-finite support. Could some one explain how the proof continues from here? I assumed that we want to try and use the regular $\sigma$-finite Fubini theorem. To do so the original spaces $(X,\mu_1)$ and $(Y,\mu_2)$ should be $\sigma$-finite, and this could be done if we could in some way approximate the support of $f$ by countable disjoint unions of rectangles or something of the sort. However I couldn't show that this can indeed be done. – user30275 Dec 29 '12 at 15:58

You do not need $\sigma$-finiteness of the measure in Fubini theorem, although it is an hypothesis that can be assumed with no loss of generality, in that the support of an integrable function is, of course, $\sigma$-finite.

On the opposite, Tonelli theorem deals with non-negative measurable functions, whose support may well be non-$\sigma$-finite (as in the quoted example) and $\sigma$-finiteness is really needed.

As to the hypothesis of completeness of the factor measures spaces $(X,\mathcal A,\mu)$ and $(Y,\mathcal B, \nu)$, it is not necessary, in either theorems (but, again, it could be assumed w.l.o.g., since the two measures can be completed). Notice however that some care is needed in stating the theorem if you consider a function on the product space which is measurable with respect to the $\mu\otimes\nu$ -completion of the product $\sigma$-algebra $\mathcal {A}\otimes\mathcal B$, more generally than just "measurable wrto the product $\sigma$-algebra". This generalization occurs quite naturally dealing with a Lebesgue measurable function on ${\bf R}^n\times{\bf R}^m$ (because the completion of the product measures is the Lebesgue measure on the product space). The result of taking this more general hypothesis is that you have to stuff the statement of the theorem with a sequel of "a.e.", which wouldn't be necessary in the case $f\in \mathcal{L}^1 (X\times Y, \mathcal {A}\otimes\mathcal B, \mu\otimes\nu)\, .$

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Thanks, Pietro. Finally I could find someone who assured my assertion. All books I know on measure theory assume $\sigma$-finite measures on Fubini's theorem while it's unnecessary. – Makoto Kato Apr 19 '12 at 6:28
Nice, Pietro! I was wondering how do "stuff the statement of the theorem with a sequel of "a.e.""? – Tim Feb 22 '13 at 14:30

Folland's Real Analysis: Modern Techniques and Their Applications states the Fubini-Tonelli Theorem as

Suppose that $(X, \mathcal{M},\mu)$ and $(Y, \mathcal{N},\nu)$ are $\sigma$-finite measure spaces.

a. (Tonelli) If $f \in L^+(X \times Y)$, then the functions $g(x) = \int f_x d\nu$ and $h(y) = \int f^y d\mu$ are in $L^+(X)$ and $L^+(Y)$, respectively, and $$\int f d(\mu \times \nu) = \int \left[ \int f(x,y) d\nu(y) \right] d\mu(x) = \int \left[ \int f(x,y) d\mu(x) \right] d\nu(y).$$ b. (Fubini) If $f \in L^1(\mu \times \nu)$, then $f_x \in L^1(\nu)$ for a.e. $x \in X, f^y \in L^1(\mu)$ for a.e. $y \in Y$, the a.e.-defined functions $g(x) = \int f_x d\nu$ and $h(x) = \int f^y d\nu$ are in $L^1(\mu)$ and $L^1(\nu)$, respectively, and the above equation holds.

(Page 67)

So completeness is not necessary. On the other hand, $\sigma$-finiteness is necessary. This exercise follows the Fubini-Tonelli theorem in Folland. Here $\mathcal{B}_{[0,1]}$ denotes the usual Borel $\sigma$-algebra on $[0,1]$.

46. Let $X = Y = [0,1]$, $\mathcal{M} = \mathcal{N} = \mathcal{B}_{[0,1]}$, $\mu =$ Lebesgue measure, and $\nu =$ counting measure. If $D = \{(x,x): x\in [0,1]\}$ is the diagonal in $X \times Y$, then $\iint \chi_D \; d\mu\; d\nu$, $\iint \chi_D \; d\nu \; d\mu$, and $\int \chi_D \; d(\mu \times \nu)$ are all unequal.

(Page 68)

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Adam, the measure of the diagonal is not finite. So that's not a counterexample for my assertion. – Makoto Kato Apr 19 '12 at 5:06
Ah! Thank you for pointing out my error, Makoto (and Pietro and Juan). – Adam Saltz Apr 21 '12 at 13:59
Anyway, if this is a counterexample or not depends on the definition of integrable: a function can be defined to be integrable in the extended sense if either $\int f^+ < \infty$ or $\int f^- < \infty$. In that case, the characteristic function of the diagonal is an honest integrable function (in the extended sense) and hence a counterexample to the statement, +1 indeed – Giovanni Nov 5 '13 at 21:49

This is a complement to the above answers. With which I agree.

I think the definition of the product measure is not unique. This is best seen in that the (usual definition of the) product of the complete Lebesgue measure on ${\bf R}$ give us not the complete Lebesgue measure on the product space ${\bf R}^2$.

What we want is a measure $\mu\otimes\nu$ such that $\mu\otimes\nu(A\times B)=\mu(A)\nu(B)$ for $A$ and $B$ measurables and of finite measure.

Therefore it is natural to consider here the sigma algebra $\Sigma_0$ generated by the products $A\times B$ with $\mu(A)$ and $\nu(B)$ finite. Usually we consider the sigma algebra $\sigma$ generated by the the products $A\times B$ of measurable sets. In the case of $\sigma$-finite measures the two sigma algebras coincide.

If you consider only the measure on $\Sigma_0$ (this is not what it is usually done) you obtain a product measure that is unique in $\Sigma_0$ and both theorems Fubini-Tonelli and Fubini are true, without assuming anything about the measures.

The usual example, given above by Adam Saltz, is not a counterexample because the diagonal is not measurable (that is, it is not in $\Sigma_0$).

With this definition of the product we get the same integrable function that with the usual one. This is what make the theorem of Fubini-Tonelli true
because the support of an integrable function is sigma finite.

So I propose to define the product measure always on $\Sigma_0$. We get the usual definition in the $\sigma$-finite case. In other case we get always Fubini-Tonelli and Tonelli theorems without restrictions. I have experimented this many times in my classes of Measure Theory (now dead by Bolonia reform).

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Here is another complement to the above answers:

With Fubini on non-$\sigma$-finite spaces you have to be careful about what you mean by 'product measure'.

Suppose we use the Carathéodory method to extend the premeasure from measurable rectangles to a complete measure. Let $\mu\otimes\nu$ mean this measure restricted to $\mathcal A \times \mathcal B$. (The restricted measure may no longer be complete.) Then Fubini holds for any function integrable with respect to $\mu \otimes \nu$.

Now consider the problem from Folland above (#46), except define $(\mu \times \nu) (S)$ to be the sum of the $\mu$ measures of all horizontal sections of $S$. (Note: Folland's $\mu \times \nu$ is the same as our $\mu \otimes \nu$, but even so, he presents Fubini only for $\sigma$-finite measures.) This is a product measure, since it satisfies $(\mu \times \nu)(A \times B) = \mu(A) \; \nu(B)$ for all measurable rectangles (defining $0 \cdot \infty = 0$). Then $\int \chi_D \; d(\mu \times \nu) = \iint \chi_D \; d\mu \; d\nu = 0$, but $\iint \chi_D \; d\nu \; d\mu = 1$. Thus, Fubini fails for this measure and integrable function.

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