Product Measures and Fubini-Tonelli Theorems

  In this post we collect standard materials on product measures, including a description for the Komolgorov's extension theorem. There are two pillars under the theory: the monotone convergence theorem, which is already important in developing the Lebesgue integration theory, and the monotone class lemma, which can be viewed as a set-theoretic counterpart of the former. Putting them together implies the all-important Fubini-Tonelli theorems that allow one to compute integrals, reducing higher dimensions to lower ones, leading also to tricks to calculating integrals such as the Gaussians. I have followed the Terry's notes most of the time.

1.1. Finite product measure spaces

The cartesian product $ {X\times Y}$ of two sets $ {X}$ and $ {Y}$ is characterized by the associated projection maps

$ \displaystyle \pi_{X}:X\times Y\rightarrow X\quad\pi_{Y}:X\times Y\rightarrow Y. $

We can make use of these maps to construct a product $ {\sigma}$-algebra: given two measurable spaces $ {(X,\mathcal{B}_{X})}$ and $ {(Y,\mathcal{B}_{Y})}$, we can form a collections of subsets in $ {X\times Y}$ by pulling back $ {\mathcal{B}_{X}}$:

$ \displaystyle \pi_{X}^{*}(\mathcal{B}_{X}):=\{\pi_{X}^{-1}(E):E\in\mathcal{B}_{X}\}=\{E\times Y:E\in\mathcal{B}_{X}\}. $

It is easily verifyed that $ {\pi_{X}^{*}(\mathcal{B}_{X})}$ is a $ {\sigma}$-algebra. Note that $ {\emptyset\times Y=\emptyset}$.

Definition 1 (Finite product $ {\sigma}$-algebra) The product $ {\sigma}$-algebra $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ is defined to be the $ {\sigma}$-algebra generated by the union of the pullback $ {\sigma}$-algebras

$ \displaystyle \mathcal{B}_{X}\times\mathcal{B}_{Y}=\langle\pi_{X}^{*}(\mathcal{B}_{X})\cup\pi_{Y}^{*}(\mathcal{B}_{Y})\rangle. $

By observing that $ {E\times Y=(E\times Y)\cup\emptyset}$, $ {(E\times Y)\cap(X\times F)=E\times F}$, we see that $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ is also the $ {\sigma}$-algebra generated by rectangles $ {E\times F}$, where $ {E\in\mathcal{B}_{X}}$ and $ {F\in\mathcal{B}_{Y}}$. The definition of product $ {\sigma}$-algebra automatically makes the projection maps $ {\pi_{X}}$ and $ {\pi_{Y}}$ measurable morphisms, and in fact, it is the coarsest one that makes this happen.

Proposition 2 Let $ {x\in X}$, $ {y\in Y}$ and $ {A\in\mathcal{B}_{X}\times\mathcal{B}_{Y}}$. The $ {x}$-slice $ {A_{x}:=\{y:(x,y)\in A\}}$ is measurable for every $ {x\in X}$. Similarly, the $ {y}$-slice $ {A^{y}:=\{x:(x,y)\in A\}}$ is measurable for every $ {y\in Y}$.

Proof: Let $ {\mathcal{A}}$ be the subset of $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ such that $ {A_{x}}$ is measurable for all $ {A\in\mathcal{A}}$. We are to show that $ {\mathcal{A=\mathcal{B}_{X}\times\mathcal{B}_{Y}}}$, and it suffices to have $ {\mathcal{A}}$ is a $ {\sigma}$-algebra containing all measurable rectangles.
First, since for $ {A=E\times F\in\mathcal{B}_{X}\times\mathcal{B}_{Y}}$, $ {A_{x}=F}$ if $ {x\in E}$ and $ {A_{x}=\emptyset}$ otherwise, all measurable rectangles are contained in $ {\mathcal{A}}$. So in particular $ {X\times Y\in\mathcal{A}}$.
Next, if $ {A\in\mathcal{A}}$, then $ {\left(A^{c}\right)_{x}=\left(A_{x}\right)^{c}}$, so $ {A^{c}\in\mathcal{A}}$. Furthermore, if $ {A_{i}\in\mathcal{A}}$ for $ {i=1,2,\dots}$, and $ {A=\bigcup_{i}A_{i}}$, then

$ \displaystyle A_{x}=\bigcup_{i}\left(A_{i}\right)_{x} $

This shows that $ {\mathcal{A}}$ is a $ {\sigma}$-algebra. $ \Box$

 The converse is NOT true, as the follwing example shows.

Example 1 If $ {\mathcal{B}_{1}}$, $ {\mathcal{B}_{2}}$ are the Borel $ {\sigma}$-algebras of $ {\mathbb{R}^{d_{1}}}$ and $ {\mathbb{R}^{d_{2}}}$ respectively, then $ {\mathcal{B}_{1}\times\mathcal{B}_{2}}$ is the Borel $ {\sigma}$-algebra of $ {\mathbb{R}^{d_{1}+d_{2}}}$. This is closely related to the fact that the product topology is generated by sets of the form $ {\pi_{1}^{-1}(U)\cup\pi_{2}^{-1}(V)}$ for $ {U}$, $ {V}$ open in $ {\mathbb{R}^{d_{1}}}$ and $ {\mathbb{R}^{d_{2}}}$ respectively. In contrast, the product Lebesgue $ {\sigma}$-algebras $ {\mathcal{M}_{1}\times\mathcal{M}_{2}}$ on $ {\mathbb{R}^{d_{1}+d_{2}}}$ is not the Lebesgue $ {\sigma}$-algebra on $ {\mathbb{R}^{d_{1}+d_{2}}}$. This can be seen via Proposition 2. For example, if $ {\mathcal{N}}$ is a non-measurable set in $ {\mathbb{R}}$ and $ {E}$ is a null set (say, a point) in $ {\mathbb{R}}$, then $ {\mathcal{N}\times E}$ would be Lebesgue measurable in $ {\mathbb{R}^{2}}$, as it has measure zero.

Now, suppose we have two measure spaces $ {(X,\mathcal{B}_{X},\mu_{X})}$ and $ {(X,\mathcal{B}_{X},\mu_{X})}$. For a measurable rectangle $ {E\times F}$ in the product $ {\sigma}$-algebra, we can define its measure by a product formula:

$ \displaystyle \mu_{X}\times\mu_{Y}(E\times F)=\mu_{X}(E)\mu_{Y}(F). $

The question is whether we can extend it to a measure on the product $ {\sigma}$-algebra. This will be done once we can show it is a pre-measure, since the extension will then follow from Hahn-Kolmogorov extension theorem.

Proposition 3 (Existence and uniqueness of (unsigned) product measure for $ {\sigma}$-finite spaces) Let $ {(X,\mathcal{B}_{X},\mu_{X})}$ and $ {(X,\mathcal{B}_{X},\mu_{X})}$ be $ {\sigma}$-finite measure spaces. Then there exists a unique measure $ {\mu_{X}\times\mu_{Y}}$ on $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ such that

$ \displaystyle \mu_{X}\times\mu_{Y}(E\times F)=\mu_{X}(E)\mu_{Y}(F) $

whenever $ {E\in\mathcal{B}_{X}}$ and $ {F\in\mathcal{B}_{Y}}$.

Recall that a pre-measure $ {\mu_{0}}$ is a set function defined on a Boolean algebra that satisfies $ {\mu_{0}(\emptyset)=0}$, finite additivity and countable additivity on disjoint sets in the Boolean algebra. The uniqueness of the Hahn-Kolmogorov extension holds under the assumption of $ {\sigma}$-finiteness. Note that the product measure, if exists, is $ {\sigma}$-finite if the individual measures are $ {\sigma}$-finite. It remains to establish the pre-measure.
Proof: Let $ {\mathcal{B}_{0}}$ be the collection consists of sets of the form

$ \displaystyle S=(E_{1}\times F_{1})\cup\cdots\cup(E_{k}\times F_{k}) $

where $ {E_{i}\in\mathcal{B}_{X}}$, $ {F_{i}\in\mathcal{B}_{Y}}$ for $ {i=1,\dots,k}$. In other words, $ {S}$ is a finite union of measurable rectangles, and we shall call such sets elementary sets. Since $ {\bigcup_{i}E_{i}}$ can be rewritten as a disjoint union $ {\bigsqcup_{i}\tilde{E}_{i}}$ where

$ \displaystyle \tilde{E}_{i}=\left(\bigcup_{j=1}^{i}E_{i}\right)\backslash\bigcup_{j=1}^{i-1}E_{i}, $

and similarly for $ {\bigcup_{i}F_{i}}$, we observe that $ {S}$ can be decomposed into a disjoint union of measurable rectangles. It is also easy to see that $ {\mathcal{B}_{0}}$ includes $ {\emptyset}$ as an element, and is closed under complement and finite unions, thus $ {\mathcal{B}_{0}}$ forms a Boolean algebra. Next, we define a function $ {\mu_{0}:\mathcal{B}_{0}\rightarrow[0,\infty]}$ by

$ \displaystyle \mu_{0}(S):=\sum_{j}\mu_{X}(E_{j})\mu_{Y}(F_{j}). $

To show it is a premeasure, let $ {S\in\mathcal{B}_{0}}$ and $ {S_{1},S_{2},\dots}$ be a countable disjoint family of elementary sets whose union is $ {S}$. By the previous observation, we may assume $ {S=E\times F}$, and also $ {S_{j}=E_{j}\times F_{j}}$ for each $ {j}$. Now, because of disjointness, we have a pointwise identity

$ \displaystyle \chi_{E}(x)\chi_{F}(y)=\sum_{j=1}^{\infty}\chi_{E_{j}}(x)\chi_{F_{j}}(y). $

Integrating with $ {d\mu_{Y}}$ over $ {Y}$ for each $ {x\in X}$, we find the LHS is $ {\chi_{E}(x)\mu_{Y}(F)}$. Applying the monotone covergence theorem to the evaluation of RHS, we get

$ \displaystyle \begin{array}{rcl} \int_{X}\sum_{j=1}^{\infty}\chi_{E_{j}}(x)\chi_{F_{j}}(y)d\mu_{Y}(y) & = & \sum_{j=1}^{\infty}\int_{X}\chi_{E_{j}}(x)\chi_{F_{j}}(y)d\mu_{Y}(y)\\ & = & \sum_{j=1}^{\infty}\chi_{E_{j}}(x)\mu_{Y}(F_{j}). \end{array} $

Now, integrating with $ {d\mu_{X}}$ over $ {X}$, and applying the monotone convergence theorem again, we get

$ \displaystyle \mu_{X}(E)\mu_{Y}(F)=\sum_{j=1}^{\infty}\mu_{X}(E_{j})\mu_{Y}(F_{j}). $

This shows that $ {\mu_{0}}$ is a pre-measure. $ \Box$

1.2. Infinite product measure spaces

From axiom of choice, we can form

$ \displaystyle X_{A}:=\prod_{\alpha\in A}X_{\alpha} $

for arbitrary family of sets $ {\{X_{\alpha}\}_{\alpha\in A}}$, where the elements in the space are tuples $ {x_{A}=(x_{\alpha})_{\alpha\in A}}$. Correspondingly, we have coordinate projection maps $ {\pi_{\beta}:X_{A}\rightarrow X_{\beta}}$, sending $ {x_{A}}$ to $ {x_{\beta}}$. We can also talk about ``products'' of such coordinate projection maps. For example, given $ {B\subset A}$, we can define a partial projection $ {\pi_{B}:X_{A}\rightarrow X_{B}}$ in an obvious way. More generally, if $ {C\subset B\subset A}$, we can define $ {\pi_{C\leftarrow B}:X_{B}\rightarrow X_{C}}$. Thus we have composition laws

$ \displaystyle \pi_{D\leftarrow C}\circ\pi_{C\leftarrow B}=\pi_{D\leftarrow B} $

if $ {D\subset C\subset B\subset A}$. As before, we can define pullback $ {\sigma}$-algebras for each $ {\beta\in A}$

$ \displaystyle \pi_{\beta}^{*}(\mathcal{B}_{X}):=\{\pi_{\beta}^{-1}(E):E\in\mathcal{B}_{X}\} $

and infinite product $ {\sigma}$-algebras

$ \displaystyle \mathcal{B}_{A}=\prod_{\beta\in A}\mathcal{B}_{\beta}:=\langle\bigcup_{\beta\in A}\pi_{\beta}^{*}(\mathcal{B}_{\beta})\rangle. $

Again, it is the coarsest $ {\sigma}$-algebra such that $ {\pi_{\beta}}$'s, and more generally $ {\pi_{B}}$'s are measurable morphisms. The elements in $ {\mathcal{B}_{A}}$, by definition of generation of $ {\sigma}$-algebra, are formed by countable set-theoretic operations. In particular, this means that if $ {E\in\mathcal{B}_{A}}$, there exists an at most countable set $ {B\subset A}$ and a set $ {E_{B}\in\mathcal{B}_{B}}$ such that $ {E_{A}=\pi_{B}^{-1}(E_{B})}$. Consequently, if $ {f:X_{A}\rightarrow[0,+\infty]}$ is measurable, it means that there is an at most countable set $ {B\subset A}$ and a $ {\mathcal{B}_{B}}$-measurable function $ {f_{B}:X_{B}\rightarrow[0,+\infty]}$ such that

$ \displaystyle f=f_{B}\circ\pi_{B}. $

The above formulation can accommodate the situation when we have a infinite family of random variables $ {X_{\alpha}}$, each is defined as a Borel measurable function on a (usually) locally compact Hausdorff sample space $ {\Omega_{\alpha}}$. (We changed the notation temporarily here only to follow the convention). Then we can ``extend'' the random variables to the product space $ {\prod_{\alpha\in A}\Omega_{A}}$ via the projections

$ \displaystyle \tilde{X}_{\alpha}=X_{\alpha}\circ\pi_{\alpha} $

and thus all random variables in the family can be regarded to be defined on the same space, if we can somehow turn this product space into a ``compatible'' measure space. Namely, if we have a probability measure $ {\mu_{A}}$ defined on the product $ {\sigma}$-algebra, then it will induce a pushforward measure $ {\mu_{B}}$ on $ {\Omega_{B}}$ for any $ {B\subset A}$, by

$ \displaystyle \mu_{B}(E_{B}):=\left(\pi_{B}\right)_{*}\mu_{A}=\mu_{A}(\pi_{B}^{-1}(E_{B})) $

for all $ {E_{B}\in\mathcal{B}_{B}}$. It follows then

$ \displaystyle (\pi_{C\leftarrow B})_{*}\mu_{B}=\mu_{C} $

if $ {C\subset B\subset A}$. With this we can stop worrying about the underlying space of ``statistics'' of the family of random variables, as one often studies their asymptotic behaviors. The extension will be guaranteed by Kolmogorov's extension theorem, given below.

Theorem 4 (Kolmogorov extension theorem) Let $ {((X_{\alpha},\mathcal{B}_{\alpha}),\mathcal{T}_{\alpha})_{\alpha\in A}}$ be a family of measurable spaces $ {(X_{\alpha},\mathcal{B}_{\alpha}),}$equipped with a topology $ {\mathcal{T}_{\alpha}}$. For each finite $ {B\subset A}$, let $ {\mu_{B}}$ be an inner regular probability measure on $ {\mathcal{B}_{B}}$, where the product space $ {X_{B}}$ is equipped with the product topology, and obeys the compatibility condition

$ \displaystyle (\pi_{C\leftarrow B})_{*}\mu_{B}=\mu_{C} $

whenever $ {C\subset B\subset A}$, where $ {C}$ and $ {B}$ are finite. Then there exists a unique probability measure $ {\mu_{A}}$ on $ {\mathcal{B}_{A}}$ with the property that $ {(\pi_{B})_{*}\mu_{A}=\mu_{B}}$ for all finite $ {B\subset A}$.

1.3. Theorems of Fubini and Tonelli

There will be two essential tools that to establish the Fubini-type of results: one is the monotone convergence theorem, which has already played a crucial role to establish the convergence of the intetgral the proof of Proposition 2; the other is the monotone class lemma. Roughly speaking, the monotone convergence allows one to conclude convergence result in the unsigned setting, while the monotone class lemma allows one to conclude that the set, where the convergence result holds, is a $ {\sigma}$-algebra.
We have encountered a monotone class in the very begining. A collection $ {\mathcal{B}}$ of subsets of $ {X}$ is said to be a monotone class if $ {X}$ satisfies the following two closure properties:

1. If $ {E_{1}\subset E_{2}\subset\cdots}$ is a countable ascending sequence of sets in $ {\mathcal{B}}$, then $ {\bigcup_{n=1}^{\infty}E_{n}\in\mathcal{B}}$;
2. If $ {E_{1}\supset E_{2}\supset\cdots}$ is a countable descending sequence of sets in $ {\mathcal{B}}$, then $ {\bigcap_{n=1}^{\infty}E_{n}\in\mathcal{B}}$.

Lemma 5 (Monotone class lemma) Let $ {\mathcal{A}}$ be a Boolean algebra on $ {X}$. Then the $ {\sigma}$-algebra generated by $ {\mathcal{A}}$, denoted $ {\langle\mathcal{A}\rangle}$ is the smallest monotone class that contains $ {\mathcal{A}}$.

Proof: Obviously, $ {\langle\mathcal{A}\rangle}$ is a monotone class. Let $ {\mathcal{B}}$ be the smallest monotone class containing $ {\mathcal{A}}$, obtained by intersection of all monotone class containing $ {\mathcal{A}}$. We are to show $ {\mathcal{B}\supset\langle\mathcal{A}\rangle}$, whence $ {\mathcal{B}=\langle\mathcal{A}\rangle}$. Since $ {\mathcal{B\supset\mathcal{A}}}$, it suffices to show that $ {\mathcal{B}}$ is a $ {\sigma}$-algebra.
First, $ {\mathcal{B\supset\mathcal{A}}}$ implies in particular $ {X\in\mathcal{B}}$. By replacing sets with their complements, we see a swap of the properties 1 and 2, and thus $ {\mathcal{B}}$ is closed under complement. It remains to establish the closure under countable unions. This will be done if we can show $ {\mathcal{B}}$ is a Boolean algebra. Indeed, if so, then given a countable sequence $ {\{E_{i}\}_{i=1}^{\infty}}$ of sets in $ {\mathcal{B}}$, we can form a new ascending sequence $ {\{\tilde{E}_{i}\}}$ by defining

$ \displaystyle \tilde{E}_{n}=\bigcup_{i=1}^{n}E_{i}, $

where each $ {\tilde{E}_{n}}$ is in $ {\mathcal{B}}$ and $ {\bigcup_{i}E_{i}=\bigcup_{i}\tilde{E}_{i}}$. Since $ {\mathcal{B}}$ is a monotone class, we conclude that $ {\bigcup_{i}E_{i}\in\mathcal{B}}$. Now, for any $ {E\in\mathcal{A}}$, consider the set $ {\mathcal{C}_{E}}$ of all sets $ {F\in\mathcal{B}}$ such that $ {E\cup F}$ all lie in $ {\mathcal{B}}$. It is clear that $ {\mathcal{C}_{E}\supset\mathcal{A}}$ since $ {\mathcal{A}}$ is a Boolean algebra. Let $ {F_{1}\subset F_{2}\subset\cdots}$ be an ascending sequence in $ {\mathcal{C}_{E}}$. Since $ {\mathcal{B}}$ is a monotone class, we have $ {\bigcup_{i}F_{i}\in\mathcal{B}}$. It is cleat that $ {\bigcup_{i}\left(F_{i}\backslash E\right)=\left(\bigcup_{i}F_{i}\right)\backslash E}$ is also in $ {\mathcal{B}}$, and verifications for other cases implies that $ {\mathcal{C}_{E}}$ is a monotone class, and by definition of $ {\mathcal{B}}$, we have $ {\mathcal{C}_{E}=\mathcal{B}}$.
Finally, let $ {\mathcal{D}}$ be the set of all $ {E\in\mathcal{B}}$ such that $ {E\cup F}$ all lie in $ {\mathcal{B}}$ for all $ {F\in\mathcal{B}}$. We see that $ {\mathcal{D}\supset\mathcal{A}}$ by the above argument, and it is easy to verify that $ {\mathcal{D}}$ is a monotone class, and thus $ {\mathcal{D}=\mathcal{B}}$. This implies $ {\mathcal{B}}$ is closed under finite unions, and thus finishes the proof. $ \Box$

Theorem 6 (Tonelli's theorem, incomplete measure case) Let $ {(X,\mathcal{B}_{X},\mu_{X})}$ and $ {(Y,\mathcal{B}_{Y},\mu_{Y})}$ be $ {\sigma}$-finite measure spaces, and let $ {f:X\times Y\rightarrow[0,+\infty]}$ be measurable with respect to $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$. Then:

1. The function $ {x\mapsto\int_{Y}f(x,y)d\mu_{Y}(y)}$ is well defined, and measurble with respect to $ {\mathcal{B}_{X}}$;
2. We have
   $ \displaystyle \begin{array}{rcl} & & \int_{X\times Y}f(x,y)d\mu_{X}\times d\mu_{Y}(x,y)\\ & =& \int_{Y}\left(\int_{X}f(x,y)d\mu_{X}(x)\right)d\mu_{Y}(y) \\ & = & \int_{X}\left(\int_{Y}f(x,y)d\mu_{Y}(x)\right)d\mu_{X}(x). \end{array} $

Proof: The first part holds due to Proposition 2. For the second part, we may assume that $ {\mu_{X}}$ and $ {\mu_{y}}$ are finite measures, as the general case follows by an easy application of monotone convergence theorem. We thus have

$ \displaystyle \mu_{X}\times\mu_{Y}(X\times Y)=\mu_{X}(X)\mu_{Y}(Y)<+\infty. $

Since every unsigned measurable function is the increasing limit of unsigned simple functions, by monotone convergence theorem, it suffices to prove the claim when $ {f}$ is a simple function. By linearity of integration, it suffices to prove for the indicator function $ {\chi_{S}}$ for some $ {S\in\mathcal{B}_{X}\times\mathcal{B}_{Y}}$.
Let $ {\mathcal{C}}$ be the set of all $ {S\in\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ for which the claim holds. Measurable rectangles are certainly in $ {\mathcal{C}}$. Thus $ {\mathcal{C}}$ is a Boolean algebra. Furthermore, if $ {E_{1}\subset E_{2}\subset\cdots}$ is an ascending sequence in $ {\mathcal{C}}$, then applying monotone convergence theorem, we have

$ \displaystyle \begin{array}{rcl} \int_{X\times Y}\chi_{\cup_{i=1}^{\infty}E_{i}} & = & \lim_{n\rightarrow\infty}\int_{X\times Y}\chi_{\cup_{i=1}^{n}E_{i}}d\mu_{X}\times d\mu_{Y}\\ & = & \lim_{n\rightarrow\infty}\int_{X}\left(\int_{Y}\chi_{\cup_{i=1}^{n}E_{i}}d\mu_{Y}\right)d\mu_{X}\\ & = & \int_{X}\left(\lim_{n\rightarrow\infty}\int_{Y}\chi_{\cup_{i=1}^{n}E_{i}}d\mu_{Y}\right)d\mu_{X}\\ & = & \int_{X}\left(\int_{Y}\chi_{\cup_{i=1}^{\infty}E_{i}}d\mu_{Y}\right)d\mu_{X}, \end{array} $

and in case we have a descending sequence, we use dominated convergence theorem instead since we are in a finite measure space. Thus $ {\mathcal{C}}$ is a monotone class. It follows from monotone class lemma that $ {\mathcal{C}}$ contains $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$, and we are done. $ \Box$

 Now if $ {E\in\mathcal{B}_{X}\times\mathcal{B}_{Y}}$ is a null set, applying Tonelli's theorem to $ {\chi_{E}}$ we get

$ \displaystyle 0=\int_{Y}\left(\int_{X}\chi_{E}d\mu_{X}\right)d\mu_{Y}(y)=\int_{X}\left(\int_{Y}\chi_{E}d\mu_{Y}\right)d\mu_{X}(y). $

Since $ {\int_{X}\chi_{E}d\mu_{X}}$ and $ {\int_{Y}\chi_{E}d\mu_{Y}}$ are non-negative, we conclude that they are zero for almost every $ {y}$ and $ {x}$, respectively. Hence the $ {x}$- and $ {y}$-slices of $ {E}$ are also nullsets in $ {\mathcal{B}_{X}\times\mathcal{B}_{Y}}$. By this observation, we obtain the complete measure space version for Proposition 2 and hence Tonelli's theorem, with the only difference being that the $ {x}$- and $ {y}$-slices are now only measurable for almost every $ {x}$ and $ {y}$, respectively.

The hypotheses in the theorem that measurable spaces are $ {\sigma}$-finite, $ {f}$ being measurable are all essential. For the former, let one copy of $ {A=[0,1]}$ be equipped with the usual Lebesgue $ {\sigma}$-algebra $ {\mathcal{M}}$ and Lebesgue measure, and the other copy $ {B}$ be equipped with the discrete $ {\sigma}$-algebra $ {\mathcal{P}}$ and counting measure. Let $ {E}$ be the diagonal $ {\{(x,x):x\in[0,1]\}}$. Then $ {E\in\mathcal{M}\times\mathcal{P}}$, since it can be obtained by a descending sequence of finite unions of measurable rectangles. Hence $ {\chi_{E}}$ is measurable, and

$ \displaystyle \int_{A}\left(\int_{B}\chi_{E}d\mu_{B}\right)d\mu_{A}=\int_{A}1d\mu_{A}=1\neq0=\int_{B}\left(\int_{A}\chi_{E}d\mu_{A}\right)d\mu_{B}. $

For the latter, besides that the integration on product space makes no sense, the ``iterated integrals'' also need not agree. Let $ {[0,1]^{2}}$ be equipped with the product Lebesgue measure. We will make use of the paradoxical well-ordering $ {\prec}$ of the real numbers in $ {[0,1]}$ to cook up a non-measurable function. Let $ {E\subset[0,1]^{2}}$ be defined as

$ \displaystyle E:=\{(x,y)\in[0,1]^{2}:x\prec y\}. $

We have

$ \displaystyle \int_{[0,1]}\left(\int_{[0,1]}\chi_{E}dx\right)dy=\int_{[0,1]}0dy=0 $

since there are only at most countably many $ {x}$ that precede a particular$ {y\in[0,1]}$, and a countable set has measure zero. Meanwhile, we have

$ \displaystyle \int_{[0,1]}\left(\int_{[0,1]}\chi_{E}dy\right)dx=\int_{[0,1]}1dy=1, $

since for fixed $ {x}$, there are at most countably many $ {y}$ that precede $ {x}$, and thus the set of $ {y}$ that succeeds has full measure. By Tonelli's theorem, $ {E}$ cannot be measurable.

We may now easily extend the unsigned setting to the absolutely integrable setting, leading to Fubini's theorem.

Theorem 7 (Fubini's Theorem) Let $ {(X,\mathcal{B}_{X},\mu_{X})}$ and $ {(Y,\mathcal{B}_{Y},\mu_{Y})}$ be $ {\sigma}$-finite complete measure spaces, and let $ {f:X\times Y\rightarrow[-\infty,+\infty]}$ be absolutely integrable with respect to $ {\overline{\mathcal{B}_{X}\times\mathcal{B}_{Y}}}$. Then:

1. For $ {\mu_{X}}$-almost every $ {x}$, $ {y\mapsto f(x,y)}$ is $ {\mathcal{B}_{Y}}$ measurable, and the integral $ {\int_{Y}f(x,y)d\mu_{Y}(y)}$ is defined, and the $ {\mu_{X}}$-almost everywhere defined map $ {x\mapsto\int_{Y}f(x,y)d\mu_{Y}(y)}$ is measurble with respect to $ {\mathcal{B}_{X}}$;
2. We have
   $ \displaystyle \begin{array}{rcl} & & \int_{X\times Y}f(x,y)\overline{d\mu_{X}\times d\mu_{Y}}(x,y)\\ & =& \int_{Y}\left(\int_{X}f(x,y)d\mu_{X}(x)\right)d\mu_{Y}(y)\\ & =& \int_{X}\left(\int_{Y}f(x,y)d\mu_{Y}(x)\right)d\mu_{X}(x). \end{array} $

Again, the absolute integrability of $ {f}$ is essential, as only under such case the dominated convergence theorem shall work.