Fubini’s Theorem

This is a presentation of Fubini’s theorem related to Durrett’s proof. There are some loose ends still to clear up.

Let $(X_i,A_i,\mu_i)$ be measure spaces for $i = 1,2$ . The product space is the measurable space $(X_1 \times X_2, \sigma(A_1 \times A_2))$ .

lemma: $A_1 \times A_2$ is a semi-algebra on $X_1 \times X_2$ .

Let $a \in A_1 \times A_2$ . Then $a = a_1 \times a_2$ where $a_1 \in A_1$ and $a_2 \in A_2$ . Since

$a^C = (a_1 \times a_2)^C = a_1^C \times a_2 \cup a_1 \times a_2^C \cup a_1^C \times a_2^C$

and since for $a,b \in A_1 \times A_2$ ,

$a \cap b = (a_1 \times a_2) \cap (b_1 \times b_2) = a_1 \cap b_1 \times a_2 \times b_2$ .

Elements of $A_1 \times A_2$ are called measurable rectangles.

lemma: the function $\mu$ on $A_1 \times A_2$ defined by $\mu(a) = \mu_1(a_1) \mu_2(a_2)$ satisfies the extension conditions of the Catheodory extension theorem.

Suppose $a$ is a measurable rectangle that can be expressed as a disjoint finite union of measurable rectangles $a_1,\ldots,a_n$ . Then

$\mathbb{I}_a(x_1,x_2) = \sum_{i=1..n} \mathbb{I}_{a_i}(x_1,x_2)$

But since the sets are rectangular,

$\mathbb{I}_{a_1}(x_1) \mathbb{I}_{a_2}(x_2) = \sum_{i=1..n} \mathbb{I}_{a_{i1}}(x_1) \mathbb{I}_{a_{i2}}(x_2)$

For any fixed $x_2$ , the functions are non-negative measurable functions of the space $X_1$ . Hence their measure (integral) is defined and given by

$\mu_1(a_1) \mathbb{I}_{a_2}(x_2) = \sum_{i=1..n} \mu_1{a_{i1}} \mathbb{I}_{a_{i2}}(x_2)$

the functions are non-negative measurable functions of the space $X_2$ . Hence their measure is defined and given by

$\mu_1(a_1) \mu_2(a_2) = \sum_{i=1..n} \mu_1{a_{i1}} \mu_2{a_{i2}}$

which can be written as

$\mu(a) = \sum_{i=1..n} \mu(a_i)$ .

establishing the first condition required of the semi-measure in the extension theorem. To establish the second condition, suppose that $a$ is a measurable rectangle that is a countable disjoint union of measurable rectangles. By similar analysis, one can conclude that

$\mathbb{I}_{a_1}(x_1) \mathbb{I}_{a_2}(x_2) = \sum_{i=1..} \mathbb{I}_{a_{i1}}(x_1) \mathbb{I}_{a_{i2}}(x_2)$

For any fixed $x_2$ , the functions are non-negative measurable functions of the space $X_1$ . Since the LHS is the limit of non-decreasing sequence of partial sums of non-negative measurable functions, the monotone convergence theorem can be applied to yield

$\mu_1(a_1) \mathbb{I}_{a_2}(x_2) = \sum_{i=1..} \mu_1{a_{i1}} \mathbb{I}_{a_{i2}}(x_2)$

a second application of the MCT yields

$\mu_1(a_1) \mu_2(a_2) = \sum_{i=1..} \mu_1{a_{i1}} \mu_2{a_{i2}}$

which can be written as

$\mu(a) = \sum_{i=1..} \mu(a_i)$ .

Hence the second condition of the extension theorem is true, and therefore the semi-measure can be extended to a measure on the algebra $a(A_1 \times A_2)$ . Assuming the extension is $\sigma$ -finite (a condition automatically satisfied if the component measures are finite, for example) the measure can be extended to $\sigma (A_1 \times A_2)$ .

Hence $(X_1 \times X_2, \sigma(A_1 \times A_2), \mu)$ where for all measurable rectangles $a, \mu(a) = \mu_1(a_1) \mu_2(a_2)$ defines a measure space called the product measure space of $X_1,X_2$ .

Fubini’s Theorem

Fubini’s theorem says the measure of measurable functions of the product measure space can be expressed as a measure of measures of slices of the function. Ie $\mu(f) = \mu_2(g : g(x_2) = \mu_1(f(x_2))$ . The approach to prove this will use a proof methodology called the “standard machine.” and also rely upon Dynkin’s $\pi,\lambda$ theorem.

Note that $f(x_1,x_2) = f(x_2)(x_1)$ where now the interpretation attached to evaluation operation $(x_i)$ is to say all points whose $i$ th component matches $x_i$ .

The standard machine builds up a result by extending results from indicator functions, simple functions, non-negative measurable functions, and finally measurable functions with positive and negative parts.

Fix $x_2 \in X_2$ . Let $\mathcal{E}$ be the collection of all measurable sets $E$ of the product space for which $E(x_2)$ is measurable in $X_1$ .

lemma: $\mathcal{E}$ is a $\lambda$ -system of $X_1 \times X_2$ containing the measurable rectangles.

Note that $X_1 \times X_2 \in \mathcal{E}$ since $(X_1 \times X_2) (x_2) = X_1$ . Also when $E,F\in \mathcal{E}$ and such that $E \subseteq F$ , $(F-E)(x_2) = F(x_2) - E(x_2)$ is the difference of measurable sets and therefore is measurable. Hence $E-F \in \mathcal{E}$ . Finally, let $\left<E_i\right>$ be a non-decreasing sequence of sets of $\mathcal{E}$ . Then $(\cup_i E_i)(x_2) = \cup_i E_i(x_2)$ is the countable union of measurable sets and therefore is measurable. Hence $\cup_i E_i \in \mathcal{E}$ . Hence $\mathcal{E}$ is a $\lambda$ -system.

Since for any measurable rectangle $a$ the slice is measurable since $(a_1 \times a_2)(x_2) = a_1 | \emptyset$ the collection contains the measurable rectangles.

Since the measurable rectangles are a $\pi$ -system contained in the $\lambda$ -system $\mathcal{E}$ , the $\lambda$ -system generated from the $\pi$ -system is also contained in $\mathcal{E}$ . But since a $\pi,\lambda$ -system is a $\sigma$ algebra, it follows that $\mathcal{E}$ contains the measurable sets of the product space. Hence $\mathbb{I}_E(x_2)$ is a measurable function with measure $\mu_1 (\mathbb{I}_E(x_2)) = \mu_1(E(x_2))$ .

To establish that $\mu_1(\mathbb{I}_E(x_2))$ is a function measurable in the space $X_2$ , the same approach can be taken as above. That is verify a subset of $\mathcal{E}$ for which $f: f(x_2) = \mu_1(\mathbb{I}_E(x_2))$ measurable in $X_2$ is a $\lambda$ -system containing the measurable rectangles.

Note that when $E = X_1 \times X_2$ implies $f(x_2) = \mu_1(X_1) \mathbb{I}_{X_2}(x_2)$ and so $f$ is measurable.

Suppose $E,F$ are measurable sets in the product space such that $E \subseteq F$ . Then

$f(x_2) = \mu_1(\mathbb{I}_{(F-E)(x_2)}) = \mu_1(\mathbb{I}_{F(x_2)} - \mathbb{I}_{E(x_2)}) = \mu_1(\mathbb{I}_{F(x_2)}) - \mu_1(\mathbb{I}_{E(x_2)})$ is the difference of measurable functions and so is measurable.

Finally for a non-decreasing sequence of measurable sets in the product space,

$f(x_2) = \mu_1(\mathbb{I}_{\cup_i E_i(x_2)}) = \lim_n \mu_1(\mathbb{I}_{E_i(x_2)})$

is the monotone limit of measurable functions and so is a measurable function. Hence the collection is a $\lambda$ -system.

For a rectangular set $a = a_1 \times a_2$ ,

$f(x_2) = \mu_1(\mathbb{I}_{(a_1 \times a_2)(x_2)}) = \mu_1(a_1) \mathbb{I}_{a_2}(x_2)$ is clearly a measurable function and so the collection contains the rectangular sets and therefore the measurable sets of the product space.

Now need to establish $\mu_2(f: f(x_2) = \mu_1(\mathbb{I}_{E(x_2)}))=\mu(E)$ .

The claim is true when $E$ is a rectangular set. Since the LHS is a measure and is an extension of the product measure on rectangles, and the RHS is a measure and is an extension of the product measure on rectangles, and the extension theorem says the extension is unique, it follows that the claim is true when $E$ is a measurable set of the product space.

Now, having established that indicator functions of measurable sets have measures that are iterations of measures, can apply the standard machine.

Let $s$ be a simple function of the product space. Then $s = \sum_{i = 1..n} a_i \mathbb{I}_{E_i}$ . By linearity,

$\mu_2[f: f(x_2) = \mu_1(\sum_{i=1..n} a_i \mathbb{I}_{E_i(x_2)})]$

$\mu_2[f: f(x_2) = \sum_{i=1..n} a_i \mu_1(\mathbb{I}_{E_i(x_2)})]$

$\mu_2(\sum_{i=1..n} a_i f_i : f_i(x_2) = \mu_1(\mathbb{I}_{E_i(x_2)})$

$\sum_{i = 1..n} a_i \mu_2(f_i : ) = \sum_{i = 1..n} a_i \mu(E_i)$

$\sum_{i = 1..n} a_i \mu(E_i) = \mu(s)$ .

Let $f$ be a non-negative measurable function that is the limit of measurable simple functions. Then using the MCT, can establish

$\mu_2[g: g(x_2) = \mu_1(f(x_2))] \\= \mu_2[g: g(x_2) = \mu_1(\lim_n s_n(x_2))]$

$= \mu_2[g : g(x_2) = \lim_n\mu_1(s_n(x_2))]$

$= \mu_2[ \lim g_n : g_n(x_2) = \mu_1(s_n(x_2))]$

$= \lim_n \mu_2[g_n : g_n(x_2) = \mu_1(s_n(x_2))] = \lim_n \mu(s_n) = \mu(f)$

Let $f$ be an arbitrary measurable function with positive and negative parts $f_+ = f \mathbb{I}_{f >0}$ and $f_- = -f\mathbb{I}_{f < 0}$ respectively, then

$\mu(f) = \mu(f_+) - \mu(f_-)$

$\mu(f_+) = \mu_2[g : g(x_2) = \mu_1(f_+(x_2))]$

$\mu(f_-) = \mu_2[g : g(x_2) = \mu_1(f_-(x_2))]$

$\mu_2[g : g(x_2) = \mu_1(f_+(x_2))] - \mu_2[g : g(x_2) = \mu_1(f_-(x_2))] =$

$\mu_2[g - h : g(x_2) = \mu_1(f_+(x_2)), h(x_2) = \mu_1(f_-(x_2))]$

$\mu_2[u : u(x_2) = \mu_1(f_+(x_2)) - \mu_1(f_-(x_2))]$

$\mu_2[u : u(x_2) = \mu_1(f_+(x_2) - f_-(x_2))]$

$\mu_2[u : u(x_2) = \mu_1(f(x_2))]$

where the linearity of the slice operation has been used.

The requirement that $\mu(|f|) < \infty$ or $f \ge 0$ is sufficient to assure that subtractions when they occur, are computable (not of the form $\infty - \infty$ .)

Fubini’s Theorem

Fubini’s Theorem

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