Proof of the measurability of functions in Theorem 1.10 (Product measure) in "Analysis" by Lieb and Loss

Question

I am confronted with the following problem:

Let $\left(\Omega_1,\Sigma_1,\mu_1\right)$ and $\left(\Omega_2,\Sigma_2,\mu_2\right)$ be two sigma-finite measure spaces. Let $A$ be a measurable set in $\Sigma_1\times\Sigma_2$, where $\Sigma_1\times\Sigma_2$ is the smallest sigma-algebra containing all rectangles defined by

$ A_1\times A_2 = \left\{ \left(x_1,x_2\right) {}:{} x_1\in A_1,\quad x_2\in A_2 \right\},\quad A_1\in\Sigma_1,\quad A_2\in\Sigma_2 $.

Next, for every $x_2\in\Omega_2$, set $f\left( x_2\right) :=\mu_1\left( A_1\left(x_2\right)\right)$ and, for every $x_1\in\Omega_1$, $g\left( x_1\right) :=\mu_2\left(A_2\left(x_1\right)\right)$, where

$ A_1\left( x_2\right) = \left\{x_1\in\Omega_1\mid \left( x_1,x_2\right)\in A \right\}\in\Sigma_1 $

and

$ A_2\left( x_1\right) = \left\{x_2\in\Omega_2\mid \left( x_1,x_2\right)\in A \right\}\in\Sigma_2 $.

Then $f$ is $\Sigma_2$-measurable, $g$ is $\Sigma_1$-measurable and

$\left(\mu_1\times\mu_2\right)(A) :=\int_{\Omega_2}f\left( x_2\right)\mu_2\left(\mathrm{d}x_2\right) =\int_{\Omega_1}g\left(x_1\right)\mu_1\left(\mathrm{d}x_1\right).$

Now the first part of the proof says:

The measurability of $f$ and $g$ parallels the proof of the section property in Section 1.2 and uses the Monotone Class Theorem,

while the section property says that for every $A\in\Sigma_1\times\Sigma_2$ the following applies:

$A_1\left(x_2\right)\in\Sigma_1\quad\forall x_2\in\Omega_2\quad$ and $\quad A_2\left(x_2\right)\in\Sigma_2\quad\forall x_1\in\Omega_1.$

To show the section property, you define a class of sets which have the section property and show that this class is a sigma-algebra. Then, this class is also the smallest sigma-algebra containing all rectangles by definition - as far as I understood it, at least.

Now, to show that $f$ and $g$ are measurable, I think that I have to define a class $\mathcal{M}$ of sets, for which $f$ is $\Sigma_2$-measurable and $g$ is $\Sigma_1$-measurable, and show that $\mathcal{M}$ is a monotone class. The Monotone Class Theorem then states, that this class is also a sigma-algebra and therefore I should infer that the measurability of $f$ and $g$ holds.

But my problem is that I am having a hard time to construct such a class $\mathcal{M}$. Do you have any ideas for an approach? I'd appreciate the help.

Answer 1

It might be a good idea to start with the warm-up exercise of proving this when $\mu_1$ and $\mu_2$ are finite, as opposed to $\sigma$-finite.

The class $\mathcal M$ is not something you should be "having a hard time constructing". The idea is to simply define $\mathcal M$ to be the class of all sets $A \in \Sigma_1 \times \Sigma_2$ such that $f_A: x_2 \mapsto \mu_1\left( \{ x_1 \in \Omega_1 : (x_1, x_2) \in A \}\right)$ is $\Sigma_2$-measurable and $g_A: x_1 \mapsto \mu_2\left( \{ x_2 \in \Omega_1 : (x_1, x_2) \in A \}\right)$ is $\Sigma_1$-measurable.

You want to show that:

• If $B_1 \in \Sigma_1$ and $B_2 \in \Sigma_2$, then $B_1 \times B_2 \in \mathcal M$.
• If $A_1, A_2, \dots, A_n \in \mathcal M$ and the $A_i$'s are disjoint, then $A_1 \cup \dots \cup A_n \in \mathcal M$.
• If $A_1, A_2, A_3, \dots \in \mathcal M$ and $A_1 \subset A_2 \subset A_3 \subset \dots $, then $\bigcup_{i \in \mathbb N} A_i \in \mathcal M$.
• If $A_1, A_2, A_3, \dots \in \mathcal M$ and $A_1 \supset A_2 \supset A_3 \supset \dots $, then $\bigcap_{i \in \mathbb N} A_i \in \mathcal M$. (Here, we need the assumption that $\mu_1$ and $\mu_2$ are finite.)

Once you have demonstrated the above claims, you can infer that $\mathcal M$ is a monotone class that contains all elementary sets. (An elementary set is a union of finitely many disjoint sets of the form $B_1 \times B_2$, where $B_1 \in \Sigma_1$ and $B_2 \in \Sigma_2$.)

The monotone class theorem tells you that a monotone class that contains all elementary sets must contain all the sets in the product sigma algebra $\Sigma_1 \times \Sigma_2$. So you can conclude immediately that $\mathcal M$ must be the whole of $\Sigma_1 \times \Sigma_2$.

(Note: Your original post suggests that you believe the monotone class theorem says that all monotone classes are sigma algebras. This is not true. The monotone class theorem says that any monotone class that contains all elementary sets must necessarily contain all sets in the product sigma algebra $\Sigma_1 \times \Sigma_2$.)

I'll leave you to think about how to generalise this argument for the case where $\mu_1$ and $\mu_2$ are $\sigma$-finite, rather than finite. If this proves difficult, feel free to leave a comment.

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Edit: Fleshing this out a little more.

If $B_1 \in \Sigma_1$ and $B_2 \in \Sigma_2$, then $B_1 \times B_2 \in \mathcal M$

Observe that \begin{align} f_{B_1 \times B_2}(x_2) & = \begin{cases} \mu_1(B_1) & \text{if } x_2 \in B_2 \\ 0 & \text{if } x_2 \notin B_2 \end{cases} \\ &= \mu_1(B_1) \times \mathbf 1_{B_2}(x_2).\end{align} Since $B_2$ is measurable, the indicator function $\mathbf 1_{B_2}$ is measurable, hence $f_{B_1 \times B_2}$ is measurable.

If $A_1, A_2, \dots, A_n \in \mathcal M$ and the $A_i$'s are disjoint, then $A_1 \cup \dots \cup A_n \in \mathcal M$.

We have \begin{align} f_{A_1 \cup \dots \cup A_n} (x_2) & = \mu_1\left( \bigcup_{1 \leq i \leq n} \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = \sum_{1 \leq i \leq n} \mu_1\left( \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = f_{A_1}(x_2) + \dots + f_{A_n}(x_2). \end{align} The second equality follows from the fact that the measure of a finite (or countable) union of disjoint measurable sets is equal to the sum of the measures of the individual sets. Since $f_{A_1}, \dots, f_{A_n}$ are measurable, their sum is measurable, hence $f_{A_1 \cup \dots \cup A_n}$ is measurable.

If $A_1, A_2, A_3, \dots \in \mathcal M$ and $A_1 \subset A_2 \subset A_3 \subset \dots $, then $\bigcup_{i \in \mathbb N} A_i \in \mathcal M$.

\begin{align} f_{\bigcup_{i \in \mathbb N} A_i} (x_2) & = \mu_1\left( \bigcup_{i \in \mathbb N} \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = \lim_{n \to \infty} \mu_1\left( \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = \lim_{n \to \infty} f_{A_i}(x_2). \end{align} The second equality follows from the fact that the measure of the union of an ascending sequence of nested sets is equal to the limit of the measures of the individual sets. Since the $f_{A_i}$'s are measurable, their pointwise limit is also measurable, hence $f_{\bigcup_{i \in \mathbb N} A_i}$ is measurable.

If $A_1, A_2, A_3, \dots \in \mathcal M$ and $A_1 \supset A_2 \supset A_3 \supset \dots $, then $\bigcap_{i \in \mathbb N} A_i \in \mathcal M$. (Here, we need the assumption that $\mu_1$ and $\mu_2$ are finite.)

\begin{align} f_{\bigcap_{i \in \mathbb N} A_i} (x_2) & = \mu_1\left( \bigcap_{i \in \mathbb N} \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = \lim_{n \to \infty} \mu_1\left( \{ x_1 \in \Omega_1 : (x_1, x_2) \in A_i \} \right) \\ & = \lim_{n \to \infty} f_{A_i}(x_2). \end{align} The second equality follows from the fact that the measure of a intersection of an descending sequence of nested sets is equal to the limit of the measures of the individual sets, provided that the measures of the individual sets are finite. (This last condition is true because $\mu_1$ is assumed to be a finite measure.)

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Another edit: Dealing with the case where $\mu$ and $\nu$ are $\sigma$-finite, rather than finite.

Let $\{ X_n \}_{n \in \mathbb N}$ be a disjoint measurable cover of $\Omega_1$ such that $\mu_1(X_n) < \infty$, and let $\{ Y_m \}_{m \in \mathbb N}$ be a disjoint measurable cover of $\Omega_2$ such that $\mu_2(Y_m) < \infty$.

Let $\mathcal M'$ be the class of all sets $A \in \Sigma_1 \times \Sigma_2$ such that $A \cap (X_n \times Y_m) \in \mathcal M$ for all $n,m \in \mathbb N$.

You can prove that:

• If $B_1 \in \Sigma_1$ and $B_2 \in \Sigma_2$, then $B_1 \times B_2 \in \mathcal M'$.
• If $A_1, A_2, \dots, A_n \in \mathcal M'$ and the $A_i$'s are disjoint, then $A_1 \cup \dots \cup A_n \in \mathcal M'$.
• If $A_1, A_2, A_3, \dots \in \mathcal M'$ and $A_1 \subset A_2 \subset A_3 \subset \dots $, then $\bigcup_{i \in \mathbb N} A_i \in \mathcal M'$.
• If $A_1, A_2, A_3, \dots \in \mathcal M'$ and $A_1 \supset A_2 \supset A_3 \supset \dots $, then $\bigcap_{i \in \mathbb N} A_i \in \mathcal M'$. (Here, it's vital that $\mu_1(X_n) < \infty$ for all $n$ and $\mu_2(Y_m) < \infty$ for all $m$.)

The proofs of these four bullet points are almost identical to the proofs of our previous four bullet points from our investigation of the case where $\mu_1$ and $\mu_2$ were finite measures.

From these four bullet points, we may infer that $\mathcal M'$ is a monotone class that contains all elementary sets. By the monotone class theorem, $\mathcal M'$ must be the whole of $\Sigma_1 \times \Sigma_2$.

Thus we've shown that if $A$ is any set in $\Sigma_1 \times \Sigma_2$, then $A \cap (X_n \times Y_m) \in \mathcal M$ for all $n$ and $m$. It still remains to prove that $A$ itself is in $\mathcal M$.

To prove this, we need to prove an extra bullet point, namely:

• If $A_1, A_2, \dots \in \mathcal M$ and the $A_i$'s are disjoint, then $\bigcup_{i \in \mathbb N} A_i \in \mathcal M$.

This is very similar to the second bullet point (from our original list of bullet points) - the only difference is that we're dealing with a countable union rather than a finite union. I'm sure you'll have no trouble proving this.

Anyway, once we've proved this extra bullet point, we note that $A$ can be expressed as a countable union of disjoint sets, $$ A = \bigcup_{n \in \mathbb N, m \in \mathbb N} A \cap (X_n \times Y_m),$$ and each $A \cap (X_n \times Y_m)$ is in $\mathcal M$. Hence by our extra bullet point, $A$ itself must be in $\mathcal M$, and we're done.