Product Measures — Alex Beaudin

Products of Sets and Algebras

Measuring integrals on the real line is all well and good. However, me way want to extend our definitions of size to higher dimensions. In particular, our notion of length in a single dimension will extend to notions of area and volume in two and three dimensions, respectively.

To talk about measure in product spaces, we should probably agree on what a product is. What do we mean when we take products of sets? In particular, how do combine or extend our notion of $\sigma$ -algebras? The note starts with these topics before moving onto core results.

A picture like the one you probably have in your head is fine. However, as on the real line, we can have disjoint intervals, a rectangle may include multiple true rectangles.

We’ll also discuss cross-sections needed for taking iterated integrals. In essence, one variable becomes fixed as another varies in the integrand.

Great, now we have a common language and notation.

Proof.

The core of the proof is to consider the collection of subsets where the conclusion holds. We will then show that this collection $\mathcal{E}$ actually contains $\mathcal{S} \otimes \mathcal{T}$ .

So, consider the collection $\mathcal{E}$ , the collection of sets where the result holds. Clearly, $A \times B \in \mathcal{E}$ for every $A \in \mathcal{S}, B \in \mathcal{T}$ . This collection is closed under complements, since for any $E \in \mathcal{E}$

\htmlId{eq-4}{[(X \times Y) \setminus E]_a = Y \setminus [E]_a \in \mathcal{T}.} \tag{4}

It is also closed under countable unions since

\htmlId{eq-5}{[E_1 \cup E_2 \cup \cdots]_a = [E_1]_a \cup [E_2]_a \cup \cdots.} \tag{5}

So, $\mathcal{E}$ is a $\sigma$ -algebra containing all measurable rectangles in $\mathcal{S} \otimes \mathcal{T}$ , and so it must contain $\mathcal{S} \otimes \mathcal{T}$ .

$\blacksquare$

Now, we move on to defining cross sections of functions.

The next result shows that, just as before, cross sections preserve measurability for functions.

Proof.

Suppose $D$ is a Borel subset of $\R$ and $a \in X$ . If $y \in Y$ , then

\htmlId{eq-9}{\begin{aligned} y \in ([f]_a)^{-1}(D) &\iff [f]_a(y) \in D \\ & \iff f(a, y) \in D \\ & \iff (a, y) \in f^{-1}(D) \\ & \iff y \in [f^{-1}(D)]_{a}. \end{aligned}} \tag{9}

Thus, $([f]_a)^{-1}(D) = [f^{-1}(D)]_{a}$ . Because $f$ is an $\mathcal{S} \otimes \mathcal{T}$ -measurable function, $f^{-1}(D) \in \mathcal{S} \otimes \mathcal{T}$ . By the previous proposition, that implies that for any $a \in X$ , $[f^{-1}(D)]_{a} \in \mathcal{T}$ is therefore $\mathcal{T}$ -measurable, as required.

The same logic applies to show that $[f]^b \in \mathcal{S}$ for every $b \in Y$ .

$\blacksquare$

Monotone Class Theorem

We used a two-step proof technique for proving that cross sections of measurable sets are measurable. In general, it can be used to show that every set in a $\sigma$ -algebra has a desired property. It goes vaguely like

show that every set in a collection of sets that generate the $\sigma$ -algebra has the property;
show that the collection of sets that has the property is a $\sigma$ -algebra.

This should be used when possible, but it may not always be possible. In some cases, it seems there’s no reasonable way to show that the collection of sets is a $\sigma$ -algebra. To deal with this issue, we will introduce another technique which uses what are called monotone classes.

The following result provides an example of an algebra that we will exploit.

Now we define a monotone class as a collection of sets that is closed under countable increasing unions and countable decreasing intersections.

Clearly, every $\sigma$ -algebra is a monotone class. However, counterintuitively, some monotone classes are not closed even under finite unions, as the next example shows.

The next result provides a useful tool when the standard technique for showing that every set in a $\sigma$ -algebra has a certain property does not work.

Proof.

Let $\mathcal{M}$ be the smallest monotone class that contains $\mathcal{A}$ . Because every $\sigma$ -algebra is a monotone class, $\mathcal{M}$ is contained in the smallest $\sigma$ -algebra containing $\mathcal{A}$ .

To prove the inclusion in the other direction, first suppose $A \in \mathcal{A}$ . Let

\htmlId{eq-10}{\mathcal{E} = \left\{ E \in \mathcal{M} \,:\, A \cup E \in \mathcal{M} \right\}.} \tag{10}

First, note that $\mathcal{A} \subseteq \mathcal{E}$ . A moment’s thought reveals that $\mathcal{E}$ is a monotone class. Thus, the smallest monotone class that contains $\mathcal{A}$ is contained in $\mathcal{E}$ , $\mathcal{M} \subseteq \mathcal{E}$ . Hence, $A \cup E \in \mathcal{M}$ for all $E \in \mathcal{M}$ .

Now consider the collection

\htmlId{eq-11}{\mathcal{D} = \{ D \in \mathcal{M} \,:\, D \cup E \in \mathcal{M} \text{ for all } E \in \mathcal{M}\}.} \tag{11}

By the previous paragraph, $\mathcal{A} \subseteq \mathcal{D}$ . It’s also clear that $\mathcal{D}$ is a monotone class, and so we must have $\mathcal{M} \subseteq \mathcal{D}$ . Hence, $D \cup E \in \mathcal{M}$ for all $D, E \in \mathcal{M}$ , and so $\mathcal{M}$ is closed under finite unions.

Notice now that for $E_1, E_2, \dotsc \in \mathcal{M}$ , we can write the union as an increasing union:

\htmlId{eq-12}{E_1 \cup E_2 \cup \cdots = E_1 \cup (E_1 \cup E_2) \cup (E_1 \cup E_2 \cup E_3) \cup \cdots.} \tag{12}

Which shows that $\mathcal{M}$ is closed under countable unions.

Finally, let

\htmlId{eq-13}{\mathcal{M}' = \{E \in \mathcal{M} \,:\, W \setminus E \in \mathcal{M}\}.} \tag{13}

Since $\mathcal{A}$ is closed under complementation, $\mathcal{A} \subseteq \mathcal{M}'$ . Once again, a bit if thought reveals that $\mathcal{M}'$ is a monotone class, and so we must conclude that $\mathcal{M} \subseteq \mathcal{M}'$ . $\mathcal{M}$ is closed under complementation.

So, we’ve shown that $\mathcal{M}$ is closed under countable unions and finite intersections, and therefore is a $\sigma$ -algebra. Hence $\mathcal{M}$ is a $\sigma$ -algebra that contains $\mathcal{A}$ , and so must contain the smallest $\sigma$ -algebra that contains $\mathcal{A}$ .

$\blacksquare$

Products of Measures

This next result allows us to define the product of two $\sigma$ -finite measures.

The proof is informative but omitted for brevity.

Now, we define the product of measures. The definition we give makes sense because the inner and outer integrals are both well-defined. The restriction to $\sigma$ -finite measures is not bothersome because the main results we seek are not valid without this hypothesis.

Example.

Consider two standard measure spaces as above, and consider a rectangle $A \times B \in \mathcal{S} \otimes \mathcal{T}$ , where $A \in \mathcal{S}$ and $B \in \mathcal{T}$ , then we have

\htmlId{eq-17}{\begin{aligned} (\mu \times\nu)(A \times B) &= \int_{X}\int_{Y} \chi_{A \times B}(x, y) d\nu(y) d\mu(x) \\ &= \int_{X}\int_{Y} \chi_{A}(x) \chi_{B}(y) d\nu(y) d\mu(x) \\ &= \int_{X}\chi_{A}(x)\int_{Y} \chi_{B}(y) d\nu(y) d\mu(x) \\ &= \int_{X}\chi_{A}(x) \nu(B) d\mu(x) \\ &= \nu(B) \int_{X}\chi_{A}(x) d\mu(x) \\ &= \mu(A) \nu(B). \end{aligned}} \tag{17}

So we defined the product $\mu \times \nu$ , which is a function. But we wouldn’t want any function, we’d really want this product to be a measure. Luckily, the following result show that this is indeed the case.

Proposition.A product of measures is a measure

We have to show that the measure of the empty set is zero, and that it satisfies countable additivity. Indeed, $(\mu \times\nu)(\emptyset) = 0$ from the previous example.

Suppose that $E_1, E_2, E_3, \dotsc$ is a disjoint sequence of sets in $\mathcal{S} \otimes \mathcal{T}$ . Then we must have

\htmlId{eq-18}{\begin{aligned} (\mu \times\nu) \left( \bigcup_{k=1}^\infty E_k \right) &= \int_{X} \nu \left( \left[\bigcup_{k=1}^\infty E_k\right]_x \right) d\mu(x) \\ &= \int_{X} \nu \left( \bigcup_{k=1}^\infty [E_k]_x \right) d\mu(x) \\ &= \int_{X} \sum_{k=1}^\infty \nu \left([E_k]_x \right) d\mu(x) \\ &= \sum_{k=1}^\infty \int_{X} \nu \left([E_k]_x \right) d\mu(x) \\ &= \sum_{k=1}^\infty (\mu \times\nu)(E_k), \end{aligned}} \tag{18}

as desired.