# The Unapologetic Mathematician

## Product Measurable Spaces

Now we return to the category of measurable spaces and measurable functions, and we discuss product spaces. Given two spaces $(X,\mathcal{S})$ and $(Y,\mathcal{T})$, we want to define a $\sigma$-ring of measurable sets on the product $X\times Y$.

In fact, we’ve seen that given rings $\mathcal{S}$ and $\mathcal{T}$ we can define the product $\mathcal{S}\times\mathcal{T}$ as the collection of finite disjoint unions of sets $A\times B$, where $A\in\mathcal{S}$ and $B\in\mathcal{T}$. If $\mathcal{S}$ and $\mathcal{T}$ are $\sigma$-rings, then we define $\mathcal{S}\times\mathcal{T}$ to be the smallest monotone class containing this collection, which will then be a $\sigma$-ring. If $\mathcal{S}$ and $\mathcal{T}$ are $\sigma$-algebras — that is, if $X\in\mathcal{S}$ and $Y\in\mathcal{T}$ — then clearly $X\times Y\in\mathcal{S}\times\mathcal{T}$, which is thus another $\sigma$-algebra.

Indeed, $(X\times Y,\mathcal{S}\times\mathcal{T})$ is a measurable space. The collection $\mathcal{S}\times\mathcal{T}$ is a $\sigma$-algebra, and every point is in some one of these sets. If $(x,y)\in X\times Y$, then $x\in A\in\mathcal{S}$ and $y\in B\in\mathcal{T}$, so $(x,y)\in A\times B\in\mathcal{S}\times\mathcal{T}$. But is this really the $\sigma$-algebra we want?

Most approaches to measure theory simply define this to be the product of two measurable spaces, but we have a broader perspective here. Indeed, we should be asking if this is a product object in the category of measurable spaces! That is, the underlying space $X\times Y$ comes equipped with projection functions $\pi_1:X\times Y\to X$ and $\pi_2:X\times Y\to Y$. We must ask if these projections are measurable for our choice of $\sigma$-algebra, and also that they satisfy the universal property of a product object.

Checking measurability is pretty straightforward. It’s the same for either projector, so we’ll consider $\pi_1:X\times Y\to X$ explicitly. Given a measurable set $A\in\mathcal{S}$, the preimage $\pi_1^{-1}(A)=A\times Y$ must be measurable as well. And here we run into a snag: we know that $A$ is measurable, and so for any measurable $B\in\mathcal{T}$ the subset $A\times B\subseteq A\times Y$ is measurable. In particular, if $\mathcal{T}$ is a $\sigma$-algebra then $A\times Y$ is measurable right off. However, if $Y$ is not itself measurable then $Y$ is not the limit of any increasing countable sequence of measurable sets either (since $\sigma$-rings are monotonic classes). Thus $A\times Y$ is not the limit of any increasing sequence of measurable products $A\times B$, and so $A\times Y$ can’t be in the smallest monotonic class generated by such products, and thus can’t be measurable!

So in order for $(X\times Y,\mathcal{S}\times\mathcal{T})$ to be a product object, it’s necessary that $\mathcal{T}$ be a $\sigma$-algebra. Similarly, $\mathcal{S}$ must be a $\sigma$-algebra as well. What would happen if this condition fails? Consider a measurable function $f:Z\to X\times Y$ defined by $f(z)=(f_1(z),f_2(z))$. We can write $f_1=\pi_1\circ f$, but since $\pi_1$ is not measurable we have no guarantee that $f_1$ will be measurable!

On the other hand, all is not lost; if $f_1:Z\to X$ and $f_2:Z\to Y$ are both measurable, and if $A\in\mathcal{S}$ and $B\in\mathcal{T}$, then we calculate the preimage \displaystyle\begin{aligned}f^{-1}(A\times B)&=\left\{z\in Z\vert f(z)\in A\times B\right\}\\&=\left\{z\in Z\vert (f_1(z),f_2(z))\in A\times B\right\}\\&=\left\{z\in Z\vert f_1(z)\in A\textrm{ and }f_2(z)\in B\right\}\\&=\left\{z\in Z\vert f_1(z)\in A\right\}\cap\left\{z\in Z\vert f_2(z)\in B\right\}\\&=f_1^{-1}(A)\cap f_2^{-1}(B)\end{aligned}

which is thus measurable. Monotone limits of finite disjoint unions of such sets are easily handled. Thus if both components of $f$ are measurable, then $f$ is measurable.

Okay, so back to the case where $\mathcal{S}$ and $\mathcal{T}$ are both $\sigma$-algebras, and so $\pi_1$ and $\pi_2$ are both measurable. We still need to show that the universal property holds. That is, given two measurable functions $f_1:Z\to X$ and $f_2:Z\to Y$, we must show that there exists a unique measurable function $f:Z\to X\times Y$ so that $f_1=\pi_1\circ f$ and $f_2=\pi_2\circ f$. There’s obviously a unique function on the underlying set: $z\mapsto(f_1(z),f_2(z))$. And the previous paragraph shows that this function must be measurable!

So, the uniqueness property always holds, but with the one caveat that the projectors may not themselves be measurable. That is, the full subcategory of total measurable spaces has product objects, but the category of measurable spaces overall does not. However, we’ll still talk about the “product” space $X\times Y$ with the $\sigma$-ring $\mathcal{S}\times\mathcal{T}$ understood.

A couple of notes are in order. First of all, this is the first time that we’ve actually used the requirement that every point in a measurable space be a member of some measurable set or another. It will become more important as we go on. Secondly, we define a “measurable rectangle” in $X\times Y$ to be a set $A\times B$ so that $A\in\mathcal{S}$ and $B\in\mathcal{T}$ — that is, one for which both “sides” are measurable. The class of all measurable sets $\mathcal{S}\times\mathcal{T}$ is the $\sigma$-ring generated by all the measurable rectangles.

July 15, 2010 - Posted by | Analysis, Measure Theory

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