# The Unapologetic Mathematician

## A Counterexample

I’ve been thinking about yesterday’s post about the product of measurable spaces, and I’m not really satisfied. I’d like to present a concrete example of what can go wrong when one or the other factor isn’t a total measurable space — that is, when the underlying set of that factor is not a measurable subset of itself.

Let $X=\mathbb{R}$ be the usual real line. However, instead of the Borel or Lebesgue measurable sets, let $\mathcal{S}$ be the collection of all countable subsets of $X$. We check that this is a $\sigma$-ring: it’s clearly closed under unions and differences, making it a ring, and the union of a countable collection of countable sets is still countable, so it’s monotone, and thus a $\sigma$-ring. And every point $x\in X$ is in the measurable set $\{x\}\in\mathcal{S}$, so $(X,\mathcal{S})$ is a measurable space. We’ll let $(Y,\mathcal{T})=(Z,\mathcal{U})=(X\mathcal{S})$ be two more copies of the same measurable space.

Now we define the “product” space $(X\times Y,\mathcal{S}\times\mathcal{T})$. A subset of $X\times Y$ is in $\mathcal{S}\times\mathcal{T}$ if it can be written as the union of a countable number of measurable rectangles $A\times B$ with $A\in\mathcal{S}$ and $B\in\mathcal{T}$. But if $A$ and $B$ are both countable, then $A\times B$ is also countable, and thus any measurable subset of $X\times Y$ is countable. On the other hand, if a subset of $X\times Y$ is countable, it’s the countable union of all of its singletons $\{(a,b)\}=\{a\}\times\{b\}$, each of which is a measurable rectangle. That is, if a subset of $X\times Y$ is countable, then it is measurable. That is, $\mathcal{S}\times\mathcal{T}$ is the collection of all the countable subsets of $X\times Y$.

Next we define the function $f:Z\to X\times Y$ by setting $f(z)=(z,0)$ for any real number $z$. We check that it’s measurable: given a measurable set $C\subseteq X\times Y$ we calculate $\displaystyle f^{-1}(C)=\{z\in Z\vert f(z)\in C\}=\{z\in Z\vert (z,0)\in C\}$

but if $C$ is measurable, then it’s countable, and it can contain only countably many points of the form $(z,0)$. The preimage $f^{-1}(C)$ must then be countable, and thus measurable, and thus $f$ is a measurable function.

That said, the component function $f_2=\pi_2\circ f$ is not measurable. Indeed, we have $f_2(z)=0$ for all real numbers $z$. The set $\{0\}\subseteq Y$ is countable — and thus measurable — but its preimage $f^{-1}(\{0\})=Z$ is the entire real line, which is uncountable and is not measurable. This is exactly the counterintuitive result we were worried about.

July 16, 2010