The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Analysis, Measure Theory | Leave a comment