## 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 be the usual real line. However, instead of the Borel or Lebesgue measurable sets, let be the collection of all countable subsets of . We check that this is a -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 -ring. And every point is in the measurable set , so is a measurable space. We’ll let be two more copies of the same measurable space.

Now we define the “product” space . A subset of is in if it can be written as the union of a countable number of measurable rectangles with and . But if and are both countable, then is also countable, and thus any measurable subset of is countable. On the other hand, if a subset of is countable, it’s the countable union of all of its singletons , each of which is a measurable rectangle. That is, if a subset of is countable, then it is measurable. That is, is the collection of all the countable subsets of .

Next we define the function by setting for any real number . We check that it’s measurable: given a measurable set we calculate

but if is measurable, then it’s countable, and it can contain only countably many points of the form . The preimage must then be countable, and thus measurable, and thus is a measurable function.

That said, the component function is *not* measurable. Indeed, we have for all real numbers . The set is countable — and thus measurable — but its preimage is the entire real line, which is uncountable and is *not* measurable. This is exactly the counterintuitive result we were worried about.

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