1
$\begingroup$

Let $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ be measurable spaces. Consider the product space $X\times Y$ with the product $\sigma$-algebra $\mathcal{A}\otimes \mathcal{B}$ and $\pi:X\times Y \rightarrow X$ the natural projection.

Is it true that $Z \in \mathcal{A}\otimes \mathcal{B}$ implies $\pi(Z)\in \mathcal{A}$? (that is, the projection sends measurable sets in measurable sets)

There is a topological analogue of this results («projections are open») so I wonder if we could extend it to measure theory.

$\endgroup$
  • $\begingroup$ It is true in some simple cases ($\mathcal{A} = \mathfrak{P}(X)$, for example). But I don't think it holds in general, because you don't have $\pi(\complement Z) = \complement \pi(Z)$. For the topological result, you get every open set as a union of rectangles, and you have $f(W \cup Z) = f(W) \cup f(Z)$ for all $f$, so no problem. $\endgroup$ – Daniel Fischer Jul 1 '13 at 10:42
  • $\begingroup$ I would say that $\mathcal A\times \mathcal B$ is a collection of measurable rectangles, which further generates the product $\sigma$-algebra that is usually denoted by $\mathcal A\otimes \mathcal B$. In that notation $\pi(Z)$ trivially belongs to $\mathcal A$ if $Z\in \mathcal A\times\mathcal B$ $\endgroup$ – Ilya Jul 1 '13 at 14:23
  • $\begingroup$ Yes, Ilya, you're right. I'll just edit $\endgroup$ – Helio Jul 2 '13 at 13:47
4
$\begingroup$

Lebesgue thought so in the context of real line but young Souslin, famously, constructed a Borel (G-delta is enough) subset of plane with non Borel projection.

$\endgroup$
  • $\begingroup$ What was the complexity of the non-Borel projection? $\endgroup$ – Quinn Culver Jul 1 '13 at 11:59
  • $\begingroup$ @Ilya I'd meant complexity in terms of the analytic heirarchy; I'd forgotten that just being the projection of a Borel set makes it $\Sigma^{1}_{1}$. $\endgroup$ – Quinn Culver Jul 1 '13 at 14:34
  • $\begingroup$ @Ilya Ya that's what I just said. $\endgroup$ – Quinn Culver Jul 1 '13 at 15:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.