# Measurability of projections

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.

• 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. – Daniel Fischer Jul 1 '13 at 10:42
• 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 $Z\in \mathcal A\times\mathcal B$ – Ilya Jul 1 '13 at 14:23
• Yes, Ilya, you're right. I'll just edit – Helio Jul 2 '13 at 13:47

• @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 $\Sigma^{1}_{1}$. – Quinn Culver Jul 1 '13 at 14:34