projections of analytic sets are analytic

Projections along compact paved spaces

Given sets $X$ and $K$, the projection map ${\pi }_X:X\times K\to X$ is defined by ${\pi }_X(x,y)=x$. An important property of analytic sets (http://planetmath.org/AnalyticSet2) is that they are stable under projections.

Theorem 1.

Let $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{)}$ be a paved space (http://planetmath.org/PavedSpace), $\mathrm{(}K\mathrm{,}\mathrm{K}\mathrm{)}$ be a compact (http://planetmath.org/PavedSpace) paved space and ${\pi }_X\mathrm{:}X\mathrm{\times }K\mathrm{\to }X$ be the projection map.

If $S\mathrm{\subseteq }X\mathrm{\times }K$ is $\mathrm{F}\mathrm{\times }\mathrm{K}$-analytic then ${\pi }_X\mathit{}\mathrm{(}S\mathrm{)}$ is $\mathrm{F}$-analytic.

The proof of this follows easily from the definition of analytic sets. First, there is a compact paved space $(K^{\prime },{𝒦}^{\prime })$ and a set $T\in {\left(ℱ\times 𝒦\times {𝒦}^{\prime }\right)}_{\sigma \delta }$ such that $S={\pi }_{X\times K}(T)$. Then,

$${\pi }_X(S)={\pi }_X\left({\pi }_{X\times K}(T)\right)={\pi }_X(T).$$

However, $(K\times K^{\prime },𝒦\times {𝒦}^{\prime })$ is a compact paved space (see products of compact pavings are compact (http://planetmath.org/ProductsOfCompactPavingsAreCompact)), which shows that ${\pi }_X(S)$ satisfies the definition of $ℱ$-analytic sets.

Projections along Polish spaces

Theorem 1 above can be used to prove the following result for projections from the product of a measurable space and a Polish space. For $\sigma$-algebras (http://planetmath.org/SigmaAlgebra) $ℱ$ and $ℬ$, we use the notation $ℱ\otimes ℬ$ for the product $\sigma$-algebra (http://planetmath.org/ProductSigmaAlgebra), in order to distinguish it from the product paving $ℱ\times ℬ$.

Theorem 2.

Let $\mathrm{(}X\mathrm{,}\mathrm{F}\mathrm{)}$ be a measurable space and $Y$ be a Polish space with Borel $\sigma$-algebra (http://planetmath.org/BorelSigmaAlgebra) $\mathrm{B}$.

If $S\mathrm{\subseteq }X\mathrm{\times }Y$ is $\mathrm{F}\mathrm{\otimes }\mathrm{B}$-analytic, then its projection onto $X$ is $\mathrm{F}$-analytic.

An immediate consequence of this is the measurable projection theorem.

Although Theorem 2 applies to arbitrary Polish spaces, it is enough to just consider the case where $Y$ is the space of real numbers $ℝ$ with the standard topology. Indeed, all Polish spaces are Borel isomorphic to either the real numbers or a discrete subset of the reals (see Polish spaces up to Borel isomorphism), so the general case follows from this.

If $Y=ℝ$, then the Borel $\sigma$-algebra is generated by the compact paving $𝒦$ of closed and bounded intervals. The collection $a(ℱ\times 𝒦)$ of analytic sets is closed under countable unions and countable intersections so, by the monotone class theorem, includes the product $\sigma$-algebra $ℱ\otimes ℬ$. Then, as the analytic sets define a closure operator,

$$a(ℱ\otimes ℬ)\subseteq a(a(ℱ\times 𝒦))=a(ℱ\times 𝒦).$$

Thus every $ℱ\otimes ℬ$-analytic set is $ℱ\times 𝒦$-analytic, and the result follows from Theorem 1.

Title	projections of analytic sets are analytic
Canonical name	ProjectionsOfAnalyticSetsAreAnalytic
Date of creation	2013-03-22 18:46:21
Last modified on	2013-03-22 18:46:21
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	8
Author	gel (22282)
Entry type	Theorem
Classification	msc 28A05
Related topic	MeasurableProjectionTheorem