# Projection (measure theory)

In measure theory, projection maps often appear when working with product (Cartesian) spaces: The product sigma-algebra of measurable spaces is defined to be the finest such that the projection mappings will be measurable. Sometimes for some reasons product spaces are equipped with ๐-algebra different than the product ๐-algebra. In these cases the projections need not be measurable at all.

The projected set of a measurable set is called analytic set and need not be a measurable set. However, in some cases, either relatively to the product ๐-algebra or relatively to some other ๐-algebra, projected set of measurable set is indeed measurable.

Henri Lebesgue himself, one of the founders of measure theory, was mistaken about that fact. In a paper from 1905 he wrote that the projection of Borel set in the plane onto the real line is again a Borel set.[1] The mathematician Mikhail Yakovlevich Suslin found that error about ten years later, and his following research has led to descriptive set theory.[2] The fundamental mistake of Lebesgue was to think that projection commutes with decreasing intersection, while there are simple counterexamples to that.[3]

## Basic examples

For an example of a non-measurable set with measurable projections, consider the space $\displaystyle{ X := \{0, 1\} }$ with the ๐-algebra $\displaystyle{ \mathcal{F} := \{\varnothing, \{0\}, \{1\}, \{0, 1\}\} }$ and the space $\displaystyle{ Y := \{0, 1\} }$ with the ๐-algebra $\displaystyle{ \mathcal{G} := \{\varnothing, \{0, 1\}\}. }$ The diagonal set $\displaystyle{ \{(0, 0), (1, 1)\} \subseteq X \times Y }$ is not measurable relatively to $\displaystyle{ \mathcal{F}\otimes\mathcal{G}, }$ although the both projections are measurable sets.

The common example for a non-measurable set which is a projection of a measurable set, is in Lebesgue ๐-algebra. Let $\displaystyle{ \mathcal{L} }$ be Lebesgue ๐-algebra of $\displaystyle{ \Reals }$ and let $\displaystyle{ \mathcal{L}' }$ be the Lebesgue ๐-algebra of $\displaystyle{ \Reals^2. }$ For any bounded $\displaystyle{ N \subseteq \Reals }$ not in $\displaystyle{ \mathcal{L}. }$ the set $\displaystyle{ N \times \{0\} }$ is in $\displaystyle{ \mathcal{L}', }$ since Lebesgue measure is complete and the product set is contained in a set of measure zero.

Still one can see that $\displaystyle{ \mathcal{L}' }$ is not the product ๐-algebra $\displaystyle{ \mathcal{L} \otimes \mathcal{L} }$ but its completion. As for such example in product ๐-algebra, one can take the space $\displaystyle{ \{0, 1\}^\Reals }$ (or any product along a set with cardinality greater than continuum) with the product ๐-algebra $\displaystyle{ \mathcal{F} = \textstyle {\bigotimes\limits_{t\in\Reals}} \mathcal{F}_t }$ where $\displaystyle{ \mathcal{F}_t = \{\varnothing,\{0\} ,\{1\} ,\{0, 1\}\} }$ for every $\displaystyle{ t \in \Reals. }$ In fact, in this case "most" of the projected sets are not measurable, since the cardinality of $\displaystyle{ \mathcal{F} }$ is $\displaystyle{ \aleph_0 \cdot 2^{\aleph_0} = 2^{\aleph_0}, }$ whereas the cardinality of the projected sets is $\displaystyle{ 2^{2^{\aleph_0}}. }$ There are also examples of Borel sets in the plane which their projection to the real line is not a Borel set, as Suslin showed.[2]

## Measurable projection theorem

The following theorem gives a sufficient condition for the projection of measurable sets to be measurable.

Let $\displaystyle{ (X, \mathcal{F}) }$ be a measurable space and let $\displaystyle{ (Y, \mathcal{B}) }$ be a polish space where $\displaystyle{ \mathcal{B} }$ is its Borel ๐-algebra. Then for every set in the product ๐-algebra $\displaystyle{ \mathcal{F} \otimes \mathcal{B}, }$ the projected set onto $\displaystyle{ X }$ is a universally measurable set relatively to $\displaystyle{ \mathcal{F}. }$[4]

An important special case of this theorem is that the projection of any Borel set of $\displaystyle{ \Reals^n }$ onto $\displaystyle{ \Reals^{n-k} }$ where $\displaystyle{ k \lt n }$ is Lebesgue-measurable, even though it is not necessarily a Borel set. In addition, it means that the former example of non-Lebesgue-measurable set of $\displaystyle{ \Reals }$ which is a projection of some measurable set of $\displaystyle{ \Reals^2, }$ is the only sort of such example.