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 𝜎-algebra different than the product 𝜎-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 𝜎-algebra or relatively to some other 𝜎-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 𝜎-algebra ${\displaystyle {\mathcal {F)):=\{\varnothing ,\{0\},\{1\},\{0,1\}\))$ and the space ${\displaystyle Y:=\{0,1\))$ with the 𝜎-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 𝜎-algebra. Let ${\displaystyle {\mathcal {L))}$ be Lebesgue 𝜎-algebra of ${\displaystyle \mathbb {R} }$ and let ${\displaystyle {\mathcal {L))'}$ be the Lebesgue 𝜎-algebra of ${\displaystyle \mathbb {R} ^{2}.}$ For any bounded ${\displaystyle N\subseteq \mathbb {R} }$ 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 𝜎-algebra ${\displaystyle {\mathcal {L))\otimes {\mathcal {L))}$ but its completion. As for such example in product 𝜎-algebra, one can take the space ${\displaystyle \{0,1\}^{\mathbb {R} ))$ (or any product along a set with cardinality greater than continuum) with the product 𝜎-algebra ${\displaystyle {\mathcal {F))=\textstyle {\bigotimes \limits _{t\in \mathbb {R} )){\mathcal {F))_{t))$ where ${\displaystyle {\mathcal {F))_{t}=\{\varnothing ,\{0\},\{1\},\{0,1\}\))$ for every ${\displaystyle t\in \mathbb {R} .}$ 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 𝜎-algebra. Then for every set in the product 𝜎-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 ${\displaystyle \mathbb {R} ^{n))$ onto ${\displaystyle \mathbb {R} ^{n-k))$ where ${\displaystyle k 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 ${\displaystyle \mathbb {R} }$ which is a projection of some measurable set of ${\displaystyle \mathbb {R} ^{2},}$ is the only sort of such example.