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Projection (measure theory)

## Summary

In measure theory, projection maps often appear when working with product 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 sigma-algebra different than the product sigma-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 sigma-algebra or relatively to some other sigma-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

As an example for a non-measurable projection, one can take the space ${\displaystyle X:=\left\{0,1\right\}}$  with the sigma-algebra ${\displaystyle {\mathcal {F}}:=\left\{\emptyset ,\left\{0\right\},\left\{1\right\},\left\{0,1\right\}\right\}}$  and the space ${\displaystyle Y:=\left\{0,1\right\}}$  with the sigma-algebra ${\displaystyle {\mathcal {G}}:=\left\{\emptyset ,\left\{0,1\right\}\right\}}$ . The diagonal set ${\displaystyle \left\{\left(0,0\right),\left(1,1\right)\right\}\subset X\times Y}$  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 sigma-algebra. Let ${\displaystyle {\mathcal {L}}}$  be Lebesgue sigma-algebra of ${\displaystyle \mathbb {R} }$  and let ${\displaystyle {\mathcal {L}}'}$  be the Lebesgue sigma-algebra of ${\displaystyle \mathbb {R} ^{2}}$ . For any bounded ${\displaystyle N\subset \mathbb {R} }$  not in ${\displaystyle {\mathcal {L}}}$ , the set ${\displaystyle N\times \left\{0\right\}}$  is in ${\displaystyle {\mathcal {L}}'}$ , since Lebesgue measure is complete and the product set is contained in a set of zero measure.

Still one can see that ${\displaystyle {\mathcal {L}}'}$  is not the product sigma-algebra ${\displaystyle {\mathcal {L}}\otimes {\mathcal {L}}}$  but its completion. As for such example in product sigma-algebra, one can take the space ${\displaystyle \left\{0,1\right\}^{\mathbb {R} }}$  (or any product along a set with cardinality greater than continuum) with the product sigma-algebra ${\displaystyle \textstyle {\mathcal {F}}=\bigotimes _{t\in \mathbb {R} }{\mathcal {F}}_{t}}$  where ${\displaystyle {\mathcal {F}}_{t}=\left\{\emptyset ,\left\{0\right\},\left\{1\right\},\left\{0,1\right\}\right\}}$  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 ${\displaystyle \left(X,{\mathcal {F}}\right)}$  be a measurable space and let ${\displaystyle \left(Y,{\mathcal {B}}\right)}$  be a polish space where ${\displaystyle {\mathcal {B}}}$  is its Borel sigma-algebra. Then for every set in the product sigma-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.