Projection (measure theory)

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]

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 ot ${\displaystyle \mathbb {R} ^{n}}$ onto ${\displaystyle \mathbb {R} ^{n-k}}$ where ${\displaystyle k<n}$ 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 ${\displaystyle \mathbb {R} }$ which is a projection of some measurable set of ${\displaystyle \mathbb {R} ^{2}}$, is the only sort of such example.

• Analytic set
• Descriptive set theory

• "Measurable projection theorem", PlanetMath