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 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 [math]\displaystyle{ X := \{0, 1\} }[/math] with the 𝜎-algebra [math]\displaystyle{ \mathcal{F} := \{\varnothing, \{0\}, \{1\}, \{0, 1\}\} }[/math] and the space [math]\displaystyle{ Y := \{0, 1\} }[/math] with the 𝜎-algebra [math]\displaystyle{ \mathcal{G} := \{\varnothing, \{0, 1\}\}. }[/math] The diagonal set [math]\displaystyle{ \{(0, 0), (1, 1)\} \subseteq X \times Y }[/math] is not measurable relatively to [math]\displaystyle{ \mathcal{F}\otimes\mathcal{G}, }[/math] 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 [math]\displaystyle{ \mathcal{L} }[/math] be Lebesgue 𝜎-algebra of [math]\displaystyle{ \Reals }[/math] and let [math]\displaystyle{ \mathcal{L}' }[/math] be the Lebesgue 𝜎-algebra of [math]\displaystyle{ \Reals^2. }[/math] For any bounded [math]\displaystyle{ N \subseteq \Reals }[/math] not in [math]\displaystyle{ \mathcal{L}. }[/math] the set [math]\displaystyle{ N \times \{0\} }[/math] is in [math]\displaystyle{ \mathcal{L}', }[/math] since Lebesgue measure is complete and the product set is contained in a set of measure zero.

Still one can see that [math]\displaystyle{ \mathcal{L}' }[/math] is not the product 𝜎-algebra [math]\displaystyle{ \mathcal{L} \otimes \mathcal{L} }[/math] but its completion. As for such example in product 𝜎-algebra, one can take the space [math]\displaystyle{ \{0, 1\}^\Reals }[/math] (or any product along a set with cardinality greater than continuum) with the product 𝜎-algebra [math]\displaystyle{ \mathcal{F} = \textstyle {\bigotimes\limits_{t\in\Reals}} \mathcal{F}_t }[/math] where [math]\displaystyle{ \mathcal{F}_t = \{\varnothing,\{0\} ,\{1\} ,\{0, 1\}\} }[/math] for every [math]\displaystyle{ t \in \Reals. }[/math] In fact, in this case "most" of the projected sets are not measurable, since the cardinality of [math]\displaystyle{ \mathcal{F} }[/math] is [math]\displaystyle{ \aleph_0 \cdot 2^{\aleph_0} = 2^{\aleph_0}, }[/math] whereas the cardinality of the projected sets is [math]\displaystyle{ 2^{2^{\aleph_0}}. }[/math] 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 [math]\displaystyle{ (X, \mathcal{F}) }[/math] be a measurable space and let [math]\displaystyle{ (Y, \mathcal{B}) }[/math] be a polish space where [math]\displaystyle{ \mathcal{B} }[/math] is its Borel 𝜎-algebra. Then for every set in the product 𝜎-algebra [math]\displaystyle{ \mathcal{F} \otimes \mathcal{B}, }[/math] the projected set onto [math]\displaystyle{ X }[/math] is a universally measurable set relatively to [math]\displaystyle{ \mathcal{F}. }[/math]^[4]

An important special case of this theorem is that the projection of any Borel set of [math]\displaystyle{ \Reals^n }[/math] onto [math]\displaystyle{ \Reals^{n-k} }[/math] where [math]\displaystyle{ k \lt n }[/math] 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 [math]\displaystyle{ \Reals }[/math] which is a projection of some measurable set of [math]\displaystyle{ \Reals^2, }[/math] is the only sort of such example.