Factorization of a map from measurability #
Consider f : X → Y and g : X → Z and assume that g is measurable with respect to the pullback
along f. Then g factors through f, which means that (if Z is nonempty)
there exists h : Y → Z such that g = h ∘ f.
If Z is completely metrizable, the factorization map h can be taken to be measurable.
This is the content of the Doob-Dynkin lemma:
see exists_eq_measurable_comp.
If a function g is measurable with respect to the pullback along some function f,
then to prove g x = g y it is enough to prove f x = f y.
If a function g is strongly measurable with respect to the pullback along some function f,
then to prove g x = g y it is enough to prove f x = f y.
If Z is not empty there exists h : Y → Z such that g = h ∘ f.
If Z is also completely metrizable, the factorization map h can be taken to be measurable
(see exists_eq_measurable_comp).
If a function g is strongly measurable with respect to the pullback along some function f,
then there exists some strongly measurable function h : Y → Z such that g = h ∘ f.
If a function g is measurable with respect to the pullback along some function f,
then there exists some measurable function h : Y → Z such that g = h ∘ f.
If a function is measurable with respect to the σ-algebra generated by the first coordinates, then it only depends on those first coordinates.
If a function is strongly measurable with respect to the σ-algebra generated by the first coordinates, then it only depends on those first coordinates.
If a function is measurable with respect to the σ-algebra generated by the first coordinates, then it only depends on those first coordinates.
If a function is strongly measurable with respect to the σ-algebra generated by the first coordinates, then it only depends on those first coordinates.