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.