- #1
P3X-018
- 144
- 0
I have 2 maps f and h such
[tex] f :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Y}, \mathbb{K}) [/tex]
[tex] h :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Z}, \mathbb{G}) [/tex]
where [itex] \mathbb{K} [/itex] and [itex] \mathbb{G} [/itex] are [itex] \sigma[/itex]-algebras on the spaces Y and Z respectively, and [itex] \mathbb{E} = \sigma(f) [/itex] is the [itex] \sigma[/itex]-algebra generated by the map f.
f is assumed to be surjective and [itex] \mathbb{G} [/itex] is assumed to separate points in Z. That is for any 2 different point a and b in Z we can find a set [itex] A\in \mathbb{G} [/itex] such that [itex] a\in A [/itex] and [itex] b \notin A [/itex].
The problem is that, assuming h is [itex]\mathbb{E}[/itex]-[itex]\mathbb{G} [/itex] measurable, to show that there is a map [itex] g:\, \mathcal{Y} \rightarrow \mathcal{Z} [/itex] such that [itex] h = g\circ f [/itex].
I don't even know how to attack this problem. What is even meant by showing that there IS such a map? A hint could be helpful.
[tex] f :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Y}, \mathbb{K}) [/tex]
[tex] h :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Z}, \mathbb{G}) [/tex]
where [itex] \mathbb{K} [/itex] and [itex] \mathbb{G} [/itex] are [itex] \sigma[/itex]-algebras on the spaces Y and Z respectively, and [itex] \mathbb{E} = \sigma(f) [/itex] is the [itex] \sigma[/itex]-algebra generated by the map f.
f is assumed to be surjective and [itex] \mathbb{G} [/itex] is assumed to separate points in Z. That is for any 2 different point a and b in Z we can find a set [itex] A\in \mathbb{G} [/itex] such that [itex] a\in A [/itex] and [itex] b \notin A [/itex].
The problem is that, assuming h is [itex]\mathbb{E}[/itex]-[itex]\mathbb{G} [/itex] measurable, to show that there is a map [itex] g:\, \mathcal{Y} \rightarrow \mathcal{Z} [/itex] such that [itex] h = g\circ f [/itex].
I don't even know how to attack this problem. What is even meant by showing that there IS such a map? A hint could be helpful.
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