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Measurable maps

  1. Sep 23, 2007 #1
    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.
    Last edited: Sep 23, 2007
  2. jcsd
  3. Sep 23, 2007 #2


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    Can you go from (Y,K) to (X,E)? If you can, then you know how to go from (X,E) to (Z,G).
  4. Sep 23, 2007 #3
    So if I could go from (Y,K) to (X,E) that is if there was a inverse map of f, that would be enough to conclude that THERE IS a map g?
    But f would have to be bijective inorder to go from Y -> X.
    Can f be injective under the given assumption? Because it's already surjective, so if it were injective we would be done.
  5. Sep 23, 2007 #4

    matt grime

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    You do not assume injectivity.

    Clearly the hypotheses have been given for a reason. So start to see what you can deduce from them. Let's see what you need to do. You need, given a point y in Y to find a way to associate it to a point g(z) in Z. And we need to do it in such a way that gf(x)=h(x). Wow, what do you know, from just writing that down I've essentially solved the question for you....
  6. Sep 23, 2007 #5
    I didn't mean that I assumed f to be injective, only that wether it should (or could) be shown that f is injective, and if so would that solve the problem, since we could go from Y to Z by h(f^(-1)(y)).
    Hmm this sounds too 'simple'...
    So basically the argument is, that since f is surjective then for every y in Y there is a x in X, such that y = f(x), and since z = h(x), z is the element in Z that is mapped into by y = f(x), with g(f(x)) = h(x), that is z is mapped by z = g(y).

    Another question is to show that g is K-G measurable.
    Now if I could look at the map from Y -> X that is [itex] f^{-1}[/itex], then I could use the theorem that says that g is K-G measurable if and only if [itex] f^{-1}[/itex] is K-E measurable, assuming that h is E-G measurable.
    I'm pretty certain of that I need to use this theorem, somehow, to show the measurability of g, since obviously I can't investigate wether [itex] g^{-1}(A) \in \mathbb{K},\,\, \forall A\in\mathbb{G} [/itex].
    Last edited: Sep 23, 2007
  7. Sep 24, 2007 #6

    matt grime

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    What was the other assumption in the original question that you've not used? The one abotu G separating points. PErhaps that is useful.
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