# Measurable maps

1. Sep 23, 2007

### P3X-018

I have 2 maps f and h such

$$f :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Y}, \mathbb{K})$$
$$h :\, (\mathcal{X}, \mathbb{E}) \rightarrow (\mathcal{Z}, \mathbb{G})$$

where $\mathbb{K}$ and $\mathbb{G}$ are $\sigma$-algebras on the spaces Y and Z respectively, and $\mathbb{E} = \sigma(f)$ is the $\sigma$-algebra generated by the map f.
f is assumed to be surjective and $\mathbb{G}$ is assumed to separate points in Z. That is for any 2 different point a and b in Z we can find a set $A\in \mathbb{G}$ such that $a\in A$ and $b \notin A$.

The problem is that, assuming h is $\mathbb{E}$-$\mathbb{G}$ measurable, to show that there is a map $g:\, \mathcal{Y} \rightarrow \mathcal{Z}$ such that $h = g\circ f$.

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. Sep 23, 2007

### EnumaElish

Can you go from (Y,K) to (X,E)? If you can, then you know how to go from (X,E) to (Z,G).

3. Sep 23, 2007

### P3X-018

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.

4. Sep 23, 2007

### matt grime

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....

5. Sep 23, 2007

### P3X-018

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).

EDIT:
Another question is to show that g is K-G measurable.
Now if I could look at the map from Y -> X that is $f^{-1}$, then I could use the theorem that says that g is K-G measurable if and only if $f^{-1}$ 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 $g^{-1}(A) \in \mathbb{K},\,\, \forall A\in\mathbb{G}$.

Last edited: Sep 23, 2007
6. Sep 24, 2007

### matt grime

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