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Here is a restatement of a problem in a textbook I encountered. I'm well beyond the age of doing homework and this is purely for self-study.

Exercise: Let $f : (X,\Sigma_1) \to (Y, \Sigma_2)$ and $h:(X,\Sigma_1) \to (\mathbb R, \mathcal B)$ be measurable maps where in the latter case $\mathcal B$ denotes the Borel $\sigma$-algebra over $\mathbb R$. Let $\Sigma_f = \sigma(f)$. Show that $h$ is $\Sigma_f$-measurable if and only if there exists $g : (Y,\Sigma_2) \to (\mathbb R, \mathcal B)$ such that $h(x) = g(f(x))$ for all $x \in X$.

One direction of the proof is easy. Suppose such a $g$ exists. Then, for all $B \in \mathcal B$, $h^{-1}(B) = f^{-1}( g^{-1}(B) )$ and so $h^{-1} \in \Sigma_f$.

There seem to be some holes in the opposite direction which I can't quite fill.

For all $z \in \mathbb R$, I defined

$ A_z = \{x: h(x) = z\}. $

Then $A_z \in \Sigma_1$ since the singletons $\{z\}$ are Borel-measurable. Also, for $z \neq z'$, it is true that $A_z \cap A_{z'} = \emptyset$. Now, if $h$ is $\Sigma_f$-measurable, then $A_z = f^{-1}(B_z)$ for some $B_z \in \Sigma_2$. But then, for $z \neq z'$, we have that $B_z \cap B_{z'} = \emptyset$ as well, so the $\{B_z\}_{z \in \mathbb R}$ sets partitions $Y$ modulo the portion not in the image of $f$.

Now, set $g(y) = z$ on $B_z$ and set $g(y) = 0$ on $y \in N_0 := Y \setminus \cup_{z \in \mathbb R} B_z$. It seems reasonable to claim that $N_0$ is a measurable set by considering that $N_0 = Y \setminus \cup_n C_n$ where $f^{-1}(C_n) = h^{-1}((-\infty,n))$ and $C_n \in \Sigma_2$ by assumption.

But, this only seems to show that we can construct a well-defined $g$. It doesn't seem to prove that it is measurable! To get measurability we need to show something in addition to this, like $\{y: g(y) \leq z\} \in \Sigma_2$ for all $z \in \mathbb R$.

For z < 0 it seems we should be able to get a correspondence between $\{y: g(y) \leq z\}$ and $C_z$ where $C_z \in \Sigma_2$ satisfies $f^{-1}(C_z) = h^{-1}((-\infty,z))$. For $z \geq 0$, I think it would be something like $\{y : g(y) \leq z\} = C_z \cup N_0$, I think.

I can't quite seem to make the argument go through.

Questions:

  1. Is this on the right track? If so, how do we finish it off? (It seems a little "too constructive" for a typical measure-theoretic argument.)
  2. Is there some other more clever or direct argument? If so, what is it?
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    Related question on MO: http://mathoverflow.net/questions/3912/question-on-sigma-fields2012-04-12

1 Answers 1

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This proof goes by what probabilists and other practitioners of measure theory sometimes call the "standard mantra": first indicator functions, then simple functions (those with finite range), then nonnegative measurable functions, then all measurable functions.

  1. As a preliminary step, show that $\Sigma_f = \{ f^{-1}(B) : B \in \Sigma_2\}$. ($\supset$ is obvious. For $\subset$, show that the right side (call it $\Sigma'$) is a $\sigma$-algebra and $f$ is $(X, \Sigma'),(Y,\Sigma_2)$ measurable.)

  2. Now suppose $h = 1_A$ for some $A \in \Sigma_f$. Find a $g$ such that $h = g \circ f$. (Use the previous step).

  3. Suppose next that $h = \sum_{i=1}^n c_i 1_{A_i}$ is a simple function. Again, find a $g$. (It's not hard to guess what it is, given the previous step.)

  4. If $h$ is a nonnegative $\Sigma_f$-measurable function, then recall that $h = \sup_n h_n$ for some sequence $h_n$ of $\Sigma_f$-measurable simple functions. (This is about the only "constructive" way to describe measurable functions.) Again, find a $g$.

  5. Finally, if $h$ is any $\Sigma_f$-measurable function, write $h = h^+ - h^-$ and use the previous step.

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    @bpt: Thanks for the correction, I fixed it. I don't see how to do it without some limiting or density-type argument. One could use a functional monotone class or $\pi$-$\lambda$ argument, but those might not be in your book either. What book is it?2012-04-12