Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


  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?
share|cite|improve this question
Show it for functions with countable range first. Then, take limits. – Michael Greinecker Apr 12 '12 at 17:26
Related question on MO: – cardinal Apr 12 '12 at 19:21
up vote 5 down vote accepted

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.

share|cite|improve this answer
Thanks, Nate. I will try this general idea. In Step 1, did you mean to have $\Sigma_f = \{f^{-1}(B):B \in \Sigma_2\}$ instead? Is there any chance there is another way to show this result? The particular text I'm looking at doesn't introduce the "standard mantra" until almost 100 pages after this exercise appears, making me wonder if it might be possible to do the exercise without it. – bpt Apr 12 '12 at 18:42
@bpt You certainly need to make a limiting argument. Maybe your book contains a result such as every real-valued measurable function being the uniform limit of simple functions? – Michael Greinecker Apr 12 '12 at 18:46
@Michael: Yes, it does...100 pages later. At this point in the book, one has only seen the most basic facts of measurable maps, like closure under composition, limits, summation, etc. and the fact that we can check measurability by looking at the pullbacks of a generating class of sets on the range space. The definition of simple function has been introduced, but the development of approximation by discretizing the range has not. Maybe the exercise is just misplaced? – bpt Apr 12 '12 at 18:51
Every proof of the result I've seen uses some limiting result. The result is fairly standard and often referred to as the Doob-Dynkin lemma. – Michael Greinecker Apr 12 '12 at 18:55
@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? – Nate Eldredge Apr 12 '12 at 19:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.