# Equivalent conditions for a measurability restriction?

1. Jul 29, 2013

### economicsnerd

Let $Z:= A^T$, where $T$ is a countable set and $A$ is a finite set. Under the product topology, $Z$ is a compact metrizable space. (As a special case, notice that $Z$ could be the Cantor set).

Given a closed set $X \subseteq Z$, I'm interested in answering the question, "Does there exist some partition $\mathcal P$ of $T$ such that $X = \{f \in Z: \enspace f \text{ is } \mathcal P\text{-measurable}\}$?"

Is anybody here aware of any alternative formulations of the above question? For example, is there a topological characterization?

2. Jul 29, 2013

### micromass

Staff Emeritus
What do you mean with measurable with respect to a partition?

3. Jul 29, 2013

### economicsnerd

Sorry, I could have been clearer...

I meant measurable functions $$(T, \Sigma)\to (A, 2^A),$$ where $\Sigma$ is the $\sigma$-algebra generated by $\mathcal P$.

4. Jul 29, 2013

### micromass

Staff Emeritus
Given a single function $f$, this is a closed set $\{f\}$. Clearly the coarsest $\sigma$-algebra that makes $f$ measurable is

$$\sigma(f) := \{f^{-1}(E)~\vert~E\subseteq A\}$$

This is clearly generated by the partition $\{f^{-1}(a)~\vert~a\in A\}$.
But any function constant on the partition is also measurable. In particular, the globally constant functions are measurable. So there is no $\sigma$-algebra which only makes $f$ measurable. So the answer is negative because it fails for $\{f\}$.

5. Jul 29, 2013

### economicsnerd

Right. So (given nontrivial $A$) it's not true for singleton $X$, and therefore not true for arbitrary closed $X.$

What I'm after is a proposition of the form:

$$Given a closed set $X \subseteq Z$:

$$\Phi(X) \iff \exists\mathcal P\text{ such that } X = \{f \in Z: \enspace f \text{ is } \mathcal P\text{-measurable}\} \text{''}$$
where $\Phi(X)$ is some other axiom about $X$, possibly some topological statement.

Said differently, I'm wondering what statements are true of members of $$\bigg\{ \{f \in Z: \enspace f \text{ is } \mathcal P\text{-measurable}\}: \enspace \mathcal P \text{ partition of } T\bigg\}$$ and fail for every other closed subset of $Z$.

Again, apologies for any lack of clarity (and for being long-winded).

6. Jul 29, 2013

### economicsnerd

I have some vague notion that there's an equivalent condition involving retractions.