Equivalent conditions for a measurability restriction?

[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?

 

[micromass]

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

 

[economicsnerd]

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

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.

 

[micromass]

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 $\{f\}$.

 

[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).

 

[economicsnerd]

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