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Equivalent conditions for a measurability restriction?

  1. Jul 29, 2013 #1
    Let [itex]Z:= A^T[/itex], where [itex]T[/itex] is a countable set and [itex]A[/itex] is a finite set. Under the product topology, [itex]Z[/itex] is a compact metrizable space. (As a special case, notice that [itex]Z[/itex] could be the Cantor set).

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

    Is anybody here aware of any alternative formulations of the above question? For example, is there a topological characterization?
     
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  3. Jul 29, 2013 #2

    micromass

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    What do you mean with measurable with respect to a partition?
     
  4. Jul 29, 2013 #3
    Sorry, I could have been clearer...

    I meant measurable functions [tex](T, \Sigma)\to (A, 2^A),[/tex] where [itex]\Sigma[/itex] is the [itex]\sigma[/itex]-algebra generated by [itex]\mathcal P[/itex].
     
  5. Jul 29, 2013 #4

    micromass

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

    [tex]\sigma(f) := \{f^{-1}(E)~\vert~E\subseteq A\}[/tex]

    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\}##.
     
  6. Jul 29, 2013 #5
    Right. So (given nontrivial [itex]A[/itex]) it's not true for singleton [itex]X[/itex], and therefore not true for arbitrary closed [itex]X.[/itex]

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

    [itex]``[/itex]Given a closed set [itex]X \subseteq Z[/itex]:

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

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

    Again, apologies for any lack of clarity (and for being long-winded).
     
  7. Jul 29, 2013 #6
    I have some vague notion that there's an equivalent condition involving retractions.
     
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