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Let $X,Y$ be compact metric spaces, $2^X$ the set of all closed subsets of $X$ and $f:X\to Y$ be a 1st class Borel mapping.

Im trying to check Borel class of mapping $G:2^Y\to 2^X$. I submit it in a form of 2 compositions: $\theta :2^Y \to X$ and $\omega:X \to 2^X$, $G=\theta \circ \omega$. I proved that $\omega$ has 1st Borel class, so if $\theta$ has 1st Borel class, then $G$ too.

I know the following theorem:

Let $X$ be a compact metric space, $Y$ a metric space. If a multivalued mapping $F:Y\to 2^X$ is upper (lower) semicontinuous, then $F$ is a 1st class Borel mapping.

But it works with mappings like $F:Y\to 2^X$ and only for upper (lower) semicontinuous mapping.

Q1: Is there such a theorem for my example $\theta :2^Y \to X$?

I have a stupid idea: Using method of compositions: Let $f^{-1} : Y \to X $ and $\sigma : X \to 2^X$. By condition $f$ has 1st Borel class. Correct me if I wrong, but $f^{-1}$ has 1st Borel class too. Like $\omega$, $\sigma$ has 1st Borel class. Therefore, we obtain composition of two Borel mappings $\theta=f^{-1}\circ\sigma$, so $\theta$ has 1st Borel class.

Q2:Am I mistaken in my reasoning?

Improve this question
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  • $\begingroup$ The composition of two maps of the 1-st Borel class needs not be of the first Borel class: the classical Dirichlet function (the characteristic function of rationals) can be written as a composition of two functions of the first Baire class. $\endgroup$
    – Taras Banakh
    Commented Jun 7, 2018 at 18:50
  • $\begingroup$ It's not clear what your $G$ and $\theta$ are intended to be. The only obvious map $2^Y\to2^X$ is inverse image under $f$, but that doesn't naturally factor through $X$. And the only obvious map $X\to2^X$ would be $x\mapsto\{x\}$, but your $\omega$ is apparently something else --- what is it? $\endgroup$
    – Andreas Blass
    Commented Jun 7, 2018 at 18:51
  • $\begingroup$ I thougth, the Dirichlet function the example of 2nd Borel class,where $2^X\mapsto 2^Y$ and $2^Y\mapsto 2^X$ aren't Borel mappings. $\endgroup$
    – Tony T.
    Commented Jun 8, 2018 at 10:12
  • $\begingroup$ Multivalued mapping $F:2^X \mapsto 2^Y$ has 2nd Borel class, for $f-$ 1st Borel class. So I think for G there exists a similar theorem, but I dont know it. For $\omega$ I used theorem 3.1 from M. M. Čoban, “Multi-valued mappings and Borel sets” $\endgroup$
    – Tony T.
    Commented Jun 8, 2018 at 10:29
  • $\begingroup$ A map from $X$ to $Y$ is denoted $X\to Y$. The symbol $\mapsto$ is used to denote the assignment $x\mapsto f(x)$, and precisely to distinguish it from the map itself. I edited accordingly. Also removed confusing use of minus sign, replaced by words. $\endgroup$
    – YCor
    Commented Jun 26, 2018 at 0:32

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