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I have recently read about about disintegration theorem, i.e.,

Disintegration theorem Let

  • $X$ be a Polish space, $\mathcal X$ its Borel $\sigma$-algebra, and $\mu$ a Borel probability measure on $X$.
  • $(Y, \mathcal Y)$ a measurable space, $\pi:X\to Y$ a measurable map, and $\nu := f_\sharp \mu$ the push-forward of $\mu$ by through $f$.

There is a collection $(\mu_y)_{y\in Y}$ of Borel probability measures on $X$ with the following properties.

  1. For all bounded measurable $f:X\to \mathbb C$, the map $y \mapsto \int_X f\mathrm d\mu_y$ is measurable.
  2. For all bounded measurable $f:X\to \mathbb C$ and $\nu$-integrable $g:Y\to \mathbb C$, $$ \int_X f (g\circ \pi) \mathrm d\mu = \int_Y \left(\int_X f\mathrm d\mu_y\right)g(y)\mathrm d\nu(y). \quad (\star) $$
  3. For all bounded measurable $g:Y\to \mathbb C$, for $\nu$-a.e. $y \in Y$, $$ g\circ \pi = g(y) \quad \mu_y\text{-a.e.} \quad (\star\star) $$
  4. If, moreover, $Y$ is a separable metric space and $\mathcal Y$ its Borel $\sigma$-algebra, then $\mu_y$ is supported on $\pi^{-1} (y)$ for $\nu$-a.e. $y \in Y$.

In above version, $f \in L_\infty (\mu)$ and thus $h:y \mapsto \int_X f\mathrm d\mu_y$ belongs to $L_\infty (\nu)$. On the other hand, $g \in L_1 (\nu)$. So $hg \in L_1 (\nu)$ by Hölder's inequality. It motivates me to think of a variant in which the assumptions on $f,g$ are exchanged, i.e.,

  • "$f$ is bounded measurable" is replaced by "$f$ is $\mu$-integrable", and
  • "$g$ is $\nu$-integrable" is replaced by "$g$ is bounded measurable".

Below is my failed attempt. The first difficulty is that $f$ must be $\mu_y$-integrable for all $y\in Y$. Could you elaborate if my proposed version is possible?


My attempt: Let $f$ be $\mu$-integrable. Let's prove that claim 1. holds for $f$. Let $(f_n)$ be a sequence of $\mu$-simple (and thus bounded measurable) functions such that $f_n \to f$ pointwise $\mu$-a.e. Let $$ F_n: y \mapsto \int_X f_n \mathrm d \mu_y. $$

Then $F_n$ is well-defined and measurable. We hope that for all $y \in Y$,

  • $\lim_n \int_X f_n \mathrm d \mu_y$ and $\int f \mathrm d \mu_y$ exist, and
  • $$ \lim_n \int_X f_n \mathrm d \mu_y = \int f \mathrm d \mu_y. $$

In this dream case, let $F(y) := \int f \mathrm d \mu_y$ for all $y\in Y$. Then $F_n \to F$ pointwise and everywhere. So $F$ is measurable.

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  • $\begingroup$ What is $F$ here? $\endgroup$ Nov 21, 2022 at 14:59
  • $\begingroup$ @MichaelGreinecker It's a typo. I meant $\int f \mathrm d \mu_y$, not $\int F \mathrm d \mu_y$. $\endgroup$
    – Akira
    Nov 21, 2022 at 15:00
  • $\begingroup$ I guess somlike the following should work: Take the positive and negative parts of $f$ and do the problem for both separately using monotone approximations by simple functions. Then everything works by the monotone convergence theorem and you just need to show that their averages are finite. $\endgroup$ Nov 21, 2022 at 15:58

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