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

>[Disintegration theorem](https://math.stackexchange.com/questions/4563738/disintegration-theorem-how-do-the-authors-prove-that-mu-y-is-supported-on) 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. 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](https://math.stackexchange.com/questions/4560364/hausdorff-topological-spaces-is-the-pointwise-limit-of-a-sequence-of-borel-meas).