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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.

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.

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. 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.

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.

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 the 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.

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. 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.