# Measurability of an Indexed Product-Measure

If for any fixed $\omega_1$, $P_{\omega_1}$ is a probability measure and $Q_{\omega_1}$ is a stochastic kernel and both are measurable in $\omega_1$, is the indexed product measure $K_{\omega_1}:=P_{\omega_1}\otimes Q_{\omega_1}$ measurable in $\omega_1$?

To rephrase the question more precisely, let $\Omega_1$, $\Omega_2$ and $\Omega_3$ be non-empty sets, let $\mathcal{A}_1$, $\mathcal{A}_2$ and $\mathcal{A}_3$ be $\sigma$-algebras on $\Omega_1$, $\Omega_2$ and $\Omega_3$, respectively, let $P$ be a stochastic kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$ and let $Q$ be a stochastic kernel from $\mathcal{A}_1\otimes\mathcal{A}_2$ to $\mathcal{A}_3$. For each $\omega_1\in\Omega_1$, denote by $P_{\omega_1}$ the function

$$P_{\omega_1}:\mathcal{A}_2\rightarrow\left[0,1\right],\space\space P_{\omega_1}\left(B_2\right):=P\left(\omega_1,B_2\right)$$

and by $Q_{\omega_1}$ the function

$$Q_{\omega_1}:\Omega_2\times\mathcal{A}_3\rightarrow\left[0,1\right],\space\space Q_{\omega_1}\left(\omega_2,B_3\right):=Q\left(\left(\omega_1,\omega_2\right),B_3\right)$$

It is readily seen that $P_{\omega_1}$ is a probability measure on $\mathcal{A}_2$ and that $Q_{\omega_1}$ is a stochastic kernel from $\mathcal{A}_2$ to $\mathcal{A}_3$. So $P_{\omega_1}\otimes Q_{\omega_2}$ is a product measure on $\mathcal{A}_2\otimes\mathcal{A}_3$ and we can define

$$K:\Omega_1\times\left(\mathcal{A}_2\otimes\mathcal{A}_3\right)\rightarrow\left[0,1\right],\space\space K\left(\omega_1,B_{2,3}\right):=\left(P_{\omega_1}\otimes Q_{\omega_1}\right)\left(B_{2,3}\right)$$

Is $K$ a stochastic kernel? For $K$ to be one, it needs to satisfy two conditions:

1. For every $\omega_1\in\Omega_1$, $K\left(\omega_1,B_{2,3}\right)$ is a probability measure on $\mathcal{A}_2\otimes\mathcal{A}_3$,
2. For every $B_{2,3}\in\mathcal{A}_2\otimes\mathcal{A}_3$, $K\left(\omega_1,B_{2,3}\right)$ is $\mathcal{A}_1/\mathfrak{B}$-measurable, with $\mathfrak{B}$ being the Borel field on the real line.

It is easy to see that condition #$1$ is satisfied, but what about condition #$2$?

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The answer is yes, I will write a detailed answer later. –  Michael Greinecker May 17 '13 at 4:40
@MichaelGreinecker: Thanks, Michael. Encouraged by your comment, i figured out a proof. Would you mind if i post it? –  Evan Aad May 17 '13 at 10:02
Of course not, write it down. I also have a post about how one can simplify this actually here. –  Michael Greinecker May 17 '13 at 10:34

Let $B_{2,3}\in\mathcal{A}_2\otimes\mathcal{A}_2$. We need to show that the map

$$\omega_1\mapsto K\left(\omega_1,B_{2,3}\right)$$

is $\mathcal{A}_1/\mathfrak{B}$-measurable.

Let $\omega_1^*\in\Omega_1$. Then

$$K\left(\omega_1^*,B_{2,3}\right)=\int_{\Omega_2}P_{\omega_1^*}\left(d\omega_2\right)\int_{\Omega_3}Q_{\omega_1^*}\left(\omega_2,d\omega_3\right)\mathbb{1}_{B_{2,3}}\left(\omega_2,\omega_3\right)$$

Define

$$g:\left(\Omega_1\times\Omega_2\right)\times\Omega_3\rightarrow\left[0,1\right],\space\space g\left(\left(\omega_1,\omega_2\right),\omega_3\right):=\mathbb{1}_{\Omega_1\times B_{2,3}}\left(\omega_1,\left(\omega_2,\omega_3\right)\right)$$

It is easy to see that $g$ is non-negative, $\left(\mathcal{A}_1\otimes\mathcal{A}_2\right)\otimes\mathcal{A}_3/\mathfrak{B}$-measurable and that

$$K\left(\omega_1^*,B_{2,3}\right)=\int_{\Omega_2}P\left(\omega_1^*,d\omega_2\right)\space I_g\left(\omega_1^*,\omega_2\right)$$

with

$$I_g:\Omega_1\times\Omega_2\rightarrow\left[0,\infty\right],\space\space I_g\left(\omega_1,\omega_2\right):=\int_{\Omega_3}Q\left(\left(\omega_1,\omega_2\right),d\omega_3\right)\space g\left(\left(\omega_1,\omega_2\right),\omega_3\right)$$

Now, consider Lemma 14.20 in Klenke's Probability Theory (2006), which i reproduce here with some cosmetic changes.

Let $\kappa$ be a finite transition kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$ and let $f:\Omega_1\times\Omega_2\rightarrow\left[0,\infty\right]$ be $\mathcal{A}_1\otimes\mathcal{A}_2$/$\overline{\mathfrak{B}}$- measurable ($\overline{\mathfrak{B}}$ being the Borel field on the extended real line). Then the map $$I_f:\Omega_1\rightarrow\left[0,\infty\right],\space\space I_f\left(\omega_1\right):=\int_{\Omega_2} \kappa\left(\omega_1,d\omega_2\right)\space f\left(\omega_1,\omega_2\right)$$ is well-defined and $\mathcal{A}_1$/$\overline{\mathfrak{B}}$-measurable.

Applying this lemma with $\kappa:=Q$, $f:=g$, we see that $I_g$ is $\mathcal{A}_1\otimes\mathcal{A}_2/\overline{\mathfrak{B}}$-measurable. Since $I_g$ is clearly non-negative, we can apply the lemma a second time with $\kappa:=P$, $f:=I_g$ to arrive at the desired result.

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