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 #$2$?

  • 1
    $\begingroup$ The answer is yes, I will write a detailed answer later. $\endgroup$ May 17, 2013 at 4:40
  • $\begingroup$ @MichaelGreinecker: Thanks, Michael. Encouraged by your comment, i figured out a proof. Would you mind if i post it? $\endgroup$
    – Evan Aad
    May 17, 2013 at 10:02
  • 2
    $\begingroup$ Of course not, write it down. I also have a post about how one can simplify this actually here. $\endgroup$ May 17, 2013 at 10:34

1 Answer 1


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

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


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

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

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


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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.