why is $Y(\omega_1)=\int_{\Omega_2}K(\omega_1,d\omega_2)X_{\omega_1}(\omega_2)$ a $\mathcal{B}_1$-measurable function? In the picture below, why is $$Y(\omega_1)=\int_{\Omega_2}K(\omega_1,d\omega_2)X_{\omega_1}(\omega_2)$$  $\mathcal{B}_1$-measurable?

I understand that $X_{\cdot}(\omega_2)$ is $\mathcal{B}_1$-measurable, being a section of measurable function $X$, and $K(\cdot, A_2)$ is also $\mathcal{B}_1$-measurable. But why is $Y$ ?
Any help would be appreciated.
 A: You may use a version of the functional monotone class theorem found in Durrett (Theorem 6.1.3 on page 235) to show that (1) is $\mathcal{B}_1$-measurable for all nonnegative measurable functions on $\Omega_1\times\Omega_2$. Let $\mathcal{H}$ be a collection of functions $f$ on $\Omega_1\times\Omega_2$ for which
$$
\omega_1\mapsto \int_{\Omega_2} K(\omega_1,d\omega_2)f(\omega_1,\omega_2)\tag{1}
$$
is measurable. Then for $f,g\in\mathcal{H}$ and $c\in \mathbb{R}$, $(f+g),cf\in \mathcal{H}$. Also $\mathcal{H}$ is closed under increasing pointwise limits and contains all indicators of sets of the form $A\times B$ with $A\in \mathcal{B}_1$ and $B\in\mathcal{B}_2$ because
$$
\int_{\Omega_2}K(\omega_1,d\omega_2)1_{A\times B}(\omega_1,\omega_2)=1_A(\omega_1)\int_B K(\omega_1,d\omega_2).
$$
Thus, $\mathcal{H}$ contains all bounded measurable functions. Finally, measurability holds for all nonnegative measurable functions by considering the limit of $\int_{\Omega_2}K(\omega_1,d\omega_2)(n\wedge f(\omega_1,\omega_2))$.
