Composition of markov kernels is markov kernel

Question

Let $(X,\mathcal{X}$), $(Y,\mathcal{Y})$ and $(Z,\mathcal{Z})$ be measurable spaces. Let $\kappa:X \times \mathcal{Y} \to [0,1]$ and $\kappa':Y \times \mathcal{Z} \to [0,1]$ be markov kernels (also known as probability kernels). How do I prove that $\kappa'':X \times \mathcal{Z} \to [0,1]$, defined by $$ \kappa''(x,S):= \int_{y \in Y} \kappa'(y)(S) \; \kappa(x)(dy) $$ is a markov kernel?

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I am assuming this proof has been done already, so I would particularly appreciate a source containing a detailed proof. But the proof itself or even hints on the proof could also be helpful.

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Proving that $\forall x \in X$, $\kappa''(x,\cdot)$ is a probability measure is simple (by applying properties of the Lebesgue integral). But how do I prove that $\forall S \in \mathcal{Z}$, $\kappa''(\cdot,S)$ is measurable? Based on similar questions, I am fairly certain I need to apply the monotone class theorem, but I do not know how.

Answer 1

Here is a version of the functional monotone class theorem found in Pollard's "A User Guide to Measure Theoretic Probability"

Theorem. Let $\mathcal{H}^+$ be a $\lambda$-cone of bounded, nonnegative functions, and $\mathcal{G}$ be a subclass of $\mathcal{H}^+$ that is stable under the formation of pointwise products of pairs of functions. Then $\mathcal{H}^+$ contains all bounded, nonnegative, $\sigma(\mathcal{G})$-measurable functions.

$\lambda$-cone here is a class ($\mathcal{H}$) of bounded, nonnegative functions satisfying:

• $\mathcal{H}$ is a cone containing the constant function $1$;
• If $h_1,h_2\in \mathcal{H}$ and $h_1\ge h_2$, then $h_1-h_2\in\mathcal{H}$;
• $\mathcal{H}$ is closed under increasing pointwise limits.

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Let $\mathcal{H}^+$ be the collection of all bounded, nonnegative, $(\mathcal{X}\otimes\mathcal{Y})$-measurable functions $f$ for which $\int_Y f(x, y)\kappa(x,dy)$ is measurable. Then $\mathcal{H}^+$ is a $\lambda$-cone containing the class $\mathcal{G}$ of all indicators of measurable rectangles $\{1_{A\times B}:A\in\mathcal{X},B\in\mathcal{Y}\}$. Finally, $\mathcal{G}$ is closed under pointwise products and generates $\mathcal{X}\otimes\mathcal{Y}$.

Since $\kappa'(y,S)$ is measurable for all $S\in\mathcal{Z}$, for a given $S\in\mathcal{Z}$ take $f(x,y)=\kappa'(y,S)$.