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Where can I find a treatment of the countable product of Markov kernels (or more generally transition kernels)? In all my resources only pairwise products are treated, and Google offers an impressive $0$ results.

Clearly, we can define countable product via $\bigotimes_{i=1}^\infty\kappa_i:\Omega\times\bigotimes_{i=1}^\infty\Sigma_i\rightarrow[0,1]$, $(\omega,\mathcal E)\mapsto(\bigotimes_{i=1}^\infty\kappa_i(\omega,\cdot))(\mathcal E)$, for kernels $\kappa_i:\Omega\times\Sigma_i\rightarrow[0,1]$, i.e. we simply take the product measure, or any finite product, with the remainder being independent or trivial. Thus, I would assume that somebody already discussed this. I didn't even find a result indicating that the countable product above is measurable.

Matija
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    Look at the ionescu tulcea theorem in Klenke's book. – Mason Oct 16 '22 at 16:33
  • Thanks, that's a good starting point. Unless I missed something, the Ionescu-Tulcea Theorem ensures the existence (and uniqueness) of a measure, that is a kernel over a trivial $\sigma$-algebra. I asked for the generalization of, well, actually Theorem 14.35, to kernels over any Borel product algebras. The proof was also not as straightforward as I hoped it would be. This is a follow-up to this question, where I had to avoid assuming measurability of the product. – Matija Oct 16 '22 at 18:04
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    I don't understand your question. In the Ionescu-Tulcea theorem, you have an initial measure $P_0$ on $(\Omega_0, \mathcal{A}0)$, and for each $n \in \mathbb{N}$ a kernel $\kappa : \Omega_0 \times \dots \times \Omega{n - 1} \times \mathcal{A}n \to [0, 1]$, and the theorem says that there is a measure $P$ on $(\prod{n \in \mathbb{N}0}\Omega_n, \otimes{n \in \mathbb{N}0}\mathcal{A}_n)$ such that the distribution of $(X_0, X_1, \dots, X_n)$ is $P_0 \otimes \otimes{i = 1}^{n}\kappa_i$. The measure spaces $(\Omega_i, \mathcal{A}_i)$ are allowed to be completely arbitrary. – Mason Oct 16 '22 at 18:34
  • If you define $X\sim P$, yes, that's what it says. However, I would like to see that $\bigotimes_{i=1}^{\infty}\kappa_i:\Omega_0\times\bigotimes_{i=1}^\infty\mathcal A$ is a Markov kernel. There's no initial measure here, and to my knowledge this map is not defined in Klenke's book. It's what I call projective limit in the answer. What I'm asking for is one single Markov kernel $\kappa$ such that for any choice of the initial measure $P_0$ you obtain the measure $P=P_0\otimes\kappa$ from the Ionescu-Tulcea Theorem. If this is a consequence of the theorem, I don't see the argument. – Matija Oct 16 '22 at 19:00
  • However, I do see your point regarding the arbitrary measures. I build on Theorem 14.35, but also later on I heavily rely on the fact that we work with Borel spaces. For me, it was hard enough as is. I don't know if we can get that for arbitrary measure spaces (neither with the consistent family given by kernels as in the Ionescu-Tulcea-Theorem, nor given directly as in Theorem 14.35). First step would be to understand if Theorem 14.35 only restricts to Borel algebras for convenience (to reduce to 14.32) or out of necessitiy. To be frank, Borel spaces are good enough for me. – Matija Oct 16 '22 at 19:06
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    We can define $(\bigotimes_{i = 1}^{\infty}\kappa_i)(x_0, A) := (\delta_{x_0} \otimes \bigotimes_{i = 1}^{\infty}\kappa_i)(A)$. The right hand side is defined by the Ionescu-Tulcea theorem. I think you can show that this is a measurable function of $x_0$ using the $\pi-\lambda$ theorem. We already know that if $A = A_1 \times \dots \times A_n$ with $A_i \in \mathcal{A}i$, then $(\bigotimes{i = 1}^{\infty}\kappa_i)(x_0, A) = (\bigotimes_{i = 1}^{n}\kappa_i)(x_0, A_1 \times \dots \times A_n)$ is measurable in $x_0$. – Mason Oct 16 '22 at 19:27
  • Makes sense, I used the entire topology, the rest is fairly similar. The major difference is the assumption. I consider a sequence of kernels in the answer (to derive a consistent family), while you consider kernel products (which makes sense). Thus, you can apply 14.32 directly, while I have to rely on 14.35. Problem solved. Thanks! – Matija Oct 16 '22 at 20:11

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I still haven't found a reference in the literature, but I have the following proof. We refer to Klenke's Probability Theory (2nd Edition, 2014) unless mentioned otherwise.

Let $(\Omega_0,\mathcal A_0,\mathbb P)$ be a measure space and $I=\mathbb Z_{>0}$. Let $(\Omega_i,\mathcal A_i)$, $i\in I$, be Borel spaces (Definition 8.35), $\Omega=\prod_{i\in I}\Omega_i$ and $\mathcal A=\bigotimes_{i\in I}\mathcal A_i$ (cf. Section 14.3). For $J\subseteq I$ let $\Omega_J=\prod_{i\in J}\Omega_i$ and $\mathcal A_J=\bigotimes_{i\in J}\mathcal A_i$. For $J\subseteq L\subseteq I$ let $X^L_J:\Omega_L\rightarrow\Omega_J$, $x\mapsto(x_i)_{i\in J}$, be the canonical projection. For $n\in I$ let $[n]=I\cap[1,n]$.

Consider a familiy of Markov kernels $\kappa_n:\Omega_0\times\mathcal A_{[n]}\rightarrow[0,1]$ for $n\in I$. For $\omega\in\Omega_0$ and $n\in I$ let $P_{\omega,n}$ be given by $P_{\omega,n}(\mathcal E)=\kappa_n(\omega,\mathcal E)$ for $\mathcal E\in\mathcal A_{[n]}$. Then $(\kappa_n)_n$ is consistent for all $\omega\in\Omega_0$, for all $n\in I$ and all $\mathcal E\in\mathcal A_{[n]}$ we have $P_{\omega,n}(\mathcal E)=P_{\omega,n+1}(\mathcal E\times\Omega_{n+1})$.

Let $\kappa_n$ be consistent and fix $\omega\in\Omega$. Notice that for finite $J\subseteq I$ the measure $P'_J$ on $(\Omega_J,\Sigma_J)$ given by $P'_J(\mathcal E_J)=P_{\omega,n}(\mathcal E_J\times\Omega_{[n]\setminus J})$, for any $n\ge\max J$, is well-defined, and that the family $\{P'_J:J\subseteq I,|J|<\infty\}$ is consistent (Definition 14.34). Thus, Theorem 14.35 ensures the existence of a unique measure $P_\omega$ on $(\Omega,\mathcal A)$ such that $P'_J=P_\omega\circ X^{I-1}_J$ for all $J\subseteq I$ with $|J|<\infty$ (in particular $P_{\omega,n}=P_\omega\circ X^{I-1}_{[n]}$, i.e.$P_{\omega,n}(\mathcal E)=P_\omega(\mathcal E\times\Omega_{I\setminus[n]})$), called the projective limit (Kolmogorov's Extension Theorem 14.36). Let $\kappa:\Omega_0\times\mathcal A\rightarrow[0,1]$, $(\omega,\mathcal E)\mapsto P_\omega(\mathcal E)$ be the corresponding (unique) projective limit of $(\kappa_n)_n$.

Now, we show that the projective limit is a transition kernel. First, we give the spaces structure. Since the $(\Omega_i,\mathcal A_i)$ are Borel spaces, we can pull the metric $\mathrm{d}_i:\Omega_i^2\rightarrow\mathbb R$ from the subset $B_i\in\mathcal B(\mathbb R)$ using the Borel isomorphism $\varphi:\Omega_i\rightarrow B_i$, thus $\mathrm{d}_i$ induces a topology $\mathcal T_i$ for which $\mathcal A_i=\mathcal B(\Omega_i)$ is the Borel algebra. This induces the Tychonoff topology $\mathcal T$ on $\Omega$, which is induced by the canonical product metric $\mathrm d:\Omega^2\rightarrow\mathbb R$ obtain from the $\mathrm{d}_i$, and for which the product $\sigma$-algebra $\mathcal A=\mathcal B(\Omega)$ is the Borel algebra (Lemma 1.2 in Kallenberg's Probability Theory, 3rd Edition, 2021, with separability of the countable product).

First, recall that for all $\omega\in\Omega$ the projective limit $\kappa(\omega,\cdot)=(\kappa(\omega,\mathcal E))_{\mathcal E}=P_\omega$ is a probability measure, so we only have to show that $f_{\mathcal E}:\Omega_0\rightarrow[0,1]$, $\omega\mapsto\kappa(\omega,\mathcal E)$, is measurable for all $\mathcal E\in\mathcal A$. First, assume that $\mathcal E\subseteq\Omega$ is closed. For $n\in I$ let $\mathcal E_n=\{(x_i)_{i\in[n]}:x\in\mathcal E\}$, $\mathcal E^*_n=\mathcal E_n\times\Omega_{I\setminus[n]}$, notice that $(\mathcal E^*_n)_n$ is non-increasing and let $\mathcal E^*=\lim_{n\rightarrow\infty}\mathcal E^*_n=\bigcap_{n=1}^\infty\mathcal E^*_n$. By definition we have $\mathcal E\subseteq\mathcal E^*$. On the other hand, for $x^*\in\mathcal E^*$ and all $n\in I$ there exists $x(n)\in\mathcal E$ such that $x_{[n]}(n)=x^*_{[n]}$. But the geometric series yields $$\mathrm{d}(x(n),x^*)=\sum_{i>n}\frac{\mathrm{d}_i(x_i(n),x^*_i)}{2^i(1+\mathrm{d}_i(x_i(n),x^*_i))}\le\sum_{i>n}2^{-i}=\frac{2^{-(n+1)}}{1-2^{-1}}=2^{-n}$$ and thus $\lim_{n\rightarrow\infty}x(n)=x^*$, so $x^*\in\mathcal E$ since $\mathcal E$ is closed. This shows that $\mathcal E^*=\mathcal E$. Using continuity of measures from above we obtain $$f_{\mathcal E}(\omega)=\kappa(\omega,\mathcal E)=P_\omega(\mathcal E)=\inf_{n}P_{\omega}(\mathcal E^*_n)=\inf_nP_{\omega,n}(\mathcal E_n)=\inf_n\kappa_n(\omega,\mathcal E_n)=f_{n}(\omega),$$ where $f_n:\Omega\rightarrow[0,1]$, $\omega\mapsto\kappa_n(\omega,\mathcal E_n)$. But this means that $f_n$ is measurable since $\kappa_n$ is a Markov kernel, which means that $f_{\mathcal E}$ is measurable since it is the infimum of measurable functions. This shows that $f_{\mathcal E}=1-f_{\Omega\setminus\mathcal E}$ is measurable for $\mathcal E$ open.

Now, consider the set $\mathcal D=\{\mathcal E\in\mathcal A:f_{\mathcal E}\,\mathrm{ measurable}\}$. By the above we have $\mathcal T\subseteq\mathcal D$, and $\mathcal T$ is a $\pi$-system, since it is closed with respect to pairwise intersections. Now, we show that $\mathcal D$ is a $\lambda$-system. Clearly, we have $\Omega\in\mathcal T\subseteq\mathcal D$. Also, for any $\mathcal E,\mathcal F\in\mathcal D$ with $\mathcal E\subseteq\mathcal F$ we have $f_{\mathcal F\setminus\mathcal E}(\omega)=P_\omega(\mathcal F\setminus\mathcal E)=P_\omega(\mathcal F)-P_\omega(\mathcal E)=f_{\mathcal F}(\omega)-f_{\mathcal E}(\omega)$, i.e. $f_{\mathcal F\setminus\mathcal E}=f_{\mathcal F}-f_{\mathcal E}$, so clearly $f_{\mathcal F\setminus\mathcal E}$ is measurable and thus $\mathcal F\setminus\mathcal E\in\mathcal D$. Similarly, for $\mathcal E_i\in\mathcal D$, $i\in\mathbb Z_{>0}$, pairwise disjoint we have $f_{\bigcup_i\mathcal E_i}=\sum_if_{\mathcal E_i}$, which is thereby measurable because sums and pointwise limits are (recall that $0\le f_{\mathcal G}\le 1$), hence we have $\bigcup_i\mathcal E_i\in\mathcal D$. This shows that $\mathcal D$ is a $\lambda$-system and we conclude that $\mathcal A=\sigma(\mathcal T)=\mathcal D$, since $\mathcal T$ generates $\mathcal A$ by definition. This means that $f_{\mathcal E}$ is measurable for all $\mathcal E\in\mathcal A$, which completes the proof that the projective limit $\kappa$ is a Markov kernel.

Concluding Remarks: Clearly, if we want to define the $P_n$ via kernels, we are free to do so, meaning that we obtain both the Ionescu-Tulcea Theorem 14.32 and Theorem 14.35 as special cases (using that any consistent family $P'_J$ is determined by $P'_{[n]}$ as discussed above) for $\mathcal A_0$ being trivial. Another important special case is the countable product of kernels $\kappa_i:\Omega\times\mathcal A_i\rightarrow[0,1]$, i.e. the conditional product measure.

Matija
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