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I'm trying to understand a proof in Achim Klenke's textbook Probability Theory: A Comprehensive Course (Springer, 2008). The proof in question is the one for Theorem 14.42, "Kernel via a consistent family of kernels" (pp. 289-290).

The proof proceeds in two steps: (1) Show existence of $\kappa$ using Kolmogorov's extension theorem, (2) Show that $\kappa$ is a probability transition kernel. I'm concerned with the first step.

In order to use Kolmogorov's extension theorem, a family of finite dimensional distributions $\left(P_J : J\subset I \space \mathrm{finite},\space 0\in J\right)$ is defined and shown to be consistent.

According to Klenke, Kolmogorov's extension theorem then yields a probability measure from $\left(E,\mathcal B\left(E\right)\right)$ to $\left(E^I,\mathcal B\left(E\right)^{\otimes I}\right)$ . However in my opinion, due to the condition $0\in J$ Kolmogorov's theorem yields a probability measure from $\left(E,\mathcal B\left(E\right)\right)$ to $\left(E^H,\mathcal B\left(E\right)^{\otimes H}\right)$ where $H = I\setminus\left\{0\right\}$.

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You are right. A strict application of Kolmogorov's extension theorem (14.36) needs a consistent family of probability measures $P_J$ (on $E^J$) indexed by all finite subsets $J$ of $I$.

However, Kolmogorov's theorem really only needs $P_J$ for all large subsets of $I$. To be precise, suppose that $\cal F$ is a collection of finite subsets of $I$ with the property that every finite subset $K$ is contained in some $J\in{\cal F}$. Also, suppose $(P_J, J\in{\cal F})$ is a consistent family of probability measures.

For any finite subset $K$ in $I$ we define: $P_K=P_J\circ (X^J_K)^{-1},\tag1$ where $J\in {\cal F}$ is any set that contains $K$.

We now have a family of probability measures $P_K$ (on $E^K$) defined for all finite subsets $K$ of $I$.
Its nice properties follow from a basic fact about projections: If $L\subseteq K\subseteq J$, then $X^J_L=X^K_L\circ X^J_K.\tag2$

  1. The measure $P_K$ is well-defined, that is, it doesn't matter which set $J$ is used, provided it contains $K$. To see this, suppose that $J_1,J_2\in{\cal F}$ with $K\subseteq J_1$ and $K\subseteq J_2$. Let $J\in {\cal F}$ so that $J_1\cup J_2\subseteq J$. Then, by the consistency of the original family and equation (2) we have \begin{eqnarray*} P_{J_1}\circ(X^{J_1}_K)^{-1} &=&P_{J}\circ(X^{J}_{J_1})^{-1}\circ(X^{J_1}_K)^{-1}\\ &=&P_{J}\circ(X^J_K)^{-1}\\ &=&P_{J}\circ(X^{J}_{J_2})^{-1}\circ(X^{J_2}_K)^{-1}\\ &=&P_{J_2}\circ(X^{J_2}_K)^{-1}. \end{eqnarray*}

  2. By consistency of the original family, we see that if $K\in {\cal F}$ then
    the new definition in (1) doesn't change anything.

  3. The extended family is also consistent. Suppose $L\subseteq K\subseteq J$ with $J\in{\cal F}$. Then
    \begin{eqnarray*} P_L&=&P_{J}\circ(X^{J}_L)^{-1}\\ &=&P_{J}\circ(X^J_K)^{-1}\circ(X^K_L)^{-1}\\ &=&P_K\circ(X^K_L)^{-1}. \end{eqnarray*}


Let's take a closer look at Klenke's example. He shows that, for fixed $x\in E$, the right hand side of (14.15) defines a consistent family of probability measures indexed by $\cal F=$ "all finite subsets that include $0$". As in (1) above, we may extend $(P_J, J\in{\cal F})$ to a family indexed by all finite subsets. For Klenke's construction, if $J=\{j_0,j_1,\dots,j_n\}$ with $0, this gives
$P_J=\delta_x\kappa_{\small{0},j_0}\otimes\bigotimes_{k=0}^{n-1} \kappa_{j_k,j_{k+1}}.\tag3$ Now apply Kolmogorov's extension theorem to the whole family to get the required measure $P$.


Added: This is just a long-winded version of did's nice, succinct answer.

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The family $H_0=\left(P_J : J\subset I \space \mathrm{finite},\space 0\in J\right)$ uniquely defines a family $H=\left(P_J : J\subset I \space \mathrm{finite}\right)$: either $0\in J$ and nothing needs to be done, or $0\notin J$, then project $P_{J\cup\{0\}}$ upon $J$ to get $P_J$. Furthermore, a moment of thought reveals that if $H_0$ is consistent then $H$ is. Hence, Kolmogorov's extension theorem indeed yields a probability measure from $\left(E,\mathcal B\left(E\right)\right)$ to $(E^I,\mathcal B(E)^{\otimes I})$.