Klenke's proof for "Kernel via a consistent family of kernels" 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 $H = I\setminus\left\{0\right\}$.
 A: 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 $(E^I,\mathcal B(E)^{\otimes I})$.
A: 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 
$$X^J_L=X^K_L\circ X^J_K.\tag2$$ 


*

*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*}

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

*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 $0<j_0<j_1<\cdots<j_n$, 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. 
