For each $n\geq 1$ and each choice $u_1, \cdots, u_n \geq 1$, define nonnegative real numbers $p_n(u_1,\cdots, u_n)$ such that $\sum_{k} p_1(k)=1$ and for $n\geq 1$:
$\sum_k p_{n+1}(u_1,\cdots, u_n, k) = p_n(u_1,\cdots,u_n)$.
Show that there exists a probability measure on $\mathbb{N}^{\mathbb{N}}$, a probability measure $\mu$ such that
$\mu \{ \omega \in \mathbb{N}^{\mathbb{N}}: \pi_i(\omega)=u_i, i=1,\cdots, n \} = p_n(u_1,\cdots, u_n)$, where $\pi_i (\omega)$ represents the $i^{th}$ coordinate of the sequence $\omega$.
My attempt:
I want to apply Kolmogorov's extension theorem. In this case, $p_n$ represent the probability measure and I need to check the consistency conditions with $p_n$.
I have that for $n\geq 1$, $\sum_k p_{n}(u_1,\cdots, u_n, k)=1$. It is possible to define $p_n(\emptyset)=0$ and the missing condition to verify that $p_n$ is a probability (disjoint sets).
Now, in order to use Kolmogorov's extension I need to check
First condition: for all permutation $\sigma$ of $\{1,\cdots, n\}$: \begin{align} p_n(u_{\sigma(1)}, \cdots, u_{\sigma(n)}) &= \sum_k p_{n+1}(u_{\sigma(1)}, \cdots, u_{\sigma(n)},k) \\ =& \sum_{(u_{\sigma(1)}, \cdots, u_{\sigma(n)})\in \mathbb{N}^n} \sum_k p_{n}(u_{\sigma(1)}, \cdots, u_{\sigma(n)}) \\ &= \sum_{(u_{1}, \cdots, u_{n})\in \mathbb{N}^n} \sum_k p_{n}(u_{1}, \cdots, u_{n}) \\ &= p_n(u_1,\cdots, u_n) \end{align}
Second condition: for a set $A=\{u_1,\cdots, u_n\}$ I need to verify that: $p_{n}(u_1,\cdots, u_n) = p_{n+m}(A\times \mathbb{N} \times \mathbb{N} \times \cdots \times \mathbb{N} )$
Is the second condition true? Is correct to apply the theorem for this problem? Any help with the second condition is well received.
Thanks.
You may use the Ionescu-Tulcea theorem instead. For each $n\ge 1$, define $$ \kappa_{n}((u_1,\ldots,u_n),A):=\sum_{k\in A}\frac{p_{n+1}(u_1,\ldots,u_n,k)}{p_n(u_1,\ldots,u_n)}\times 1\{p_n(u_1,\ldots,u_n)>0\},\quad A\subseteq \mathbb{N}. $$ Also let $P_0(A):=\sum_{k\in A}p_1(k)$. Then each $\kappa_{n}$ is a Markov kernel and $P_0$ is the initial distribution, and so there exists a probability measure $\mu$ satisfying the required condition.