Constructing a point $w_0$ in an arbitrary product space whose projection onto $J_k$ coordinates is $w_{J_k}$ for all $k\in N$.

Question

This is part of the proof in the construction of arbitrary product space of probability spaces given in Bauer's Probability Theory.

Let $J_k$ be a nonempty finite index subset of an arbitrary index set $I$, and $J_k \subset J_{k+1}$ for all $k\in \mathbb{N}$. Let $p_{J_k}^{J_{k+1}}$ be the projection of a point in the product space of $J_{k+1}$ sets onto $J_k$ sets. And for each $w_{J_k}\in \Omega_{J_k}$ we have $p_{J_k}^{J_{k+1}}(w_{J_{k+1}})=w_{J_k}.$ Then this property ensures that there exists an $w_0\in \Omega= \otimes_{i \in I} \Omega_i$ with $p_{J_k} (w_0)=w_{J_k}$ for every $k \in \mathbb{N}$.

How can we get this $w_0$ from $p_{J_k}^{J_{k+1}}(w_{J_{k+1}})=w_{J_k}$? I can use $p_{J_k}=p_{J_k}^{J_{k+1}} \circ p_{J_{k+1}}$, to inductively come up with $p_{J_k} (w_k, w')=w_{J_k}$ where $w'=(w_{k+1}, w'')$ for some $w'' \in \otimes_{i \in I-{J_{k+1}}}\Omega_i$ but when the index set $I$ is infinite, I don't know how to construct such a $w_0$ which has the desired property for every $k$.

I would greatly appreciate any help. Below is the excerpt from the text.

Answer 1

You are looking for a mapping $$\omega_0: I \rightarrow \bigcup_{i\in I}\Omega_i$$ such that

1) $\omega_0(i) \in \Omega_i$ for every $i \in I$ and

2) $P_{J_k}(\omega_0) = \omega_{J_k}$ for $k=1,2,\dots.$

By the axiom of choice, there exists a mapping satisfying the first condition; let us denote it by $\omega_*$. We construct $\omega_0$ from $\omega_*$ and the $\omega_{J_k}$ as follows:

• If $i \notin \bigcup_{k=1}^\infty J_k$, set $\omega_0(i) = \omega_*(i)$.
• If $i \in \bigcup_{k=1}^\infty J_k$, let $m$ be the smallest $k$ such that $i\in J_k$ and set $\omega_0(i) = \omega_{J_m}(i)$.

Let us now check that the two conditions are satisfied. By construction, 1) is clearly satisfied. To check 2), let $k \in \mathbb N$ and $i \in J_k$ be arbitrary. Since $P_{J_k}(\omega_0)$ is by definition the restriction of $\omega_0$ to $J_k$, we need to show that $\omega_0(i) = \omega_{J_k}(i)$. By definition of $\omega_0$, there exists $m \leq k$ such that $\omega_0(i) = \omega_{J_m}(i)$, so we need to show only that $$ \omega_{J_m}(i) = \omega_{J_k}(i) = p^{J_k}_{J_m} (\omega_{J_k})(i),$$ where the last inequality makes sense because $J_m \subset J_k$ and $p^{J_k}_{J_m}$ is the restriction operator. This is where the equation $p^{J_{k+1}}_{J_k}(\omega_{J_{k+1}}) = \omega_{J_k}$ must be used. Applying this equation $k - m$ times, you get precisely that $p^{J_k}_{J_m} (\omega_{J_k}) = \omega_{J_m}$, which concludes the discussion.