How to define a product space probability using a mixture of probabilities?

Question

Let $(\Omega, \mathcal{F})$ be a measurable space, and let $P, P_1, P_2,...$ be probability measures on this space with $P = \sum_{n=1}^\infty a_n P_n$, where $a_n > 0$ and $\sum_{n=1}^\infty a_n = 1$.

I want to define a probability measure on the product space $\Omega \times \mathbb{N}$.

If $\Omega$ is countable, I can easily verify that $Q$ defined by $$Q(\omega, n) = a_nP_n(\omega)$$ is a probability on $\Omega \times \mathbb{N}$.

But now suppose $\Omega$ is not countable. As usual, we equip $\Omega \times \mathbb{N}$ with the sigma-algebra $\mathcal{F} \times \mathscr{P}(\mathbb{N})$ generated by the semi-algebra $\mathcal{S} = \{F \times N: F \in \mathcal{F}, N \in \mathscr{P}(\mathbb{N}) \}$ of measurable rectangles. I want to proceed as in the countable case and define $Q$ on $\mathcal{S}$ by $$Q(F \times N) = \sum_{n \in N}a_nP_n(F).$$ Now, in order to uniquely extend $Q$ to $\mathcal{F} \times \mathscr{P}(\mathbb{N})$, I just need to show that $Q$ is countably additive on $\mathcal{S}$ (or finitely additive and countably sub additive, but I was hoping I could just show countable additivity.)

Let $F \times N \in \mathcal{S}$ be a countable disjoint union $\cup_{j \in J} F_j \times N_j$ of measurable rectangles. We need to show that $$Q(F \times N) = \sum_{j \in J}Q(F_j \times N_j).$$

I have convinced myself that we can reduce to the case where each $N_j$ is a singleton $\{ n_j\}$. But now I find myself stuck. I've tried playing around with the indices, but I can't seem to make progress.

Question. Can you please give me a hint or a suggestion for how to proceed? My feeling is that this should be very easy and I'm missing something fairly obvious.

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Somehow I think we can assume without loss of generality that $\cup_{j \in J}F_j \times N_j$ is of the form $\cup_{n \in N}F \times \{n\}$. If that's so, countable additivity follows by two applications of the definition of $Q$: $$Q(F \times N) = \sum_{n \in N}a_nP_n(F) = \sum_{n \in N}Q(F \times \{n\}).$$

Here's my idea for reducing the problem. First, as above, I think I can show that it's sufficient to assume that $N_j$ in $\cup_{j \in J}F_j \times N_j$ is a singleton $\{n_j\}$, with $n_j \in N$. Next, we can assume that each $n_j$ is unique. If not, and $n_j = n_k = n$, then we write $F_n = F_j \cup F_k$ where the union is disjoint. So we reduce to the case where $\cup_{j \in J}F_j \times N_j$ is of the form $\cup_{n \in N}F_n \times \{n\}$.

Finally, since by assumption $F \times N \in \mathcal{S}$, we have $F \times N = \cup_{n \in N} F \times \{n\}$. So by the previous paragraph $$\cup_{n \in N}F \times \{n\} = \cup_{n \in N} F_n \times \{n\},$$ which implies that $F_n = F$ for all $n \in N$ and completes the reduction.

Does this work?

Answer 1

There is no need to use something as strong as Caratheodory to prove the existence of such a $Q$.

For each $n \in \Bbb N$ there is a map $\gamma_n: \Omega \to \Omega \times \Bbb N$ given by $\omega \mapsto (\omega, n)$. This map is measurable since each component is measurable.

For each $n$ we define a measure $Q_n := \gamma_n^* P_n$. (This is a probability measure on $\Omega \times \Bbb N$ which is the pushforward of $P_n$ under the map $\gamma_n$. More precisely $Q_n(B) = P_n(\gamma_n^{-1}(B))$, for all $B \in \mathcal F \otimes \mathcal P(\Bbb N)$.)

Then define $Q = \sum_n a_n Q_n$ which is a probability measure on $\Omega \times \Bbb N$ since it is a convex combination of probability measures. And we have $Q(A \times F) = \sum_{n\in F} a_n P_n(A)$, since $\gamma_n^{-1}(A \times F) = A$ for all $n \in F$, and $\emptyset$ otherwise.