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In my probability notes, I have the following definition and theorem:

Let $(\Omega_1,\mathcal{A}_1,P_1)$ be a probability space and $(\Omega_2,\mathcal{A}_2)$ a measure space.

Definition: We say that $p:\Omega_1\times\mathcal{A}_2\rightarrow [0,1]$ is a transition probability if: $$ \forall\,\omega_1\in\Omega_1\;\;p(\omega_1,\cdot):\mathcal{A}_2\rightarrow [0,1]\text{ is a probability,} $$ $$ \forall\,A\in\mathcal{A}_2\;\;p(\cdot,A):\Omega_1\rightarrow [0,1]\text{ is measurable.} $$

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Theorem: Given the above notations, there exists a unique probability $Q$ in $(\Omega_1\times\Omega_2,\mathcal{A}_1\otimes \mathcal{A}_2)$ such that $$Q(A_1\times A_2)=\int_{A_{1}}p(\omega_1,A_2)\,P_1(d\omega_1),\;\forall\,A_1\in\mathcal{A}_1,\,A_2\in\mathcal{A}_2.$$ Moreover, for any measurable function $X:\Omega_1\times\Omega_2\rightarrow\mathbb{R}$ nonnegative or $Q$-integrable one has $$\int_{\Omega_1\times\Omega_2}X\,dQ=\int_{\Omega_1}\int_{\Omega_2}X(\omega_1,\omega_2)\,p(\omega_1,d\omega_2)\,P_1(d\omega_1).$$

I've always heard that Fubini "says" that the double integral is equal to the iterated integrals. Is the theorem that I wrote a generalization of Fubini's theorem (the particular case of $p(\omega_1,A)=p(A)=:P_2(A)$ and $Q(A_1\times A_2)=P_1(A_1)P_2(A_2)$)?

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Yes. Replacing $p(\cdot, A):= P_2(A)$ with some measure $P_2$ and forcing $X \geq 0$ reduces your theorem to Tonellis Theorem. Fubinis Theorem extends this result for $Q$-integrable functions.

However I think you forgot to require that $(\Omega_2,\mathcal{A}_2)$ is a $\sigma$-finite measure space (otherwise you can construct counterexamples).

A proof of Tonellis Theorem and Fubinis Theorem can be found in the book Measure and integration theory from Heinz Bauer (Proposition 23.6 and corollary 23.7 ).

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