Am interested in the following proposition:
Proposition. Let $(\Omega_j,\mathcal{A}_j,\mu_j)$ $j=1,\dots,n$ be measure spaces and let $f$ be an extended real-valued $\mathcal{A}_1\otimes \dots\otimes\mathcal{A}_n$ measurable function on $\Omega_1\times\dots\times\Omega_n$. Then for every permutation $j_1,\dots,j_n$ of $1,\dots,n$ we have that $f'(\omega_{j_1},\dots,\omega_{j_n}):=f(\omega_1,\dots,\omega_n)$ is $\mathcal{A}_{j_1}\otimes \dots\otimes\mathcal{A}_{j_n}$ measurable when regarded as a function on $\Omega_{j_1}\times\dots\times\Omega_{j_n}$. Moreover, if $f$ is $\mu_1\otimes\dots\otimes\mu_n$ integrable then $f'$ is $\mu_{j_1} \otimes\dots\otimes\mu_{j_n}$ integrable, and in this case their integrals are equal.
The result seems obvious on grounds of symmetry, but I want to show it rigorously. I believe the result follows from Fubini-Tonelli's theorem, but I think we can show it directly.
My idea is to proceed as follows:
To begin we can fix $n=2$. Set $\Omega:=\Omega_1 \times \Omega_2$, $\mathcal{A}:= \mathcal{A}_1\otimes \mathcal{A}_2$ and $\pi=\mu_1\otimes \mu_2$. Also let $\Omega':=\Omega_2 \times \Omega_1$, $\mathcal{A}':= \mathcal{A}_2\otimes \mathcal{A}_1$ and $\pi'=\mu_2\otimes \mu_1$. I can show that $\mathcal{A}'=\{Q': Q \in \mathcal{A}\}$ so that $\mathcal{A}$ measurability of $f$ implies $\mathcal{A}'$ measurability of $f'$. Also $\pi'(Q')=\pi(Q)$ for $Q\in\mathcal{A}$, and using this I can show that $\pi$ integrability of $f$ implies $\pi'$ integrability of $f'$ for simple functions. Then I consider positive measurable functions as monotone limits of simple functions, and finally mesurable functions by considering positive and negative parts. The case of general $n$ is exactly the same.
Is this correct?
EDIT: We need $\sigma$-finiteness of the measure spaces to have uniqueness of the product measure and to ensure that $\pi(Q)=\pi'(Q')$ for all $Q\in\mathcal{A}_1\otimes \mathcal{A}_2$.