From two measurable spaces $(\Omega_1, \mathcal A, \mu)$ and $(\Omega_2, \mathcal B, \nu)$, we can define another measurable space denoted $(\Omega_1\times \Omega_2, \mathcal A\otimes \mathcal B, \mu\otimes \nu)$ in wich we can establish double integrals and related theorems (as Fubini-Tonelli & Fubini-Lebesgue).
My question is : how to extend "product measure space" for $n$-dimensional spaces $\mathbb R^n$ defined as $\mathbb R^{\{1,..., n\}}$ ? Does an isomorphism preserve "measurability" of subsets ?
For example, suppose that $\mathbb R^2\cong \mathbb R\times \mathbb R$ through cannonical injection $\phi : (x_i)_{i\in \{1,2\}} \mapsto (x_1,x_2)$.
If these sets are equiped of the associated borelian $\sigma$-algebra, I think that $\phi$ is measurable... (sorry if I'm wrong, because I'm currently overviewing my courses on Measure and Integration).
If $\phi$ is actually measurable, what the measures on $\mathbb R^2$ and $\mathbb R \times \mathbb R$ have to satisfy to get :
$$ ``\int_{\mathbb R^2} f\,\mathrm d\mu_{\mathbb R^2} = \int_{\mathbb R \times \mathbb R} f\circ \phi^{-1} \;\mathrm d\mu_{\mathbb R \times \mathbb R} " $$
I don't know if that really makes sense :/ But... I don't agree the following definition of $\mathbb R^n$ as $\mathbb R \times \mathbb R \times \cdots\times \mathbb R$ ($n$ times), because set theory does not allow to write this... moreover I am not fond of "recursive definition" of $\mathbb R^n$ (like that : $\mathbb R^n = \mathbb R^{n-1} \times \mathbb R$), and consequently, of product measure space on $n$-dimensionals :(
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