# Multiple integrals with Lebesgue theory?

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 :(

## 1 Answer

Your fondness of recursive definition might increase by reading up on the recursion theorem (Google away), which rigorously justifies, in the context of ZF or whichever set theory you prefer, definitions by recursion.

We can define $\mathbb{R}^n$ to be the set of all functions $n\to \mathbb{R}$, where $n=\{0,\dots,n-1\}$. We can also define, recursively, another family of sets by: $$\begin{cases} E_1:=\mathbb{R}\\ E_{n+1}:=E_n\times \mathbb{R}\end{cases}$$ We can show (by induction), that $E_n$ and $\mathbb{R}^n$ are homeomorphic, and so in particular isomorphic as measurable spaces with the Borel $\sigma$-algebra associated to the product topology.

We can also recursively define $n$-dimensional Lebesgue $m_n$ on $E_n$ measure by $m_{n+1}:=m_n\otimes m$, where $m$ is one-dimensional Lebesgue measure. By the remarks, this defines a regular Borel measure on $\mathbb{R}^n$ via the natural isomorphism, which then becomes a measure preserving isomorphism (measurable with measurable inverse).

We can also show $\mathbb{R}^n \times \mathbb{R}^k \cong \mathbb{R}^{n+k}$, which then proves $E_n\times E_k\cong E_{n+k}$ (this can also be showed directly by induction, keeping one index fixed).

If you want a more direct approach, you could read up on the construction of Haar measure on arbritrary locally-compact Hausdorff group, which $\mathbb{R}^n$ certainly is. This is, however, a tad more advanced, and you're likely to run into the Riesz representation theorem (which also allows for a direct construction, look at Rudin's Real and Complex Analysis, chapter 2).