Measure of sets on $\mathbb R^{n+m}$ isomorphic to $\mathbb R^n\times \mathbb R^m$ Does the natural isomorphism between $\mathbb R^n \times \mathbb R^m$ and $\mathbb R^{n+m}$ preserve the measure of sets ?
Let $\lambda^p$ be the Lebesgue measure on $\mathbb R^{p}$, does for $\lambda^{n+m}$-measurable set $A\subseteq \mathbb R^{n+m}$ we have
$$ \lambda^{n+m}(A) = \lambda^n\otimes\lambda^m(\iota(A)) $$
where $\iota : \mathbb R^{n+m} \to \mathbb R^n\times \mathbb R^m$ denotes the natural isomorphism ?
Thanks for answers.
 A: I believe the answer is yes. Here's what I'm thinking: Let $E=(a_1,b_1)\times(a_n,b_n)\times(c_1,d_1)\times\dots(c_m,d_m)$ be an open rectangle of $\mathbb{R}^{n+m}$. It is quite obvious that $\iota(E)=E_1\times E_2$, where $E_1=(a_1,b_1)\times\dots\times(a_n,b_n)\subset\mathbb{R}^n$ and $E_2=(c_1,d_1)\times\dots\times(c_m,d_m)\subset\mathbb{R}^m$. 
Remember that the product measure $\mu\times\nu$ for sets of the form $A\times B$ with $A,B$ being measurable (with respect to each measure space) is simply $\mu(A)\nu(B)$. Since $\iota(E)$ is of that form, $\lambda_n\otimes\lambda_m(\iota(E))=\lambda_n(E_1)\lambda_m(E_2)=\lambda_{m+n}(E)$, by definition.
Now for an arbitrary measurable set $A\subset\mathbb{R}^{n+m}$, we have $\lambda_{n+m}(A)=\displaystyle{\inf\{\sum_{j}\lambda_{n+m}(E_j): A\subset\bigcup_{j}E_j}$, where $E_j$ are open rectangles $\}=$ $=\displaystyle{\inf\{\sum_{j}\lambda_n\otimes\lambda_m(\iota(E_j)): A\subset\bigcup_jE_j\}=\inf\{\sum_{j}\lambda_n\otimes\lambda_m(\iota(E_j)): \iota(A)\subset\bigcup_j\iota(E_j)\}=}$ $\lambda_n\otimes\lambda_m(\iota(A))$
