A question from a dumb phycisist. I found this similar question but it is about space filling curves, not bijective maps.
I understand the answer here as: there are always bijective measurable maps between $\mathbb{R}$ and $\mathbb{R}^n$ for any $n \in \mathbb{N}$.
Consider a friendly function
$\displaystyle f\colon \ \mathbb{R}^2\to Z, \ (x_1,x_2)\mapsto z=f(x_1,x_2)$.
Riemann integrals
$\displaystyle I(a_1,a_2,b_1,b_2) = \int_{a_1}^{a_2}dx_1\int_{b_1}^{b_2} dx_2 \ f(x_1,x_2)$
shall exist for the function. Now the bijective measurable map $m$ comes into play.
$\displaystyle m\colon \ \mathbb{R}^2\to \mathbb{R}, \ (x_1,x_2)\mapsto x=m(x_1,x_2)$.
We use it to define another function $\tilde{f}(x) \equiv f(x_1,x_2)$ depending only on 1 variable $x$.
Is it possible, to define an integral $J$ for $\tilde{f}(x)$ that has the same values as the Riemann integral $I$?
I.e.
$\displaystyle J(\Omega) = \int_\Omega \tilde{f}(x)\ d\mu(x) = I(a_1,a_2,b_1,b_2)$
with $\Omega$ beeing the image set, i.e. $[a_1,a_2]\times[b_1,b_2]\to \Omega$.
And if yes, which type of integral is sufficient to do the job?
