When and why is it true that $ \int_{\mathbb{R}^m } f(\pi(x)) dx = \int_{\mathbb{R}^m } f(u_{\leq k}) du $

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Let $k < m$ and consider an arbitrary projection $\pi: \mathbb{R}^m \rightarrow \mathbb{R}^k$ with $$\pi(x_1,\ldots,x_m)=(x_{i_1},\ldots,x_{i_k}),$$ where $1 \leq i_1 < \ldots < i_k \leq m$ and the specific projection $\pi_1 \colon \mathbb{R}^m \rightarrow \mathbb{R}^k$ with $$\pi_1(x_1,\ldots,x_m)=(x_1,\ldots,x_k).$$ Furthermore let $f: \mathbb{R}^k \rightarrow \mathbb{R}$ denote an arbitrary integrable function.

If possible, I want to do some change of variables such that

$$ \int_{\mathbb{R}^m } f(\pi(x)) dx = \int_{\mathbb{R}^m } f(\pi_1(u)) du, $$ or in order words apply a permutation (and possibly apply Fubini's theorem as well to change order of integration) so that the integrand becomes $f$ applied to the $k$ first entries of an $m$-dimensional vector. Specifically it should be that $u=(u_1, \ldots u_m)=(x_{i_1}, \ldots,x_{i_k},x_{i_{k+1}},\ldots,x_{i_m})$.

My questions are 1) if it is always possible to perform this change of variables and 2) if so how to justify it using for instance measure theory or some other transformation theorems?

Edit: It would for instance suffice that $u=\sigma(x)$ satisfies that the permutation $\sigma$ is continuous differentiable and the determinant of the Jacobian is $1$ or $-1$. Although I believe $\sigma$ to be continuous, I am however not sure it is differentiable.