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My question concerns an application of the standard integration by substitution tecnique. I am aware of the following result.

Let $V \subseteq \mathbb{R}^d$ be an open set and $\varphi \colon V \mapsto \mathbb{R}^d$ be a one-to-one $\mathcal{C}^1(V)$ mapping with non-vanishing Jacobian $J_\varphi$. Then it holds $$ \int_{\varphi (V)} h d\lambda_d = \int_V (h \circ \varphi) \lvert J_\varphi \rvert d\lambda_d $$ for a continuous function $h$.

My question is the follwing. Can we apply a similar technique, if $\varphi$ is a linear injective function of the form $\varphi \colon \mathbb{R}^n \rightarrow \mathbb{R}^m$, with $m \neq n$? Note that in this case, the Jacobian $J_\varphi$ is simply its corresponding matrix, which is non-square.

Fq00
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  • In principle you can, but you have to be careful with the meaasures. The measure on $\varphi(V)$ must be an appropriate $n$-dimensional measure. Since you assume $\varphi$ is injective and linear, it follows that $m>n$. In that case, the $m$-dimensional measure vanishes on any $n$-dimensional subset. – uniquesolution Apr 15 '19 at 08:30
  • Take a case of $m=3$, $n=2$, on the left you basically want to integrate something over a plane in $\mathbb{R}^3$. 3D Lebesgue measure will not give you anything non-trivial. There are ways to induce a measure on the plane from a measure in $\mathbb{R}^3$, but they are more involved than mere change of variables with a Jacobian factor, see A rigorous meaning of “induced measure”? – Conifold Apr 15 '19 at 08:40
  • Thank you for your reply. What can we say about the case of $\varphi$ not being injective? – Fq00 Apr 15 '19 at 08:57

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