This question is at its core a confusion regarding the classical substitution rule in integration and the transformation theorem, I feel:
I had the excercise in measure theory to prove the commutativity of the convolution.
I tried to first show it in $\mathbb{R}$ (with $\lambda$ lebesgue measure and $f,g \in \mathscr{L}^{1}(\lambda)$):
$$(f*g)(x) = \int_{\mathbb{R}} f(x-y)g(y) \; d\lambda(y)$$
substitute $\xi = x-y$ , in other words, $d\xi = -dy$ and we arrive at
$$\int_{\mathbb{R}} -f(\xi)g(x-\xi) \; d\lambda(\xi) = -(g*f)(x)$$
(which would mean they don't commute in general)
Using the transformation theorem which should generalize this, however, I should arrive, with the same transformation $y = x-\xi = \phi(\xi)$, at the same result, but since we have $$|det(D\phi)|=1$$ we don't have the issue with the negative sign, and the world is fine again. What is my mistake?
edit： so while reading this again, I thought about my high school math classes and I think I forgot to transform the integration ends like always, since my substitution would also "flip" $\mathbb{R}$? Can someone confirm?