Faddeev-Popov Haar measure invariance for $U(1)$

Question

I am studying the Faddeev-Popov procedure for quantizing Gauge fields. I am stuck in the step where it says that the measure is gauge invariant for the $U(1)$ case.

I came across this question on stackexchange: How to apply the Faddeev-Popov method to a simple integral

Here OP says in the question that ${\cal D}\omega\omega' = {\cal D}\omega$, for fixed $\omega'$, which follows from the product rule, but I don't see how. I figured:

$D\omega\omega' = \omega'{\cal D}\omega + \omega {\cal D}\omega'$, where the second term goes to zero as $\omega'$ is just a fixed gauge transformation. But then ${\cal D}\omega\omega' = \omega'{\cal D}\omega$.

So, what am I missing here?

Answer 1

${\cal D}$ is not a differentiation; ${\cal D}\omega=\prod_x d\omega(x)$ is a path integral measure, and the right invariance of the Haar measure is being used.