Showing that the integration measure is preserved under gauge transformation in the non-Abelian case

Question

I am trying to show that the integration measure we use in the Fadeev-Popov method of quantisation of non-Abelian gauge theory is invariant under a gauge transformation.

I am using Peskin & Schroeder chapter 16.2. The gauge transformation of the gauge field is given by $$ (A^\alpha)^a_\mu=A^a_\mu+\frac{1}{g}D_\mu\alpha^a $$ which is in the adjoint representation as shown by the transformation. Now the integration measure we use in the functional integral is given by $$ \mathcal{D}A=\prod_x\prod_{a.\mu}dA^a_\mu $$ So when we take the gauge transformed measure we have $$ \mathcal{D}A^\alpha=\prod_x\prod_{a,\mu}d(A^\alpha)^a_\mu=\prod_x\prod_{a,\mu}\left( dA^a_\mu+\frac{1}{g}d(\partial_\mu\alpha^a)+f^{abc}d(A^b_\mu\alpha^c)\right) $$ This looks like a more complicated shift in our integration but I don't quite understand how they leave the measure invariant. The autors mention that this is a shift followed by a rotation of the components of $A_\mu^a$ but how can we see this explicitly?

Some of my (maybe incorrect) reasoning

The second term in the transformed measure is just a shift and since we are integrating over fields $A^a_\mu(x)$ it indeed leaves the measure invariant. It's the third term that I really struggle to make sense of.

Answer 1

What the authors means is this: When you disregard the shift all that's left is $$ (\delta^{ab} + f^{abc}\alpha^c)\mathrm{d}A^b_\mu,$$ which is the infinitesimal version of a linear transformation generated by the matrix $M^{ab} = f^{abc}\alpha^c$. Since the structure constants are anti-symmetric, $M^{ab}$ is, too, and so it is the generator of a rotation.