I am confused about the uniqueness of Haar integrals. A Haar integral is a Radon integral determined by the corresponding Haar measure.
Mostly I am interested in the following quite special integral, which shows up in Cauchy's theorem and the residue formula, and which features integration over the boundary of the boundary of the complex disc :
$$ λ f(z) ↦ \int_{S^1} \frac{f(z)}{z} dz $$
If $f$ arises from a holomorphic function on the disc, then $\int_{S^1} f(z)dz = 0$, and yet $\int_{S^1} \frac{f(z)}{z} dz = 0$ remains interesting and satisfies the properties of a Haar integral.
My first thought was "why isn't $\int_{S^1} f(z) z^{-n} dz/z = 0$ a Haar integral as well"?
I don't fully understand the uniqueness of Haar measure in the complex case. I had thought that the measure would be unique, but these units for the complex case are really confusing me.
Is $λ f ↦\int_{S^1} f(z) z^{-n} (dz/z)$ a Haar integral?