The integral of a character is $0$

Question

Let $G$ be an additive compact topological abelian group group and fix a Haar measure $\mu$ on it. Moreover let $\psi\in \hat G$ a nontrivial character, this means that $\psi:G\to S^1\subset \mathbb C^\ast$ is a continuous homomorphism.

Why do we have that $$\int_G \psi d\mu=0\;?$$

Can you give at least a hint for the proof?

Answer 1

Use the translation invariance after a change of variables. Let $h\in G$ so that $\psi(h^{-1})\ne 1$ (which exists by non-triviality)

$$\int_G\psi(g) d\mu(g) = \int_G\psi(h^{-1}g) d\mu(h^{-1}g)$$ $$=\psi(h^{-1})\int_G\psi(g)d\mu(g)$$

so the integral must be zero.