Let $G$ be a compact group with Haar measure $m$, then $m$ is left-invariant, in the sense that $\int_{G} f(x) \ dm(x) = \int_{G} f(s^{-1}x) \ dm(x)$ for all $s\in G$ and for all $f\in C(G)$, and $m$ is also right-invariant.
Why does $m$ satisfies the relation $\int_{G} f(x) \ dm(x) = \int_{G} f(x^{-1}) \ dm(x)$ for all $f\in C(G)$?
My idea was to show that $f\mapsto m(\tilde{f})$ with $\tilde{f}(x) = f(x^{-1})$ is also a haar measure on $G$ and then the equality would follow from the uniqueness of the haar measure. But I don't know how to show the left and right invariance.
So, I think I figured it out. Let $f\in C(G)$ and $\tilde{f}(x) = f(x^{-1})$ for $x\in G$. Then $\tilde{f}\in C(G)$ and we get $\int_{G} (L_{s}\tilde{f})(x) dm(x) = \int_{G} \tilde{f}(sx) dm(x) = \int_{G} \tilde{f}(x) dm(x) = \int_{G} f(x^{-1}) dm(x)$, where I used the Left-Invariance of $m$ in the second equality. Then $f \mapsto m(\tilde{f})$ is also a Haar Measure and the uniqueness gives $m(\tilde{f}) = m(f)$. Does anybody see any mistakes?