Normalized Haar measure on a compact group

Remarks

It exists and is unique. A Haar measure $\nu$ is finite on compact sets, so $0 < \nu(G) < \infty$, where $\nu(G) > 0$ because $G$ is a nonempty open set. Then $\mu = \nu / \nu(G)$ has total mass $1$. Uniqueness follows from the existence and uniqueness of Haar measure: two Haar measures differ by a constant $c > 0$, and equal total mass forces $c = 1$.

The normalized Haar measure is also right-invariant and invariant under $x \mapsto x^{-1}$, because compact groups are unimodular. One often writes $\int_G f(x) \, dx$ for the integral against it.

Example: for a finite group $G$ of order $n$ with the discrete topology, $\mu(A) = \abs{A}/n$. It is left-invariant, since $\abs{gA} = \abs{A}$, and $\mu(G) = 1$.

Non-example: on $(\R, +)$ every Haar measure is $c\lambda$ with $c > 0$ and $\lambda$ Lebesgue measure, so it has infinite total mass. No Haar measure of $\R$ can be normalized; the hypothesis that $G$ is compact fails. A locally compact group has a finite Haar measure exactly when it is compact.