Haar measure on the circle

Example

Let $\mathbb{T} = \set{z \in \C : \abs{z} = 1}$ be the circle group with complex multiplication. Let $p \colon [0, 2\pi) \to \mathbb{T}$, $p(\theta) = e^{i\theta}$, which is a bijection. For a Borel set $A \subseteq \mathbb{T}$ put $$\mu(A) = \frac{1}{2\pi} \lambda\left(p^{-1}(A)\right),$$ where $\lambda$ is Lebesgue measure. In integral form, $\int_{\mathbb{T}} f \, d\mu = \frac{1}{2\pi} \int_0^{2\pi} f(e^{i\theta}) \, d\theta$.

Check that $\mu$ is a Haar measure. It is a finite Borel measure on a compact metric space, hence a Radon measure, and $\mu(\mathbb{T}) = 1$, so it is nonzero. For $g = e^{i\alpha}$, multiplication by $g$ corresponds to $\theta \mapsto \theta + \alpha \bmod 2\pi$ on $[0, 2\pi)$. This map cuts $[0, 2\pi)$ into two intervals and translates each of them, so it preserves Lebesgue measure. Hence $\mu(gA) = \mu(A)$ for every $g \in \mathbb{T}$ and every Borel set $A$.

Since $\mu(\mathbb{T}) = 1$, it is the normalized Haar measure of $\mathbb{T}$. Every other Haar measure of $\mathbb{T}$ is $c\mu$ for some $c > 0$, for instance arc length $2\pi\mu$. An arc of angle $\beta$ with $0 \le \beta \le 2\pi$ has measure $\beta / (2\pi)$.