Showing there is a unique invariant measure on the unit circumference 
Exercise:Assume that $\Omega$ is a circumference of radius $1$ and centred at the origin of $\mathbb{R}^2$. Show that there exists a unique measure $\mu$ defined on $\mathscr{B}_{\Omega}$ such that $\mu(\Omega)=1$ and $\mu$ is invariant for all rotations centred at the origin.

I tried to solve the question the following way:
I can define de measure using the Lebesgue measure $\mu(A)=\frac{\lambda(A)}{\lambda(\Omega)}=\frac{\lambda(A)}{2\pi}$ for $A \subseteq\Omega$
If $A_j$ is a disjoint sequence of sets so that $\bigcup_{j\in\mathbb{N}}A_j=\Omega$
If there were two measures $\mu_1$ and $\mu_2$ then since by assumption $\mu_1(\Omega)=1$ and $\mu_2(\Omega)=1$ then  $\mu_2(\bigcup_{j\in\mathbb{N}}A_j)=\sum_{j\in\mathbb{N}}\mu_2(A_j)=\mu_2(\Omega)=1=\mu_1(\Omega)=\mu_1(\bigcup_{j\in\mathbb{N}}A_j)=\sum_{j\in\mathbb{N}}\mu_1(A_j)$
Since the measure is invariant to the rotations I think it does not matter the measure attributed to each single $A_j$, so what it needs to be assured in order to have uniqueness is that both $\mu_1$ and $\mu_2$ attribute the same set of values to the different sets. Otherwise if one was the Dirac measure and the other was not the uniqueness would fail.
Question:
How should I solve the problem?
Thanks in advance!
 A: $\Omega=\{(\cos \theta, \sin \theta): 0\leq \theta<2\pi\}$, and $\mathcal{A}=\{A(B):=\{(\cos \theta, \sin \theta):\theta\in B\}:B\in\mathcal{B}_{[0,2\pi)}\}$. For any $2\times 2$ orthogonal matrix $P_\alpha=\left[\begin{array}{cc}\cos \alpha & \sin \alpha\\-\sin \alpha&\cos\alpha\end{array}\right]$ we have $\mu \circ P_\alpha=\mu.$ Now $\mu(A([0,2\pi)))=1$ and hence $1=\mu(A([0,2\pi)))=\sum_{i=1}^n \mu(A([2\pi\frac{i-1}{n},2\pi\frac{i}{n})))=\sum_{i=1}^n \mu \circ P_{-2\pi\frac{i-1}{n}} (A([2\pi\frac{i-1}{n},2\pi\frac{i}{n})))=n \mu(A([0,\frac{2\pi}{n})))\Rightarrow \mu(A([0,\frac{2\pi}{n})))=1/n.$ 
Now for any interval $B\subset [0,2\pi)$ of length $2\pi/n$, we have $\mu(A(B))=1/n$ and  for any interval $B\subset [0,2\pi)$ of length $2\pi m/n$ with $m/n< 1$, we have $\mu(A(B))=m/n$. Hence, by continuty of measure for any $0<x<1$  any interval $B\subset [0,2\pi)$ of length $2\pi x$ with $x\leq 1$, we have $\mu(A(B))=x.$ As $\mathcal{C}=\{A(B):B\subset [0,2\pi)\text{ is an interval of length }2\pi x\text{ with }0\leq x< 1\}$ is a field which generates $\mathcal{A}.$ By extension of measure we get $\mu(A(B))=\frac{Leb(B)}{2\pi}.$
