invariant measure under irrational rotation on $S^1$ 
Prove that if $T:S^1 \to S^1$ is an irrational rotation, then the only  probability measure on $S^1$ that is $T-$invariant is the lebesgue measure or a multiple or it. We are considering the lebesgue measure times $\frac{1}{\pi}$ (in this way it's a probability measure) and we are considering the $\sigma-$algebra of the borel sets restricted to $S^1$. 

$S^1=\{u\in \mathbb R^2:||u||=1\}$. If $T$ is said to be an irrational rotation, if there exist an irrational number $\alpha\in \mathbb R-\mathbb Q$ such that if $u=(cos t,sin t)$ then $Tu=(cos (t+\alpha),sin ((t+\alpha)))$.
I have no idea how to attack this problem :S. Clearly Lebesgue measure is invariant under rotations, but I don't know how to proceed.
EDITED: Now all the assumptions are right. Thanks!
 A: Notice that the Lebesgue measure works. So, our problem sums up to proving it is unique.
If you prove that any such measure $\mu$ must coincide with the Lebesgue measure $\lambda$ on the "intervals", for instance, then the uniqueness follows from Carathéodory's extension theorem.
First, let us show that such a measure must be invariant by EVERY rotation. Let $\alpha \in (0,1]$. Take $r_k \in T^n(0)$ with $r_k \uparrow \alpha$ ($r_k$ increases and converges to $\alpha$). Now, notice that for an "interval" $I = \frac{[a,b)}{\mathbb{Z}} \subset S^1$,
$$
  \alpha + I
  =
  \liminf (r_k + I)
  =
  \limsup (r_k + I).
$$
Since the probability measures are continuous...
$$
  \mu(\alpha + I)
  =
  \lim \mu(r_k + I)
  =
  \lim \mu(I)
  =
  \mu(I).
$$
That implies that $\mu$ is invariant by any rotation, because an interval is a countable union of intervals like the above. Alternatively, we could have used the fact that the measure of any singleton is null. (why?)
Now, take an interval of the shape $I = \left[a,a+\frac{1}{n}\right)$. Since its $\mu$ measure is invariant under any rotation, the sets $I + \frac{m}{n}$ partition $S^1$ and have all the same measure $\mu$. Therefore, $\mu(I) = \frac{1}{n} \lambda(I)$. Since any interval is a countable uninon of such intervals, $\mu$ and $\lambda$ coincide over the intervals. QED
A: this is equivalent to show that
If $\int_{S^1}z^nd{\mu}=0,\forall n\neq 0,\int_{S^1}d{\mu}=1$, then $\mu $ is lebesgue measure. And this statement is easy to get.    
A: Let $T_{\alpha}(z) = \exp(2\pi i\alpha)$ be where $\alpha \notin\mathbb{Q}$ and $[v,w)\subset S^1$, with $v = exp(2\pi ip)$ y $w = \exp(2\pi iq)$, a semi-opened interval in $S^1$ that has taken counterclockwise such that $l([v,w)) = q-p = \frac{1}{m}$ for some $m \in\mathbb{N}$.
We take $\mu\in M_{T_{\alpha}}(X)$, by regularity, given $\epsilon > 0$ there exists $F\subset(v,w)$ closed subset and $U\supset[v,w]$ opened subset such that $\mu((v,w)\backslash F) < \epsilon$ and $\mu(U\backslash [v,w]) < \epsilon$. Since $v, w\in U\backslash F$ there exists $\epsilon > \delta > 0$ such that $[v-\delta,v+\delta]\cup[w-\delta,w+\delta]\subset U\backslash F$ and we take $u\in [v-\delta,v)$. We know that the orbit of $u$ is dense in $S^1$, then there exist $n_1,n_2,n_3\in\mathbb{N}$ with $n_1<n_2<n_3$ such that $T_{\alpha}^{n_1}u\in(v,v+\delta], T_{\alpha}^{n_2}u\in[w-\delta,w)$ and $T_{\alpha}^{n_3}u\in(w,w+\delta]$. Then we have
        $$\displaystyle\bigcup_{k=0}^{m-1}[T_{\alpha}^{n_1+k(n_3-n_1)}u,T_{\alpha}^{n_2+k(n_3-n_1)}u]\subset \bigcup_{k=0}^{m-1}[v.e^{2\pi ik(q-p)},v.e^{2\pi i(k+1)(q-p)})\subset\bigcup_{k=0}^{m-1}[T_{\alpha}^{kn_2}u,T_{\alpha}^{n_3+kn_2}u]$$
        Notice that the family $\{[v.e^{2\pi ik(q-p)},v.e^{2\pi i(k+1)(q-p)})\}_{0\leq k\leq m-1}$ form a partition of $S^1$. Thus we have
        $$\displaystyle\sum_{k=0}^{m-1}\mu([T_{\alpha}^{n_1+k(n_3-n_1)}u,T_{\alpha}^{n_2+k(n_3-n_1)}u])\leq 1\leq\sum_{k=0}^{m-1}\mu([T_{\alpha}^{kn_2}u,T_{\alpha}^{n_3+kn_2}u])$$
        Since $\mu$ is $T_{\alpha}-$invariant it follows that
        $$m\mu([T_{\alpha}^{n_1}u,T_{\alpha}^{n_2}u]) \leq 1\leq m\mu([u,T_{\alpha}^{n_3}u])$$
        Notice that $F\subset[T_{\alpha}^{n_1}u,T_{\alpha}^{n_2}u]$ and $[u,T_{\alpha}^{n_3}u]\subset U$. Then from the regularity condition it follows that
        $$m\mu([v,w))-m\epsilon < m\mu(U)-m\epsilon < m\mu(F)\leq1\leq m\mu(U) < m\mu(F) + m\epsilon < m\mu([v,w))+m\epsilon$$ 
        So
        $$\dfrac{1}{m} - \epsilon < \mu([v,w)) < \dfrac{1}{m} + \epsilon$$
        From this $\mu([v,w)) = \dfrac{1}{m}$.
        When $l([v,w)) = \dfrac{s}{m}$ for some $s,m \in\mathbb{N}$ we divide the interval in $s$ semi-opened intervals of equal length and we conclude by the above that $\mu([v,w)) = \frac{s}{m}$.
        To the last, if $[v,w)$ is any semi-opened interval, we can write it as embedded union of semi-opened intervals with rational length. By continuity we have $\mu([v,w)) = l([v,w))$.
Finaly by Caratheodory theorem it follows the $\mu$ must be the Haar measure.
If we have a finite measure, we use the probability measure obtained from divide this measure by the measure of the total space.
