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!
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