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.
