Circle rotation is ergodic for Lebesgue measure exactly at irrational angles

If $\alpha =p\mathrm{/}q$ with integers p and q>=1, the set $A={\bigcup }_{k=0}^{q-1}[k\mathrm{/}q,k\mathrm{/}q+1\mathrm{/}(2q))$ is Borel, has measure $1\mathrm{/}2$, and is permuted by R_alpha. Thus $R_{\alpha }^{-1}A=A$ and the rotation is not ergodic for either measure. This includes integral alpha with q=1.

Now let alpha be irrational and let A be a strictly invariant measurable set. Suppose both A and its complement have positive measure. Choose density-one points a in A and c in its complement, different from the cut point zero. For sufficiently small $0<r<1\mathrm{/}10$, the circle balls I=B(a,r), J=B(c,r) have length 2r and satisfy $\lambda (I\setminus A)<r\mathrm{/}5$ and $\lambda (J\cap A)<r\mathrm{/}5$. The ordinary density theorem applies because these small balls at the chosen points do not cross the cut.

Choose an integer m with $d(R_{\alpha }^{m}a,c)<r\mathrm{/}10$. Strict invariance and invertibility give $R_{\alpha }^{m}A=A$, and the rotated interval I has a portion outside A of measure less than r/5. Its intersection with J has length greater than $2r-r\mathrm{/}10=19r\mathrm{/}10$: both are radius-r arcs whose centers are less than r/10 apart. Subtracting the two exceptional portions, this intersection would contain points in both A and its complement on a set of measure at least $19r\mathrm{/}10-2r\mathrm{/}5=3r\mathrm{/}2>0$, impossible. Thus every strict invariant A has measure zero or one, proving ergodicity. Together with step 1.1 this proves both directions.