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**Opalg** This result is essentially the fact that irrational rotations are ergodic. The best way to prove it is by using Fourier series.

Let f be the characteristic function of E, $\displaystyle f(t) = \begin{cases}1&(t\in E),\\ 0&(t\notin E).\end{cases}$

Extend the domain of f from [0,1] to $\displaystyle \mathbb{R}$ by making it periodic with period 1. The Fourier coefficients of f are given by $\displaystyle \hat{f}(n) = \int_0^1\!\!\!f(t)e^{-2\pi int}dt$. The fact that E is invariant under a translation through *a* tells you that $\displaystyle e^{2\pi ina}\hat{f}(n)=\hat{f}(n)$ for all n. But *a* is irrational, so $\displaystyle e^{2\pi ina}$ can never be equal to 1 unless n=0. Therefore $\displaystyle \hat{f}(n)=0$ for all $\displaystyle n\ne0$, which tells you that f is (almost everywhere equal to) a constant function. Since $\displaystyle f^2 = f,$ the constant can only be 0 or 1. Thus m(E) must be 0 or 1.