More Lebesgue Measure Theory fun

Let E be a measurable subset of [0,1]. Let f:[0,1] --> [0,1] be defined as:

f(x) = (x+a)(mod 1) where a is irrational

Suppose that f(E) = E

Show that either E or its compliment has measure 0.

My answer so far:

Let $\displaystyle O_x = $ = {$\displaystyle f^z(x)$; for all integers z} be defined as the "orbit" of x. Where $\displaystyle f^z(x)$ means f(f(f(f...(f(x)) n times.

It is not hard to see then that both E and $\displaystyle E^c $ are made of disjoint orbits.

If E or its compliment contains countable many orbits, then the entire set is countable and thus has measure 0.

Suppose m(E) > 0 and E contains uncountably many orbits.

Let S = {one point from each unique orbit}, then S is not measurable. So $\displaystyle S^c $ is not measurable. But $\displaystyle E^c \subset S^c $, so does it follow that m($\displaystyle E^c$) = 0?