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Math Help - More Lebesgue Measure Theory fun

  1. #1
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    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  O_x = = {  f^z(x); for all integers z} be defined as the "orbit" of x. Where  f^z(x) means f(f(f(f...(f(x)) n times.

    It is not hard to see then that both E and  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  S^c is not measurable. But  E^c \subset S^c , so does it follow that m(  E^c) = 0?
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  2. #2
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    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, f(t) = \begin{cases}1&(t\in E),\\ 0&(t\notin E).\end{cases}
    Extend the domain of f from [0,1] to \mathbb{R} by making it periodic with period 1. The Fourier coefficients of f are given by \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 e^{2\pi ina}\hat{f}(n)=\hat{f}(n) for all n. But a is irrational, so e^{2\pi ina} can never be equal to 1 unless n=0. Therefore \hat{f}(n)=0 for all n\ne0, which tells you that f is (almost everywhere equal to) a constant function. Since f^2 = f, the constant can only be 0 or 1. Thus m(E) must be 0 or 1.
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    Quote Originally Posted by Opalg View Post
    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, f(t) = \begin{cases}1&(t\in E),\\ 0&(t\notin E).\end{cases}
    Extend the domain of f from [0,1] to \mathbb{R} by making it periodic with period 1. The Fourier coefficients of f are given by \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 e^{2\pi ina}\hat{f}(n)=\hat{f}(n) for all n. But a is irrational, so e^{2\pi ina} can never be equal to 1 unless n=0. Therefore \hat{f}(n)=0 for all n\ne0, which tells you that f is (almost everywhere equal to) a constant function. Since f^2 = f, the constant can only be 0 or 1. Thus m(E) must be 0 or 1.
    Unfortunately, I am not familiar with the Fourier series or the term "ergodic," so I cant really relate to this at all. Thanks for the time and effort though.

    Let S = {one point from each unique orbit}, then S is not measurable. So  S^c is not measurable. But  E^c \subset S^c , so does it follow that  m(E^c) = 0 ?
    Yes, it does follow, and I think I now know why:

    So, I proved that  E^c is made up of a bunch of unique orbits.

    Now, look at  \sum_{n=-\infty}^\infty m(f^n(E^c))

    By translation invariance, we have:

     \sum_{n=-\infty}^\infty m(f^n(E^c))=  \sum_{n=-\infty}^\infty m((E^c)), which is 0 if it is a 0 set and  \infty if not. The latter is obviously inadmissible.
    Last edited by southprkfan1; April 18th 2010 at 06:26 PM.
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