More Lebesgue Measure Theory fun

**southprkfan1** · April 17th 2010, 02:13 PM

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?

**Opalg** · April 18th 2010, 12:28 PM

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.

**southprkfan1** · April 18th 2010, 04:57 PM

Originally Posted by 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, $f(t) = \begin{cases}1&(t\in E),\\ 0&(t\notin E).\end{cases}$
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.

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