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I am trying to prove the following exercise (exercise 3, chapter 7 of Rudins Book "Real and Complex Analysis"):

Suppose that $ E $ is a measurable set of real numbers with arbitrarily small periods. Explicitly, this means that there are positive numbers $ p_i $, converging to $ 0 $ as $ i\rightarrow \infty $, so that \begin{align*} E + p_i = E \ \ (i = 1, 2, 3, . . . ). \end{align*} Prove that then either $ E $ or its complement has measure $ 0 $.

I have seen the following answer: Measure zero sets, but I tried it by my own and followed the hints Rudin give in his book. That is what I have:

Given $ \alpha\in\mathbb{R}$, we define $F(x)=m(E\cap [\alpha, x])$, where $x>\alpha$. Then we get: \begin{align*} F(x+p_i)-F(x-p_i) &=m(E\cap [\alpha, x+p_i])-m(E\cap [\alpha, x-p_i])\\ &=m((E\cap [\alpha, x+p_i])-p_i)-m((E\cap [\alpha, x-p_i])+p_i) \\ &=m(E\cap [\alpha-p_i, x)-m(E\cap [\alpha+p_i, x]) \\ &=m(E\cap [\alpha-p_i,\alpha+p_i]). \end{align*} This implies that for every points $\alpha+p_i<x<y$, we get that \begin{align} F(x+p_i)-F(x-p_i)=F(y+p_i)-F(y-p_i). \end{align}

So the hint he gives now is to think about what does this imply for $F'(x)$ if $m(E) > 0$. I just have in mind that we could apply the Lebesgue differentiation theorem for $f=1_E$:

\begin{align*} \lim_{m(I)\rightarrow 0}\frac{m(E\cap I)}{m(I)}=1_E(x) \end{align*}

but this do not give enough information to conclude the density of $E$. Nevertheless, by having $m(E)>0$, we now that it must have some density point.

I would appreciate any help!

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Patricio
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1 Answers1

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If $F $ is differentiable at $x $, then $$ \frac {F (x+p)- F(x-p)}{2p} = \frac {1}{2} \left [ \frac {F (x+p)-F (x)}{p} + \frac {F (x-p) -F (x)}{-p}\right] \to \frac {1}{2} [F'(x)+F'(x)] = F'(x). $$

Use this together with what you have shown to conclude that $F'$ is constant (on the set where it is differentiable).

Using Lebesgue differentiation theorem, the claim follows.

PhoemueX
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  • Thank you very much. I get the point that like $ F (x+p)- F(x-p)=F (y+p)- F(y-p)$ and for the reason that the limit has to be unique the limits $ F'(x)$ and $F'(y)$ have to coincide. Like $x,y$ were arbitrary $F'(X)$ is constant for every point where $F$ is differentiable. But now, how I know that it is almost everywhere differentiable? And how do you use Lebesgue differentiation theorem, to conclude that $m(E^c)=0$, suposing that $m(E)>0$? I am sorry, i am not not so much into that topic. – Patricio Nov 21 '15 at 16:18
  • @Patricio: You have $F(x) = \int_\alpha^x \chi_E, dx$. Now a form of Lebesgues differentiation theorem tells you that $F$ is differentiable at a.e. $x >\alpha$ with derivative $F'(x)=\chi_E (x)$ a.e. – PhoemueX Nov 21 '15 at 21:47
  • Yes, thank you very much. But now like $F'(x)$ is constant, how do we can use the fact that $F'(x)=\chi_E (x)$ to conclude that E is dense? Thats the point I do not get. If almost every point $ x$ on the real line would satisfy $F'(x)=1$ I am done, but I do not see this :( – Patricio Nov 21 '15 at 21:53
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    @Patricio: We have just seen that $\chi_E (x) = F'(x) =c =c_\alpha$ for a.e. $x>\alpha$. Using this, the following are not hard to see 1) In fact, $c=c_\alpha$ does not depend on $\alpha$ (use that even after the excluding null-sets, you can find $x>\max{\alpha,\beta}$ for arbitrary $\alpha,\beta$) 2) We have $c \in {0,1}$ (since $\chi_E$ can only attain these values) 3) If $c=0$, the $E$ is a null-set and if $c=1$, then $\chi_E =1$ a.e. which implies $\chi_{E^c}=0$ a.e. so that $E^c$ has measure zero. – PhoemueX Nov 21 '15 at 21:59