Measurable set of real numbers with arbitrarily small periods

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!

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.