# Big Rudin Chapter 7, Q3 and about symmetric derivative

I'm getting extremely confused with this problem given in Big Rudin, Chapter 7 (Real and Complex Analysis) :

Q3. 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

$$E+p_i=E$$ ($$i = 1,2,...)$$

Show that $$E$$ or its complement has measure 0.

What I tried :

Taking $$F(x) = m(E \cap [\alpha, x])$$ (for $$x > \alpha$$) as the hint, I was able to show that if $$F$$ is differentiable at $$x$$, then $$F'$$ is constant. That much easily follows from the hint.

Now, if it so happens that $$F$$ is differentiable a.e., then we can immediately get that $$m(E^c) = 0$$, that much is clear. And I was stuck - what if $$F$$ is not differentiable a.e?

From here, I've searched google, and have seen the following two posts :

Measurable set of real numbers with arbitrarily small periods

Measure zero sets

Looking at the first link, one of the comments says that $$F$$ is differentiable a.e. (by a form of Lebesgue Differentiation theorem) - but I can't see how! What if measure of $$E$$ is infinity? Then $$\chi_{E}$$ is not in $$L^1$$ so we can't apply the Lebesgue Differentiation Theorem. (Is he looking at the metric density? But even so, metric density is just the symmetric derivative in this case - which happens to be $$\chi_{E}$$ a.e. But we can't really say that this is THE derivative, as having symmetric derivative does not mean differentiability)

So, here comes my second question:

If $$f:R \rightarrow R$$ is measurable, and symmetric derivative exists for all $$x$$, then is $$f$$ differentiable a.e.?

If this statement is false, how do I go on about solving the original problem? (Q3 that is) Also, if true, how do I prove this?

$$F$$ is finite valued and monotone. Any monotone real valued function is differentiable almost everywhere.