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.

  • $\begingroup$ Oh wow, thank you very much! I didn't know about this result - looking at Rudin, I thought we required absolute continuity for this, but looks like we don't. :-) $\endgroup$ – wanderer May 1 at 8:34

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