Continuous functions are differentiable on a measurable set? I came across the following challenging problem in my self-study:
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Then the set of points where $f$ is differentiable is a measurable set.
I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.

You can write the set of points where $f$ is differentiable as the set of points where a particular limit exists (namely, the one that defines the derivative). You can codify a limit (using the $\epsilon,\delta$ definition) in terms of intersections/unions of measurable sets.
What you want to say is that the function $f$ is differentiable at $x$ (with value $y$) if for all $\epsilon$ there exists a $\delta$ such that the difference quotient with parameter $\delta$ is $\epsilon$-close to $y$. First fix an $\epsilon$. Then the set of points where this is true for that particular $\epsilon$ is is the union (over the possible $\delta$) of the sets such that the difference quotient is $\epsilon$-close with that fixed $\delta$. This must be true for all $\epsilon$, so we take the intersection of those sets over all $\epsilon$. 
The net result is that we are taking 
$$\cap_\epsilon \cup_\delta V(\epsilon, \delta)$$
where $V(\epsilon, \delta)$ is the set of points such that the difference quotient is less than $\epsilon$ from $y$. Of course, this was just for one possible value of the derivative, so we need to take the union over all $y$ as well, giving us
$$\cup_y \cap_\epsilon \cup_\delta V(\epsilon, \delta).$$
For everything involved to be countable, you'd want to choose the $y$, $\epsilon$, $\delta$ to be rational, which is fine because the rationals are dense in $\mathbb{R}$.
