Does the derivative of a differentiable function have to be Lebesgue integrable in some interval?

The derivative is the (pointwise) limit of a sequence of continuous functions, e.g. $$g_n(x) = \frac{f(x + h_n(x)) - f(x)}{h_n(x)}$$ where we can take $h_n(x) = \frac{1}{n+1}$ if $b = +\infty$, and if $b < +\infty$ we can take $h_n(x) = \frac{b-x}{n+1}$. It follows that the family $\{ \lvert g_n\rvert : \in \mathbb{N}\setminus \{0\}\}$ is pointwise bounded.

Take an arbitrary nonempty interval $(u,v) \subset (a,b)$. For each $k \in \mathbb{N}$ the set $$A_k = \bigl\{ x \in (u,v) : \lvert g_n(x)\rvert\leqslant k \text{ for all } n\bigr\}$$ is relatively closed, and since the family is poinwise bounded we have $$(u,v) = \bigcup_{k \in \mathbb{N}} A_k\,.$$ Furthermore, $(u,v)$ is a Baire space (it's completely metrisable), hence there is a $k \in \mathbb{N}$ such that $$V = \operatorname{int} A_k \neq \varnothing\,.$$ Then $\lvert f'(x)\rvert \leqslant k$ for all $x \in V$

Thus every nonempty open interval in $(a,b)$ contains a nonempty open interval on which $f'$ is bounded. This means the set of points $x$ such that $f'$ is Lebesgue integrable on some neighbourhood of $x$ is a dense open subset of $(a,b)$, hence topologically very large.

However, the measure of this set would be the more important type of size. I don't know whether it can be arbitrarily small (of course it's nonzero), but I suspect it can.