Are “most” bounded derivatives not Riemann integrable?

Question

Given $a,b\in\mathbb R$ with $a<b$. Let $$X=\{f\in C([a,b]): f \text{ is differentiable on } [a,b] \text{ with }f' \text{ bounded }\},$$ and $$A=\{f\in X: f' \text{ is Riemann integrable}\}. $$ It is known that $A\subsetneq X$, see this post.

I wonder, is there any result concerning the question that what proportion of $A$ is in $X$? I mean, I'm looking for results similar to "A generic continuous function on $[0,1]$ is nowhere differentiable". Does a "generic" element of $X$ lie in $A$?

If we want to use the Baire category theorem, maybe we need to be careful to define an appropriate metric on $X$ to make it become a complete metric space (but I don't know how to find this appropriate metric).

Answer 1

In 1977 Clifford E. Weil showed that $A$ is a first Baire category set (i.e. a meager set) in $X$ (sup norm) -- see The space of bounded derivatives. So the situation, at least with respect to one specific metric, is about as opposite as possible from the generic element of $X$ lies in $A$. Weil showed that the generic element of $X$ does NOT lie in $A.$

In the same paper Weil also proved a stronger result in which $X-A$ (discontinuity sets of the functions have positive Lebesgue measure) is replaced by the smaller set of functions whose discontinuity sets have full Lebesgue measure. Thus, all but "infinitesimally many" bounded derivatives have Lebesgue measure zero continuity sets. Others have further strengthened this last result by replacing "Lebesgue measure zero" with, for example, (a) Hausdorff dimension zero; (b) Hausdorff $h$-measure zero for any Hausdorff measure function $h$; (c) $\mu$-measure zero for any finite Borel measure $\mu.$ See this MSE answer for more details.