On the Riemannian integrability of the bounded derivative

Question

Let $f:[a,b]\to\mathbb R$ be a differentiable function with $f'$ bounded. According to this post, $f'$ is not necessarily Riemann integrable on $[a,b]$, see also Volterra's function.

I wonder, if $f'$ is not Riemann integrable on $[a,b]$, does there exist a Riemann integrable function $g:[a,b]\to\mathbb R$ such that $g=f'$ a.e. and $$f(x)-f(a)=\int_a^xg(t)\,dt,\qquad \forall x\in[a,b] ?\tag{$*$}$$ In $(*)$ we use Riemann integration.

There are some trivial observations.

• Fistly, the boundedness of $f'$ implies that $f$ is absolutely continuous, hence $f'$ is Lebesgue integrable and $$f(x)-f(a)=\int_a^xf'(t)\,dt,\qquad \forall x\in[a,b]$$ holds in the sense of Lebesgue integration.

• There exists a bounded Lebesgue integrable function $h:[0,1]\to\mathbb R$ such that we cannot find a Riemann integrable function $g:[0,1]\to\mathbb R$ with $g=h$ almost everywhere: Take $h$ to be the characteristic function of a fat Cantor set $F$ with $0<\lambda(F)<1$, where $\lambda$ denotes the Lebesgue measure, then $h$ is discontinuous at every point in $F$, since $F$ does not contain any open interval.

• We note that if $(*)$ holds for $g$, then we have $g=f'$ almost everywhere. Indeed, $(*)$ implies that if $g$ is continuous at $x_0$, then $f'(x_0)=g(x_0)$, hence $$\{x\in[a,b]: f'(x)\neq g(x)\}\subset \{x\in[a,b]: g \text{ is not continuous at }x\},$$ and hence $g=f'$ a.e. since $g$ is Riemann integrable.

• Also, it is well known that if $g:[a,b]\to\mathbb R$ is Riemann integrable and $h:[a,b]\to\mathbb R$ is bounded such that $h=g$ a.e., then $h$ is not necessarily Riemann integrable. For example, take $g\equiv0$ on $[0,1]$ and $h$ the characteristic function of $\mathbb Q\cap[0,1]$.

Note that derivatives of everywhere differentiable functions cannot be arbitrarily badly behaved. For example, they satisfy the conclusion of the intermediate value theorem.

Answer 1

The answer to the question in the body of your post is no, for the reason that if $f' = g$ almost every where and $g$ is Riemann integrable and such that (*) holds, then $f'$ must be Riemann integrable. I sketch the proof below.

The fundamental theorem of calculus has the following slight strengthening (proof is almost the same as the classical case):

FTC1 Suppose $g$ is Riemann integrable on $[a,b]$ and $G(x) = \int_a^x g$. If $g$ is continuous at $x_0$, then $G$ is strongly differentiable at $x_0$.

Your hypothesis implies then $f$ is not only differentiable, but strongly differentiable at every point that $g$ is continuous. Note that by Lebesgue's criterion the set $\{g \text{ is cont.}\}$ is full measure.

Next, we have the standard characterization of strong differentiability:

Thm If $f$ is differentiable on an open set, then $f$ is strongly differentiable at $x$ if and only if $f'$ is continuous at $x$.

Together, this shows that $f'$ is continuous on a set of full measure, and hence necessarily $f'$ is Riemann integrable.

(Boundedness of $f'$ is not necessary for this argument.)

IIRC both results should be available somewhere in Schechter's Handbook of Analysis and its Foundations; I also have them in a set of not-yet-public lecture notes that I can share on request.

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Let me write out the direct proof by unwrapping the results above.

Let $\Sigma$ be the subset of $(a,b)$ where $g$ is continuous. I claim that $f'$ is continuous at every point of $\Sigma$ too. The result then follows from Lebesgue's criterion.

Take $x\in \Sigma$

1. For every $\epsilon > 0$, there exists $\delta > 0$ such that $|y-x| < \delta$ implies $|g(y) - g(x)| < \epsilon$.
2. Hence for $w,z\in (x-\delta, x+\delta)$, we have $$ |\int_w^z g - (z-w) g(x)| < \epsilon |z-w| $$
3. Which we unwrap as $$ \left|\frac{f(z) - f(w)}{z-w} - g(x)\right| < \epsilon $$ (Remark: this statement is basically the definition of strong differentiability, so steps 1-3 here gives the proof of FTC1 as described above the cut.)
4. Take the limit $z\to w$, the LHS converges to $|f'(w) - g(x)|$. So we have proven that for every $\epsilon > 0$, there exists a $\delta > 0$ such that $|w-x| < \delta$ implies $|f'(w) - g(x)| \leq \epsilon$.
5. This in particular implies that $f'(x) = g(x)$ and hence $f'$ is continuous at $x$.
   (Remark: steps 4 and 5 can be used to reconstruct the proof of "strong differentiability implies continuity of derivatives" direction of the Thm mentioned above the cut, for one dimensional domains.)