Riemannian Integrability – Bounded Derivative Analysis

integrationreal-analysis

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.

Best Answer

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.

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.)
Related Question