For an a.e. differentiable continuous function $f$ with $f'$ integrable, is it true in general that $f(b)-f(a) \geq \int_a^b f'(x) dx$?

Question

I am aware that the fundamental theorem of calculus fails for functions which are a.e. differentiable but not absolutely continuous, such as the Cantor function.

However, I wonder if a general inequality still holds in one direction. That is, for any continuous $f : [a,b] \to \mathbb{R}$ that is differentiable almost everywhere on the interval $[a,b]$ with $f'$ being integrable, I wonder if we have in general \begin{equation} f(b)-f(a) \geq \int_a^b f'(x)dx. \end{equation}

Or are there counterexamples even for this inequality? Could anyone please help me?

Answer 1

The Cantor function allows you to go up or down without being accounted by the derivative.

Let $g$ be the Cantor function and define $$ f(x)=\begin{cases} x,&\ 0\leq x \leq \frac12\\[0.2cm] 1-g(x),&\ \frac12<x \leq1 \end{cases} $$ Then $f$ is continuous, it is differentiable almost everywhere, $f(1)=f(0)$, and $$ \int_0^1f'=\int_0^{1/2}1=\frac12. $$

Answer 2

If $f$ is decreasing, this doesn’t hold, but it’s true if $f$ is increasing. Indeed, the difference quotients $f_h(x) = \frac1h(f(x+h)-f(x))$ converge a.e. to $f’(x)$ and are nonnegative, so by Fatou’s lemma,

$$\int_a^b f’(x)dx \leq \liminf_{h\to 0} \int_a^b f_h(x) dx$$

But

\begin{align} \int_a^b f_h(x) dx &= \frac1h \int_{a+h}^{b+h} f(x)dx - \frac1h \int_a^b f(x)dx \\ &= \frac1h \int_{b}^{b+h} f(x)dx - \frac1h \int_a^{a+h} f(x)dx\end{align}

Since $f$ is continuous, the above converges to $f(b)-f(a)$.