Counterexample of the almost-inverse of the Fundamnetal Theorem of Calculus(Lebesgue).

Question

Can anyone give me a counterexample to the following statement:

Suppose $F \colon [0,1] \to \mathbb{R}$ is continuous and differentiable almost everywhere, then $F(b)-F(a)=\int_a^b F'(t)\, \text{d}t$.

I guess let the discontinuity be $\mathbb{Q} \cap[0,1]$ may help.

Answer 1

The Cantor function is a counter-example. It's continuous and has derivative $0$ almost everywhere, yet $F(1)-F(0)=1-0=1 \neq 0=\int_0^10 \, dt=\int_0^1F'(t) \, dt$.

Answer 2

I harass Calculus students with this example (which one can easily scale and translate to $[0,1]$):

Let $f(x) = \tan^{-1}(x) + \tan^{-1}(1/x)$. Compute $f(1)$, $f(-1)$, and $f'(x)$. How is that possible?

They find $f(1) = \pi/2$, $f(-1) = -\pi/2$, and $f'(x) = 0$ (almost) everywhere. Which should be a good enough hint.

Worth noting that this $f$ is continuous (constant!) a.e. and differentiable a.e. I like this $f$ because the explanation of constancy is simple if one remembers one's Trig: $f$ is the sum of two (directed) complementary angles. The direction just reverses as we pass through $0$.