$f:[0,1] \to \mathbb R$ be a differentiable function with bounded derivative, then is $\int_0^1 f'(x)=f(1)-f(0)$

Question

Let $f:[0,1] \to \mathbb R$ be a differentiable function such that $\sup_{x\in [0,1]} |f'(x)|$ is finite. Then since $f'(x)=\lim_{n\to \infty} \dfrac{f(x+1/n)-f(x)}{1/n}$, so $f'(x)$ is measurable and also the Lebesgue integral $\int_0^1|f'(x)|dx$ is finite, thus $f' \in L^1([0,1])$.

My question is, is it true that $\int_0^1 f'(x)=f(1)-f(0)$ ?

Note that fundamental theorem of calculus does not apply here since $f'(x)$ is not continuous (not even known to be Riemann integrable)

Answer 1

Lebesgue integrability is enough. You need the following version of FTC:

Let ${[a,b]}$ be a compact interval of positive length, let ${F: [a,b] \rightarrow {\bf R}}$ be a differentiable function, such that ${F'}$ is absolutely integrable. Then the Lebesgue integral ${\int_{[a,b]} F'(x)\ dx}$ of ${F'}$ is equal to ${F(b) - F(a)}$.

This is a standard result in real analysis. See for instance this excellent set of lecture notes by Terry Tao.