Continuous function of bounded variation

Question

Let $f$ be a continuous function with bounded variation on $[0,1]$.

Question: Is it true that \begin{equation*} \int_{[0,1]}f'(x)dx=f(1)-f(0) ? \end{equation*}

Answer 1

No, it is false. The Cantor function is continuous, increasing (and so it has bounded variation) and $f'(x)=0$ for $\mathcal{L}^1$ a.e. $x\in [0,1]$, so $$f(1)-f(0)=1-0>\int_0^1 f'(x)\,dx=0.$$ For the fundamental theorem of calculus to hold it is necessary and sufficient that $f$ is absolutely continuous.