If $f$ is differentiable almost everywhere and $f'$ is abolutely integrable, then $f$ is of bounded variation? Let $f:[a,b]\to\mathbb{R}$ be of bounded variation. Then it follows that $f$ is differentiable almost everywhere in $[a,b]$, and
$$
\int_a^b|f'(x)|\ dx<\infty. 
$$
Conversely, suppose $f:[a,b]\to\mathbb{R}$ is differentiable almost everywhere and 
$$
\int_a^b|f'(x)|\ dx<\infty. 
$$
Can one conclude that $f$ is of bounded variation on $[a,b]$?

I'm not sure if the stronger condition of "differentiable everywhere" is needed, but I don't have a counterexample. 
 A: There is a continuous function $f : [0, 1] \to \Bbb{R}$ with infinite variation such that $f'(x)$ exists and is zero for a.e. $x \in [0, 1]$. For instance, let $\phi : [0, 1] \to \Bbb{R}$ be the Cantor-Lebesgue function and let $a_n = 1 - \frac{1}{2} + \frac{1}{3} + \cdots + (-1)^{n-1}\frac{1}{n}$ with $a_0 = 0$. Then we define $f$ piecewise by
$$ f(x) = a_{n-1} + \frac{(-1)^{n-1}}{n} \phi \left(1 - \tfrac{1}{2^{n-1}} + \tfrac{1}{2^n}x\right) \quad \text{if} \quad x \in [1 - \tfrac{1}{2^{n-1}}, 1 - \tfrac{1}{2^n}]$$
for each $n = 1, 2, \cdots$ and $f(1) = \log 2$.
$\hspace{7em}$
This function is continuous and has infinite variation. But since $\phi'$ is zero a.e., it follows that $f'$ is also zero a.e.
On the other hand, if we assume that $f:[a,b]\to\Bbb{R}$ is everywhere differentiable on $[a, b]$ and $f'$ is integrable, then $f$ is of bounded variation. This is because the following FToC-type result holds:
$$ f(x) - f(a) = \int_{a}^{x} f'(t) \, dt, \qquad x \in [a, b] $$
The proof of this result can be found, for instance, in Rudin RCA, Theorem 7.25.
