Integrable function has bounded variation

Let $f:[0,1] \to \mathbb{R}$ be Lebesgue integrable. Define $F(x) = \int_0^x f$. Prove $F(X)$ has bounded variation on $[0,1]$. What does this imply for differentiability of $F$ and why?

Attempt

$F$ is monotone (increasing) very obviously. So if $F$ is of bounded variation and is monotone it is differentiable a.e. (I think bounded variation also implies bounded on a compact interval?) INCORRECT

EDIT

Since $F$ is of bounded variation it can be written as the differnece of two increasing (monotone) functions on $[0,1]$. Since monotone functions are differentiable a.e. and the derivative behaves linearly we have that $F$ is differentiable a.e.

Notice $F(a)-F(b) = \int_a^b f$. Further, $|F(a)-F(b)| \leq \int_a^b f$. Let $P$ be any partition of $[0,1]$ with $0=x_0 < x_1 < \dots < x_k = 1$. We have

$$\sum |F(x_i)-F(x_{i-1})| \leq \int_0^1 |f| < \infty$$

since $f$ is Lebesgue integrable. Thus, $F$ is of bounded variation.

• $F$ is only monotonic if $f$ (almost) always has the same sign. – Daniel Fischer Jan 5 '18 at 20:49
• Good point, I wasn't thinking about that. – Aaron Zolotor Jan 5 '18 at 20:50

Hint: If $0\le a < b\le 1,$ then $|F(b)-F(a)| \le \int_a^b|f|.$