# 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. Jan 5, 2018 at 20:49
• Good point, I wasn't thinking about that. Jan 5, 2018 at 20:50
• A little comment: using your notation $F\left(b\right)-F\left(a\right)=\int_{a}^{b}f,$ and not $F\left(a\right)-F\left(b\right)=\int_{a}^{b}f.$ Sep 26, 2021 at 23:05

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