# Proof: $V_a^b(f) = \int_a^b |f'(x)|dx$

Suppose $f: [a,b] \rightarrow \mathbb{R}$ is differentiable on all $[a,b]$, and $f'$ is Riemann integrable on $[a,b]$. Denote the total variation of $f$ on $[a,b]$ by $$V_a^b(f) = \text{ sup } \{\sum_P |f(x_k) - f(x_{k-1})| : P \text{ is a partition on } [a,b] \}$$ Prove that $V_a^b(f) = \int_a^b |f'(x)|dx$.

My attempt:

We know $f'$ exists everywhere, therefore $f$ is continuous. Also, $f'$ is bounded since $f'$ is Riemann integrable. Therefore $f$ is a function of bounded variation ($f \in BV[a,b]$) which means the total variation is certainly finite.

Since $f \in BV[a,b]$ we can write $f$ as a difference of two strictly increasing functions. Let $f = u - v.$ Then $f' = u' - v'$ and $|f'| \leq |u'| + |v'|$. Since $u,v$ are strictly increasing this implies $|f'| \leq u' + v'$.

Then by comparison theorem for integrals, $$\int_a^b |f'|dx \leq \int_a^b (u' + v')dx = \int_a^b u'dx + \int_a^b v'dx$$

I don't know how to progress from here and conclude $\int_a^b|f'|dx \leq V_a^b(f)$. Do we know that $u', v'$ are Riemann integrable?

Then after that I don't see how you would show $\int_a^b|f'|dx \geq V_a^b(f)$. Could anyone please suggest a better way to go about this?

• Hint: $$|f(x_k) - f(x_{k-1})| = \left | \int_{x_{k-1}}^{x_k} f'(x) \right| \le \int_{x_{k-1}}^{x_k} |f'(x)| .$$ Commented Mar 16, 2017 at 19:52

I doubt your approach for the first part, too. Anyway, I would try to show that $\sum_P |f(x_k) - f(x_{k-1})|$ for a given partitions corresponds to a particular Rieman sum for $|f'|$. To do so, use the Mean Value Theorem to write $f(x_k) - f(x_{k-1}) = f'(c_k)(x_k-x_{k-1}$.
Then refining partitions, we approach variation on one side, and integral of $|f'|$ on the other.
• I guess I'm just struggling on how we come up with an actual value for the integral. I know that, in general, the definition says $\int_a^b f(x)dx$ is the number such that for all $\epsilon > 0$ there is a $P_\epsilon$ so that if a partition $P$ refines $P_\epsilon$ then $|S(P,f) - \int_a^b f(x)dx| < \epsilon$ for all Riemann's sums $S(P,f)$. In this proof, is there a certain theorem that I need to make use of? Commented Mar 16, 2017 at 20:10