Proof. Let the partition $\{ t_i \}$ of $[0,1]$ be arbitrary. Then
Hence $L \leq \int_0^1 \norm{\gamma'(t)} \, dt$ . (By the way, this also shows that $\gamma$ is rectifiable in the first place.)

The inequality in the other direction is more tricky. Given $\epsilon > 0$ , we know that $\int_0^1 \norm{\gamma'(t)} \, dt$ can be approximated up to $\epsilon$ by a Riemann sum of the form$$ \sum_{i=1}^n \norm{ \gamma'(t_{i-1}) } (t_i - t_{i-1})$$ provided the partition $\{ t_i \}$ is fine enough, i.e. has mesh width $\leq \Delta$ for some small $\Delta > 0$ . We want to approximate $\gamma'(t_{i-1})$ with $[\gamma(t_i) - \gamma(t_{i-1})]/(t_i - t_{i-1})$ , but this only works if $t_i - t_{i-1}$ is small.

To get the precise estimates, use uniform continuity of $\gamma'$ on $[0,1]$ to obtain a $\delta > 0$ such that $\norm{\gamma'(\tau) - \gamma'(t)} \leq \epsilon$ whenever $\abs{\tau - t} \leq \delta$ . Then for all $0 < h \leq \delta$ and $t \in [0,1]$ ,$$ \normW{ \frac{\gamma(t+h) - \gamma(t)}{h} - \gamma'(t) } \leq \frac{1}{h} \int_t^{t+h} \norm{\gamma'(\tau) - \gamma'(t)} \, d\tau \leq \frac{h}{h} \, \epsilon = \epsilon\,.$$

Let the partition $\{t_i\}$ have a mesh width less than both $\delta$ and $\Delta$ . Then setting $h = t_i - t_{i-1}$ successively in each summand, we have

Taking $\epsilon \to 0$ yields $\int_0^1 \norm{\gamma'(t)} \, dt \leq L$ .