Definition 1:

Let $r: [a,b] \to \Bbb R^d$ be a continuous differentiable function. Then the arc length is given by $$L(r) = \int_a^b || r'(t) || \, dt$$

Definition 2:

Let $r: [a,b] \to \Bbb R^d$ be a continuous function. Then the arc length is given by $$ V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$ where the supremum is taken over all partitions $P = \{a=x_0 \lt x_1 \lt \ldots \lt x_n = b \}$ of $[a,b]$.

How can I show that for a continuous differentiable $r(t)$ the two definitions are equivalent, i.e. $L(r)=V(r)$?

What I've done so far:

I found this question, which shows that I can convert the supremum to a limit

$$V(r) = \sup_P \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) ||$$

by choosing an appropriate sequence of partitions $P_n$ of which I take the $x_k$'s. This gives

$$ \lim_{n \to \infty} \sum_{k=1}^n || r(x_k)-r(x_{k-1}) || = \lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1})$$

Now I somehow need to show that

$$\lim_{n \to \infty} \sum_{k=1}^n || \frac{r(x_k)-r(x_{k-1})}{x_k-x_{k-1}} || (x_k-x_{k-1}) = \int_a^b ||r'(t)|| \, dt$$

How can I justify this step of converting the sum to an intergral and taking the limit of the inside simultaneously?

  • 1
    $\begingroup$ You should show that $\|r(a)-r(b)\|\le V(r)\le L(r)$. Now, define $s(t)$ to be the length of $r$ restricted to $[a,t]$, by the supremum definition. Show that $s'(t)=\|r'(t)\|$, through the definition of the derivative. Finish with the fundamental theorem of calculus. $\endgroup$ Jan 28, 2018 at 18:10
  • $\begingroup$ Alternatively: I think you can use the Mean Value Theorem somehow? $\endgroup$ Jan 28, 2018 at 18:13
  • $\begingroup$ See related answer math.stackexchange.com/a/3072835/72031 which deals with $d=2$, but the same argument can be used for higher dimensions. $\endgroup$
    – Paramanand Singh
    Feb 1, 2020 at 2:05

1 Answer 1


Claim. For a parametrized curve $\gamma \in \mathcal C^1([a,b], \Bbb R^d)$, we have $$\bbox[5px,border:2px solid #C0A000]{ \lim_{n\to\infty} \sum_{i=1}^n \| \gamma(x_{i,n})-\gamma(x_{i-1,n})\| = \int_a^b \|\gamma'(x)\| \,\mathrm dx, }$$

where $x_{i,n} := a \cdot (1-\frac in) + b\cdot\frac in$.

Proof. Note that the left hand side equals \begin{equation} \lim_{n\to\infty} \underbrace{ \sum_{i=1}^n \left\| \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\right\|\cdot {\frac{b-a}n} }_{=: \kappa_n} \end{equation} and observe that $x_{i,n}-x_{i-1,n}=\frac{b-a}n$.
By definition of the derivative, which is continuous on a compact interval and thus also uniformly contiuous, there exists a $\delta$ for every $\varepsilon > 0$ such that whenever $\frac{b-a}n < \delta$, $$\left\|\bigg\|\frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n}\bigg\|-\bigg\|\gamma'(x_{i,n})\bigg\|\right\| < \varepsilon.$$

In particular, for any $\varepsilon > 0$, we have \begin{equation}\tag{*} \label{*} \left\|\rule{0cm}{1cm} \underbrace{ \sum_{i=1}^n \left\| \frac{\gamma(x_{i,n})-\gamma(x_{i-1,n})}{\frac{b-a}n} \right\| \cdot\frac{b-a}n }_{=\kappa_n} -\underbrace{ \sum_{i=1}^n \Big\|\gamma'(x_{i,n})\Big\|\cdot\frac{b-a}n }_{=:\rho_n} \right\| < \varepsilon \cdot (b-a) \end{equation} whenever $n>\frac{b-a}\delta$ (note that $\delta$ depends on $\varepsilon$).

Since $\rho_n$ are just Riemann approximation sums, we have $\lim_{n\to\infty} \rho_n = \int_a^b \|\gamma'(x)\| \,\mathrm dx$. By \eqref{*}, we can conclude that $\lim_{n\to\infty} \kappa_n = \lim_{n\to\infty} \rho_n$, which proves our claim. $\square$

  • 1
    $\begingroup$ I think you should note the fact that you are using that continuous on a compact impliea uniformly continuous $\endgroup$ Jan 29, 2020 at 22:54
  • $\begingroup$ @LucioTanzini Thanks, done $\endgroup$ Feb 1, 2020 at 1:00

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