Arc length definition: integral vs partition

Question

Let $M$ be a connected Riemannian manifold. Let $\alpha:[a,b]\to M$ be a piecewise smooth curve segment. Then the arc length of $\alpha$ is defined as \begin{equation} L(\alpha)=\int_a^b|\alpha'(t)|\,dt\text. \end{equation} The Riemannian distance function on $M$ is defined as \begin{equation} d(p,q)=\inf\{L(\gamma):\text{$\gamma$ is a piecewise smooth curve segment from $p$ to $q$}\}\text. \end{equation} In any metric space $X$, the arc length of a (continuous) curve $\alpha:[a,b]\to X$ is defined as \begin{equation} |\alpha|=\sup\left\{\sum_{i=1}^md(\alpha(s_{i-1}),\alpha(s_i)):a=s_0<\cdots<s_m=b\right\}\text. \end{equation}

Is $L(\alpha)=|\alpha|$? I proved that $L(\alpha)\geq|\alpha|$, but could not prove the opposite inequality.

Answer 1

To prove that $\alpha \le L(\alpha)$ it suffices to prove for each partition $0=s_0<s_1<\cdots<s_m=1$ that $$\sum_{i=1}^md(\alpha(s_{i-1}),\alpha(s_i)) \le L(\alpha) $$ because then the supremum of all of the left hand sides is also $\le L(\alpha)$.

Using the equation $$L(\alpha) = \sum_{i=1}^m \int_{s_{i-1}}^{s_i} |a'(t)| \, dt $$ and it therefore suffices to prove for each $i$ that $$d(\alpha(s_{i-1}),\alpha(s_i)) \le \int_{s_{i-1}}^{s_i} |a'(t)| \, dt $$

Finally, using the inequality $$\biggl| \int_{s_{i-1}}^{s_i} a'(t) \, dt \biggr| \le \int_{s_{i-1}}^{s_i} |a'(t)| \, dt $$ it suffices to prove that $$d(\alpha(s_{i-1}),\alpha(s_i)) \le \biggl| \int_{s_{i-1}}^{s_i} a'(t) \, dt \biggr| $$ which is true, in fact the two sides are equal, because both sides are equal to $|\alpha(s_{i}) - \alpha(s_{i-1})|$.

Answer 2

Here is an argument for smooth curves $\alpha$. The piecewise smooth case follows easily. At any point $x_0 = \alpha(s_0)$ on the curve, smoothness implies that $\alpha$ is locally tangent to the geodesic. In particular, for every $\epsilon > 0$ there exists $\delta > 0$ such that $$ \left |\int_{s_0}^{s_1} |\alpha'(t)| \, dt \right| \le (1+\epsilon) d(\alpha(s_0), \alpha(s_1)) $$ whenever $|s_0-s_1| < \delta$. (I.e., the length of the segment of $\alpha$ from $t=s_0$ to $t=s_1$ does not exceed the geodesic distance by more than a factor of $1+\epsilon$.) A priori, $\delta$ will depend on both $s_0$ and $\epsilon$, but compactness implies that one can choose it independently of $s_0$. Now for any choice of partition $a=s_0< \ldots< s_n = b$ such that $s_i-s_{i-1}<\delta$ for all $i$, you get that the sum in the definition of $|\alpha|$ is $\ge \frac{L(\alpha)}{1+\epsilon}$, and letting $\epsilon$ go to zero, you get your desired inequality.

Details for this argument should not be too hard to fill in, assuming that $\alpha$ and the metric are at least continuously differentiable.