Proof of Arc-Length Formula in $\mathbb{R}^n$

Question

Let $\gamma$ denote a $C_1$ curve, sending an interval $I$ to $\mathbb{R}^n$. We define the length of the curve to be: $$ L (\gamma) = \sup_{P}L(\gamma,P) = \sum_{i =1}^k\|\gamma(t_i)- \gamma(t_{i-1})\| $$ where each $t_i$ is an element of partition $P$ of the interval, with $k$ elements in the partition. Now, I would like to prove that $$ L (\gamma) = \int_{I} \|\gamma'(t)\| \mathrm{d}t $$ It seems the easiest way to do this is to use the Riemann sum definition of the integral and use the mean value theorem to create Riemann sums in a sense. By this I mean the following: we have for any partition $P$ that the length approximation is given by: $$ \sum_{i =1}^k\|\gamma(t_i)- \gamma(t_{i-1})\| = \sum_{i =1}^k\left [\sum_{j = 1}^{n} (\gamma_j(t_i) -\gamma_j(t_{i-1}))^2\right]^{1/2} $$ Where $\gamma_j$ denotes the $j$th component of $\gamma$. Now, because each component function is a map from $\mathbb{R} \to \mathbb{R}$, we cna apply the mean value theorem to find points so that: $$ \gamma_j (t_i) - \gamma_j(t_{i-1}) = \gamma'(c^j_i) (x_i - x_{i-1}) $$ And substituting this into the above sum for length gives us: $$ \sum_{i =1}^k\|\gamma(t_i)- \gamma(t_{i-1})\| = \sum_{i =1}^k\left [\sum_{j = 1}^{n}(\gamma'_j(c^j_i) (x_i - x_{i-1}))^2\right]^{1/2} $$ And pulling out terms this simplifies to: $$ \sum_{i =1}^k\left[\sum_{j = 1}^{n}(\gamma'_j(c^j_i))^2\right]^{1/2}(x_i - x_{i-1}) $$ Now, if each point $c_i^j$ was the same point, $c^*$, then the approximation would be given by: $$ \sum_{i =1}^k \|\gamma'(c*)\|(x_i - x_{i-1}) $$ which is a Riemann sum of $\|\gamma'(t)\|$. My question is, how can turn these multiple points furnished by the MVT into a single point? And how then do I prove that the limit of the Riemann sums is the supremum of the approximations (I am not entirely concerned with the second part, I think I can figure it out but am just leaving it here as a general sort of question). I believe this has to do with the continuity of $\gamma'$, but I am not sure .

Answer 1

You can't use the same $c^*$ for every interval. For each $i$ the sample point must be selected from the interval $[t_{i-1},t_i]$. On the other hand you can use consistent sample points for each index $j$. Compare $$ \sum_{i =1}^k\left[\sum_{j = 1}^{n}(\gamma'_j(c^j_i))^2\right]^{1/2}(t_i - t_{i-1}) $$ with $$ \sum_{i =1}^k\left[\sum_{j = 1}^{n}(\gamma'_j(t_{i-1}))^2\right]^{1/2}(t_i - t_{i-1}) $$ [Note: you accidentally passed from $t_i$ to $x_i$ in your labeling of the partition.]

Each of the components $\gamma_j$ is continuously differentiable, so in particular each $\gamma_j'$ is uniformly continuous. Provided that the maximum value of $t_i - t_{i-1}$ is small enough, you can make $\gamma_j'(c_i^j)$ arbitrarily close to $\gamma_j'(t_{i-1})$ because $0 \le c_i^j - t_{i-1} \le t_i - t_{i-1}$.