$C_1$ curve length proof deficiency

Question

On page 5 of these notes is this theorem:

"If $\alpha: I \rightarrow \Bbb{R}^n$ is a $C^1$ curve, then

$\operatorname{length}[\alpha] = \int_I ||\alpha'(t)||dt$ "

The proof begins "It suffices to show that (i) $\operatorname{length}[α, P]$ is not greater than the above integral, for any $P \in \operatorname{Partition}[a, b]$, and (ii) there exists a sequence $P_{N}$ of partitions such that $\lim_{N\to \infty} \operatorname{length}[\alpha, P_{N}]$ is equal to the integral."

Why do those two truths suffice? I do not understand what each of them contributes.

Moreover, why does the theorem need its own proof? To me it seems a normal integral.

Answer 1

From the defintions

$$\text{length}[\alpha,P]= \sum_{i}\|\alpha(t_i)-\alpha(t_{i-1})\|$$

$$\text{length}[\alpha]= \sup_{P\in\text{partition}([a,b])}\text{length}[\alpha,P]$$

you get from $(i)$

$$\text{length}[\alpha] \leq \int_{I}\|\alpha'\|\ \mathrm{d}x$$

and from $(ii)$

$$\forall \epsilon>0\ \exists P\in \text{partition}([a,b]): \left|\text{length}[\alpha,P]-\int_{I}\|\alpha'\|\ \mathrm{d}x\right|<\epsilon$$

Answer 2

The definition of the length of $\alpha$ is $$\sup_P \operatorname{length}[\alpha,P].$$ Why does the theorem need proof? Because the length is not defined by that integral. Why do (i) and (ii) suffice? That's just how supremums work!

Say $E\subset \Bbb R$ is bounded. Saying $\sup E=B$ is equivalent to (i) every element of $E$ is less than or equal to $B$ and (ii) there exists a sequence $x_n\subset E$ with $x_n\to B$.