Suppose $f:[0,1]\to\mathbb R^n$ is continuous. For a partition $P=\{t_0=0,t_1,t_2,\cdots,t_{m-1},t_m=1\}$, with norm $|P|=\max_i(t_i-t_{i-1})$, define
$$\sum_P\lVert df\rVert=\sum_{i=1}^m\lVert f(t_i)-f(t_{i-1})\rVert,$$ $$S=\sup_P\sum_P\lVert df\rVert,$$ $$L=\lim_{|P|\to0}\sum_P\lVert df\rVert.$$
We want to prove that if either $S$ or $L$ exists, then they both exist and $S=L$.
To be clear, the definition of $L$ is: for any $\varepsilon>0$, there exists $\delta>0$ such that, for all partitions $P$ with $|P|\leq\delta$, we have $|\sum_P\lVert df\rVert-L|\leq\varepsilon$.
I can see that if $L$ exists then $S$ exists: Take any $\varepsilon$ and $\delta$ as above. For any partition $Q$, there is a refinement $P\supseteq Q$ with $|P|\leq\delta$ (e.g. repeatedly bisect the intervals until they're small enough), so $\sum_Q\lVert df\rVert\leq\sum_P\lVert df\rVert\leq L+\varepsilon$, where the first inequality is the triangle inequality. Thus the set of all sums $\sum_Q\lVert df\rVert$ is bounded, and has a supremum $S\leq L+\varepsilon$. Since $\varepsilon>0$ was arbitrary, we get $S\leq L$. And it's obvious that $L\leq S$.
It remains to be shown that if $S$ exists then $L$ exists. That is, we must show that $S$ satisfies the definition of $L$.
