The length of a rectifiable curve may be phrased as a filtered limit. To do this, we will define a filter of partitions of an interval $[a,b]$ . Let ${\bf P}$ be the set of all ordered tuplets of distinct elements of $[a,b]$ whose entries are increasing:$${\bf P} = \{ (t_1, \ldots t_n) \mid ( a \le t_1 < t_2 < \cdots < t_n \le b) \wedge (n \in \mathbb{Z}) \wedge (n > 0) \$$ We shall refer to elements of ${\bf P}$ as partitions of the interval $[a,b]$ . We shall say that $(t_1, \ldots ,t_n)$ is a refinement of a partition $(s_1, \ldots, s_m)$ if $\{t_1, \ldots ,t_n\} \supset \{s_1, \ldots, s_m\}$ . Let ${\bf F} \subset \mathcal{P} ({\bf P})$ be the set of all subsets of ${\bf P}$ such that, if a certain partition belongs to ${\bf F}$ then so do all refinements of that partition.

Let us see that ${\bf F}$ is a filter basis. Suppose that $A$ and $B$ are elements of ${\bf F}$ . If a partition belongs to both $A$ and $B$ then every one of its refinements will also belong to both $A$ and $B$ , hence $A \cap B \in {\bf F}$ .

Next, note that, if a partition of $B$ is a refinement of a partition of $A$ then, by the triangle inequality, the length of $\Pi (B)$ is greater than the length of $\Pi (A)$ . By definition, for every $\epsilon > 0$ , we can pick a partition $A$ such that the length of $\Pi(A)$ differs from the length of the curve by at most $\epsilon$ . Since the length of $\Pi(B)$ for any partition $B$ refining $A$ lies between the length of $\Pi(A)$ and the length of the curve, we see that the length of $\Pi(B)$ will also differ by at most $\epsilon$ , so the length of the curve is the limit of the length of polygonal lines according to the filter generated by ${\bf F}$ .