Let $M$ be a riemannian manifold. Then $M$ is complete iff each divergent curve in $M$ has no finite length (a curve $\gamma : [0 , 1) \to M$ is divergent if for each compact $K \subset M$ there exists $t \in [0 , 1)$ such that $\gamma([0 , 1)) \cap K = \emptyset$.
I have shown that if $M$ is complete then each divergent curve in $M$ has no finite length. But how can I prove the converse?
For the converse, assume that $M$ is not complete, and let's find a divergent curve with finite length. By incompleteness, there is an inextendible geodesic $\alpha\colon [0,b) \to M$, with $b<+\infty$. We claim that $\alpha$ is already divergent: indeed, if there is a compact $K\subseteq M$ with $\alpha(t) \in K$ for all $t \in [0,b)$, we may take an increasing sequence $(t_n) \subseteq [0,b)$ with $t_n \to b$ and pass to a subsequence to assume that $\alpha(t_n) \to p$ and $\alpha'(t_n) \to v$ for some $p \in M$ and $v \in T_pM$. With this in hands, we may pick $\epsilon > 0$ small enough and extend $\alpha$ to a geodesic $\widetilde{\alpha}\colon [0,b+\epsilon) \to M$ by $$ \widetilde{\alpha}(t) = \begin{cases} \alpha(t), &\mbox{ if }t\in[0,b) \\ \exp_p((t-b)v), &\mbox{ if }t \in [b,b+\epsilon)\end{cases}, $$contradicting the inextendability of $\alpha$. And if $\alpha$ has constant speed $c$, clearly $L[\alpha] = cb<+\infty$, concluding the proof. Note that we may reparametrize this $\alpha$ to something whose domain is $[0,+\infty)$, but the length and the property of being divergent remain both unchanged.
