I'm reading the book A Course in Metric Geometry by Dmitri Burago, Yuri Burago Sergei Ivanov and I don't understand the proof of Proposition 2.5.17 (Page 48).
More precisely, I don't know why the length is lower semi-continuous. I know the length induced by metric is semi-continuous, but I don't know why it is true for general length.
Proposition 2.5.17. If shortest path $\gamma_i $ in a length space $ (X,d) $ converge to a path $ \gamma $ as $ i\rightarrow\infty $, then $ \gamma $ is also a shortest path.
Proof. Since the endpoints of $ \gamma_i $ converge to endpoints of $ \gamma $ and the length of each $ \gamma_i $ is equal to the distance between its points, we conclude that $ L(\gamma_i) \rightarrow d(x,y) $, where $ x, y $ are the endpoints of $ \gamma $. By the lower semi-continuity of length $$ L(\gamma) \leq \lim_{i\rightarrow\infty} L(\gamma_i) = d(x,y). $$
Thanks in advance.
Given a metric space $(X,d)$ we can measure the length $L(c)$ of a path $c:[a,b]\to X$ by the formula
$$L(c)=\sup_{a=t_0<\ldots<t_k=b}\sum_{i=1}^kd(c(t_{i-1}),c(t_i))$$
where the supremum is taken over all possible (finite) partitions of the interval $[a,b]$.
This length satisfies the lower semi-continuity property: for every sequence $(c_n)_{n\in\mathbb{N}}$ of paths $c_n:[a,b]\to X$ which converges pointwise to a path $c:[a,b]\to X$, we have that $L(c)\leq\liminf L(c_n)$.
On the other hand, if $d$ is an intrinsic metric induced by a length structure $(\mathcal{C},\mathcal{L})$, then $L=\mathcal{L}$ (on the set $\mathcal{C}$ of admissible paths) if and only if $\mathcal{L}$ is lower semi-continuous (on $\mathcal{C}$).
Thus, in your quote, the authors call $L$ to the length induced by $d$.