A new metric involving curves Let $(X, d)$ be a metric space. The inner metric or length metric associated with
$d$ is the function $d_i : X \times X \to [0,\infty]$ defined by
$$d_i(x, y) := \inf L(\sigma)$$
where the inﬁmum is taken over all rectifiable curves $\sigma : [0, 1] \to X $ from x to y,
i.e $ \sigma(0) = x, \sigma(1) = y$ 
In Wikipedia it is asserted that assuming
$$
d_i \left( {x,y} \right) \lt \infty 
$$
then $(d_i)_i = d_i $ but I have no idea how to prove it. 
And how can I prove that every geodesic space is a length space :/? I have problems of that two assertions. I read it in this PDF on length spaces
Please help me two prove this. The PDF don´t prove it (it could be that it's very trivial)
 A: 
Let $\sigma: [0,1] \to (X,d)$ be a continuous and rectifiable curve. Then $\sigma: [0,1] \to (X,d_i)$ is also a continuous and rectifiable curve and $L_{d_i}(\sigma) = L_{d}(\sigma)$.

Let $\varepsilon \gt 0$. Since $[0,1]$ is compact, $\sigma: [0,1] \to (X,d)$ is uniformly continuous, so there is $\delta \gt 0$ such that $|t-t'| \lt \delta$ implies $d(\sigma(t),\sigma(t')) \leq \varepsilon/2$. On the other hand, as $\sigma$ is rectifiable, $L(\sigma)$ is finite, so we can find a partition $0 = t_0 \lt t_1 \lt \cdots \lt t_{n-1} \lt t_{n} = 1$ such that
$$
L(\sigma) - \varepsilon /2 \lt \sum_{k=1}^n d(\sigma(t_k),\sigma(t_{k-1})) \leq L(\sigma).
$$
The triangle inequality yields that refining the partition only increases the sum in this estimate, so we may assume that $t_k \lt t_{k+1} \lt t_{k} + \delta$ to begin with. But $L(\sigma) = \sum_{k=1}^n L(\sigma|_{[t_{k-1},t_k]})$, so
$$
0 \leq \sum_{k=1}^n \left( L(\sigma|_{[t_{k-1},t_k]}) - d(\sigma(t_k),\sigma(t_{k-1}))\right) \leq \varepsilon/2.
$$
Each summand in the last sum is non-negative, in particular
$$
L(\sigma|_{[t_{k-1},t_k]}) \leq d(\sigma(t_k),\sigma(t_{k-1})) + \varepsilon /2 \lt \varepsilon.
$$
This tells us two things:


*

*$t \mapsto L(\sigma|_{[0,t]})$ is continuous (and non-decreasing)

*$d_i(\sigma(t_{k}),\sigma(t_{k-1}))\lt \varepsilon$, in particular $\sigma:[0,1]\to (X,d_i)$ is uniformly continuous.


Finally,
$$L_{d_i}(\sigma) = \sup_{0=t_0 \lt \cdots \lt t_n=1} \sum_{k=1}^n d_i(\sigma(t_k),\sigma(t_{k-1}) \leq \sup_{0=t_0 \lt \cdots \lt t_n=1} \sum_{k=1}^n L_d(\sigma|_{[t_{k-1},t_k]}) = L_d(\sigma) \leq L_{d_i}(\sigma),$$
where the first inequality follows from the definition of $d_i$ and the second inequality from the fact that $d \leq d_i$.

To finish your question up, if $d_i(x,y) \lt \infty$ for all $x,y \in X$ we have $(d_i)_i = d_i$ by the above:
For every $\varepsilon \gt 0$ we may find a continuous and rectifiable curve $\sigma: [0,1] \to (X,d)$ such that $L_{d}(\sigma) \leq d_i(x,y) + \varepsilon$ but then $L_{d}(\sigma) = L_{d_i}(\sigma)$ gives that $(d_i)_i(x,y) \leq d_i(x,y)+\varepsilon$. Thus, $d_i = (d_i)_i$.
