Theorem: Suppose $(X,d)$ is a metric space and $f:[0,1] \rightarrow X$ is a path in $X$ with no-zero finite length $L$. Then, there exists a path $g:[0,1] \rightarrow X$ from $f(0)$ to $f(1)$ that has the same image as $f$ and satisfies $lth_t(g) = tL ~\forall ~t \in [0,1]$. In particular $g$ is Lipschitz with Lipschitz constant $L$.
Some background: Definition of length: Suppose $(X, d)$ is a metric space and $f: [0 , 1] → X$ is a path in $X.$ For each $t ∈ [0 , 1]$, let $P_t$ denote the set of all tuples $a ∈ \{\mathbb R^n~|~ n ∈ \mathbb N\setminus \{1\}\}$ for which $a_1 = 0, a_{ν(a)} = t$ and $a_i ≤ a_{i+1}$ for all $i ∈ \mathbb N_{ν(a)−1}$, where $ν(a)$ is the unique $n ∈ \mathbb N$ such that $a ∈ \mathbb R^n$. We define the length $ltht(f)$ of $f_{|[0 ,t]}$ to be $\sup \{\sum\limits_{i=1}^{v(a)-1} d(f (a_i), f(a_{i+1}) \}$ and the length $lth(f)$ of $f$ to be $lth_1(f)$.
Now, we return back to the proof of our theorem.
Point of Confusion: In the later part of the proof, in order to prove that $lth_t(g) = tL$, the author has proved that $lth_t(g) \le tL$ in both the forward and backward parts. Has the author erred or am I making a mistake in understanding?
As can be seen, the author has proved in the same direction twice. Did the author make a mistake? Thanks a lot for reading through! Appreciate it!