Continuous length curve

Question

Let $(X,d)$ be a metric space. A curve in $X$ is a continuous mapping $\sigma:[0,1]\to X$. Denote $P[0,1]$ the set of partitions of $[0,1]$. For a partition $Y=\{y_0,y_1,\ldots,y_n\}\in P[0,1]$, where $0=y_0<y_1<\cdots<y_n=1$, we define $\Sigma Y=\sum_{i=1}^n d(\sigma(y_{i-1}),\sigma(y_i))$. The length of $\sigma$ is defined as $$L(\sigma)=\sup_{Y\in P[0,1]}\Sigma Y.$$ For a curve $\sigma$ of finite length, define $p(t)=L(\sigma_{[0,t]})$, where $\sigma_{[0,t]}$ is the restriction of $\sigma$ on the interval $[0,t]$. The question is: how to prove that $p:[0,1]\to\mathbb{R}$ is continuous?

I've tried it using many properties of partitions, triangular inequalities and definition of supremum. Thanks for any help, solution or hint.

Answer 1

Hint.

For a partition $Y=\{y_0,y_1,\ldots,y_n\} \in P[0,1]$, define: $$\vert Y \vert = \max\limits_{1 \le i \le n} (y_i - y_{i-1})$$ The main idea is to prove that $$\lim\limits_{\vert Y \vert \to 0}\Sigma Y = L(\sigma) \tag{1}$$

It is not difficult to prove that if (1) holds, then $p(t)$ is continuous.

Regarding the proof of (1), the ideas are the following ones:

• For $\epsilon > 0$, you can find $X \in P[0,1]$ such that $$L(\sigma) - \epsilon < \Sigma X \le L(\sigma).$$
• Pick-up a modulus of continuity $\eta$ for $\sigma$, which is possible as $\sigma$ is continuous on the compact $[0,1]$, hence uniform continuous. Then take $\delta=\min(\eta,X_i-X_{i-1})$.
• Now for $Y \in P[0,1]$ with $\vert Y \vert < \delta$ and $\widetilde{Y} = X \cup Y$, you have $$L(\sigma) - \epsilon < \Sigma X \le \Sigma \widetilde{Y} \le L(\sigma).$$