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Let $(X, d)$ be a metric space, and $\gamma: [a,b]\to X$ be a curve. For any partition $P=\{a=y_0, one can associate to it the lengh of the "inscribed polygon" $\Sigma(P)=\sum_id(\gamma(y_i), \gamma(y_{i+1}))$ Then we define the length of the curve to be $L(\gamma)=\sup_{P\in \mathcal P}\Sigma(P)$ where $\mathcal P$ is the collection of all partitions of $[a,b]$. If the supremum is finite then we call the curve is rectifiable.

We denote: $\|P\|=\max_i|y_i-y_{i+1}|$

Now my question is the proof of the following statement: $\lim_{\|P\|\to 0}\Sigma(P)=L(\gamma)$

The hard part for me to prove the statement is if $P$ and $Q$ are two partitions, with $\|P\|\le \|Q\|$, we only know $\Sigma(P\cup Q)\ge\max(\Sigma(P), \Sigma(Q))$ and this won't give me any contradiction when we assume there is a sequence of partitions say $P_i$ with $\|P_i\|\to 0$ and $\Sigma(P_i)\le L(\gamma)-\varepsilon_0$ for some fixed $\varepsilon_0>0$. Anybody can help?

(btw. I thought that this may be similar with the proof of the Riemann sum for integrable function, but there one has the Osilation)

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    Thanks Davide, I've corrected the typo. We can assume the $\gamma$ is continuous and rectifible.2012-06-04

1 Answers 1

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  1. Take a partition $Q=\{y_0,\dots, y_n\}$ such that $\Sigma(Q)>L(\gamma)-\epsilon$.
  2. By the uniform continuity of $\gamma$, there exists $\delta>0$ such that $d(\gamma(t),\gamma(s))<\epsilon/n$ whenever $|t-s|<\delta$
  3. Let $P=\{x_0,\dots,x_m\}$ be any partition with $\|P\|<\delta$. For each $y_j$ there exists $x_{k(j)}$ such that $|x_{k(j)}-y_j|<\delta$.
  4. Use uniform continuity to estimate $\Sigma(\{x_{k(j)}\colon j=0,\dots,n\})$ from below.

There are some things to tidy up here, but this being homework, I'll leave the rest to you.