Length of curve in metric space Let $(X, d)$ be a metric space, and $\gamma: [a,b]\to X$ be a curve. For any partition $P=\{a=y_0<y_1<\cdots<y_n=b\}$, 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)

*

*Take a partition $Q=\{y_0,\dots, y_n\}$ such that $\Sigma(Q)>L(\gamma)-\epsilon$.

*By the uniform continuity of $\gamma$, there exists $\delta>0$ such that $d(\gamma(t),\gamma(s))<\epsilon/n$ whenever $|t-s|<\delta$

*Let $P=\{x_0,\dots,x_m\}$ be any partition with $\|P\|<\delta$. For each $y_j$ there exists $x_{k(j)}$ such that $|x_{k(j)}-y_j|<\delta$. 

*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. 
