$\int_{a}^b \Vert \gamma'(t)\Vert dt=\lim_{\delta(P)\to0}\sum_{k=0}^n\Vert \gamma(t_{k+1})-\gamma(t_k)\Vert$ The length of a smooth curve can be defined as $$lg(\gamma)=\int_{a}^b \Vert \gamma'(t)\Vert dt$$
Denote $P=\{(t_0,\ldots,t_{n+1}): a=t_0<t_1<\ldots<t_{n+1}= b\}$
Now I would like to prove that $$lg(\gamma)=\lim_{\delta(P)\to0}\sum_{k=0}^n\Vert \gamma(t_{k+1})-\gamma(t_k)\Vert$$
 where $\delta(P)$ is the norm of the partition.
I can prove that $\lim_{\delta(P)\to0}\sum_{k=0}^n\Vert \gamma(t_{k+1})-\gamma(t_k)\Vert\le lg(\gamma)$ using the identity $\gamma(t_{k+1})-\gamma(t_k)=\int_{t_k}^{t_{k+1}}\gamma'(t)dt.$
The other direction seems harder, I tried to use the fact that $\gamma'$ is uniformly continuous to have somewhat like $$\Vert \gamma'(y)-\gamma'(x)\Vert\le \varepsilon\quad\mbox{for}\quad \vert x-y\vert\le\eta$$ 
How can I continue ?
 A: I assume that smothness means that $ \gamma \in C^{1}(( a, b )) $. Then, we
have by dominated convergence and a first order Taylor approximation that
\begin{align*}
  \int_{a}^{b} \| \gamma '(t) \| \, dt
  & =   \lim_{\delta(P) \to 0} 
        \sum_{j = 1}^{n} \| \gamma '(t_{j - 1}) \| ( t_{j} - t_{j - 1} ) 
    =   \lim_{\delta(P) \to 0} 
        \sum_{j = 1}^{n} \| \gamma '(t_{j - 1}) ( t_{j} - t_{j - 1} ) \| \\ 
  & =   \lim_{\delta(P) \to 0} 
        \sum_{j = 1}^{n} \| \gamma(t_{j}) - \gamma(t_{j - 1}) \|
        + 
        o( | t_{j} - t_{j - 1} |) 
    =   \lim_{\delta(P) \to 0} 
        \sum_{j = 1}^{n} \| \gamma(t_{j}) - \gamma(t_{j - 1}) \|,
\end{align*}
since 
\begin{align*}
  \sum_{j = 1}^{n} o(| t_{j} - t_{j - 1} |)
  & \le \sup_{j \le n} \frac{
          o(| t_{j} - t_{j - 1} |)
        }{
          | t_{j} - t_{j - 1} |
        }  
        \sum_{j = 1}^{n} | t_{j} - t_{j - 1} |  
    \stackrel{\delta(P) \to 0}{\to} 0 . 
\end{align*}
A: HINT:
The idea is to approximate $||\gamma(s)- \gamma(t)||$ with $|s-t|\cdot \|\gamma'(c)\|$ with $c$ between $s$ and $t$. Apply Lagrange to each component difference $\gamma_i(s) - \gamma_i(t)$, $i=1,2,3$. Each equals to $(s-t)\cdot \gamma_i'(c_i)$. You may take the same $c$ and this will not differ too much, since the $\gamma_i$'s are uniformly continuous. So, as the norm of the division tends to $0$, the sum will approach the integral of $\|\gamma'\|$. You can do this in one shot, not need for the other inequality. 
A: We use the Taylor approximation to write $\gamma(t_{k+1})-\gamma(t_{k})=\gamma'(t_{k})(t_{k+1}-t_{k})+o((t_{k+1}-t_{k})^{2}).$ Then we have \begin{align*}
\sum_{k=0}^{n}\|\gamma(t_{k+1})-\gamma(t_{k})\|&\geq\sum_{k=0}^{n}\|\gamma'(t_{k})\||t_{k+1}-t_{k}|-o((t_{k+1}-t_{k})^{2})\\
&\geq \sum_{k=0}^{n}\|\gamma'(t_{k})\||t_{k+1}-t_{k}|-(\max_{0\leq j\leq n}|t_{j+1}-t_{j}|)|t_{k+1}-t_{k}|\\
&=\left(\sum_{k=0}^{n}\|\gamma'(t_{k})\||t_{k+1}-t_{k}|\right)-(\max_{0\leq j\leq n}|t_{j+1}-t_{j}|)(b-a).
\end{align*}
The sum on the left is exactly the Riemann sum that we obtain when we choose the left endpoint of each subinterval of $[a,b]$ while approximating $\int_{a}^{b}\|\gamma'(t)\|\mathrm{d}t.$ Thus, given $\varepsilon>0,$ when $\delta(P)=\max_{0\leq j\leq n}|t_{j+1}-t_{j}|$ is sufficiently small (and we will also suppose that this is smaller than $\varepsilon/2(b-a)$), we have that $$\sum_{k=0}^{n}\|\gamma'(t_{k})\||t_{k+1}-t_{k}|\geq \int_{a}^{b}\|\gamma'(t)\|\mathrm{d}t-\varepsilon/2.$$ Combining this with the above, we get that for such small values of $\delta(P),$ $$\sum_{k=0}^{n}\|\gamma(t_{k+1})-\gamma(t_{k})\|\geq\ell(\gamma)-\varepsilon.$$ This completes the proof.
