Calculating the lenght of a curve "piecewise" Let $ a,b\in \mathbb R $, $ a < b $, and let $ F $ be a normed space (you can take $ F = \mathbb R^n $, if you want). For a function $ \kappa\colon [a,b]\to F $ define
$$
L(\kappa) = \sup{\left\{\sum_{i = 0}^n{\lVert\kappa(t_i) - \kappa(t_{i - 1})\rVert} : \text{$ t_0,\dots,t_n\in [a,b] $, $ t_0 = a < \cdots < t_n = b $}\right\}}\text{,}
$$
and call $ \kappa $ rectifiable if $ L(\kappa) < +\infty $.
I'm trying to show that, if $ \gamma\colon [a,b]\to F $ is a rectifiable function, then for any strictly increasing sequence $ (t_n)_{n\in \mathbb N} $ of points $ t_n\in [a,b] $,
$$
L(\gamma{\restriction_{[t_0,t]}}) = \sum_{n = 0}^\infty L(\gamma{\restriction_{[t_{n - 1},t_n]}})
$$
where $ t $ is the limit of the sequence $ (t_n)_{n\in \mathbb N} $. I believe this fact more than my own existence, but so far I haven't succeeded.
It is easy to see that
$$
L(\gamma{\restriction_{[t_0,t_N]}}) = \sum_{n = 0}^N L(\gamma{\restriction_{[t_{n - 1},t_n]}})
$$
for any $ N\in \mathbb N $, $ N\geqq 1 $, so maybe it is sufficient to show that
$$
\lim_{N\to \infty}L(\gamma{\restriction_{[t_0,t_N]}}) = L(\gamma{\restriction_{[t_0,t]}})\text{.}
$$
So I noticed that for $ N\in \mathbb N $
$$
\left\lvert L(\gamma{\restriction_{[t_0,t_N]}}) - L(\gamma{\restriction_{[t_0,t]}})\right\rvert = \left\lvert L(\gamma{\restriction_{[t_N,t]}})\right\rvert
$$
but I don't know how much this can help.
 A: As @mechanodroid suggested, if $\gamma$ is assumed to be continuous at $t$, then you are in the clear.
I will start from the part where you might need continuity.
Let $M$ be arbitrary and let $\{s_0,s_1, s_2,..., s_M\}$ be an arbitrary set of points that partition $[t_0, t]$ into $M$ intervals. Also, select $N$ such that $t_N>s_{M-1}$, and create another partition of $[t_0,t]$ using the (union of) points $\{t_1,...,t_N\}\cup \{s_0,...,s_M\}$. Label these points $\{\tau^{M}_{0},...,\tau^M_{M+N}\}$ with allowing duplicated points.
Now, due to the triangle inequality (performed $N$ times)
$$
\begin{align}
\sum_{r=1}^{M}||\gamma(s_r)-\gamma(s_{r-1})|| &\leq \sum_{r=1}^{M+N}||\gamma(\tau^M_r)-\gamma(\tau^M_{r-1})||\\
&= ||\gamma(t)-\gamma(t_N)||+\sum_{r=1}^{M+N-1}||\gamma(\tau^M_r)-\gamma(\tau^M_{r-1})||.
\end{align}
$$
The sum on the right can be bounded by
$$
\sum_{r=1}^{M+N-1}||\gamma(\tau^M_r)-\gamma(\tau^M_{r-1})||\leq \sum_{n=1}^{N}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)})\leq \sum_{n=1}^{\infty}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)}).
$$
So, we proved that for any partition $\{s_0,s_1, s_2,..., s_M\}$, we can find an $N$ such that
$$
\sum_{r=1}^{M}||\gamma(s_r)-\gamma(s_{r-1})||\leq ||\gamma(t)-\gamma(t_N)|| + \sum_{n=1}^{\infty}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)}).
$$
If we have continuity at $t$, we can also select $N$ such that $||\gamma(t)-\gamma(t_N)||<\epsilon$ for arbitrarily small $\epsilon$, and obtain
$$
\sum_{r=1}^{M}||\gamma(s_r)-\gamma(s_{r-1})||\leq \epsilon + \sum_{n=1}^{\infty}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)});
$$
hence,
$$
L(\gamma\!\!\upharpoonright_{(t_{0},t)})\leq \sum_{n=1}^{\infty}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)}).
$$
The other inequality is easier. For partitions $\{s^n_0,...,s^n_{M_n}\}$ of the intervals $[t_{n-1},t_n]$, $n=1,...,N$, we have that the union is a partition for the interval $[t_0,t_N]$ and
$$
\sum_{n=1}^N\sum_{r=1}^{M_n}||\gamma(s^n_r)-\gamma(s^n_{r-1})||\leq L(\gamma\!\!\upharpoonright_{(t_0,t_N)})\leq L(\gamma\!\!\upharpoonright_{(t_0,t)}).
$$
By taking the supermum for all partitions and taking $N\to \infty$, we have that
$$
\sum_{n=1}^{\infty}L(\gamma\!\!\upharpoonright_{(t_{n-1},t_n)})\leq L(\gamma\!\!\upharpoonright_{(t_0,t)}).
$$
A: I don't think it's true as stated. Consider $\gamma : [0,1] \to \Bbb{R}$ defined as
$$\gamma(t) = \begin{cases}
0, &\text{ if }t \in [0,1\rangle,\\
1, &\text{ if }t =1.\\
\end{cases}$$
Then $\gamma$ is rectifiable, moreover $L(\gamma) = 1$. Consider the strictly increasing sequence $t_n = 1 - \frac1{n+1}, n \in \Bbb{N}_0$. We have $t_n \to t = 1$ and notice that for all $n \in \Bbb{N}_0$ holds
$$\gamma{\restriction_{[t_n,t_{n+1}]}} \equiv 0 \implies L(\gamma{\restriction_{[t_n,t_{n+1}]}}) = 0$$
and therefore
$$L(\gamma{\restriction_{[t_0,t]}}) = L(\gamma) = 1 \ne 0 = \sum_{n=0}^\infty L(\gamma{\restriction_{[t_n,t_{n+1}]}}).$$
Now if you require $\gamma$ to be continuous, I would expect the answer to be yes.
