Proof regarding function giving the length of a curve restricted to an interval Consider a parametric curve $\gamma(t): [a,b] \subset \mathbb{R} \to \mathbb{R}^n$. 
Let the interval $[a,b]$ be divided by a partition $\mathscr{P}=\{ t_1=a, t_2,..,t_{N-1},t_N=b \}$. 
The lenght of the curve is defined as 
$$L(\gamma):=\sup_{\mathscr{P}} \sum_{i=1}^{N}|\gamma(t_i)-\gamma(t_{i-1})|\tag{1}$$
Where $\mathscr{P}$ may vary among all the possible partition of $[a,b]$. 

Now consider a function $S(t):[a,b] \to [0,L(\gamma)]$ defined as 
$$S(t) := L(\gamma |_{[a,t]})$$
($\gamma |_{[a,t]}$ is the restriction of $\gamma$ to the interval $[a,t]$).

Given an $h>0$ how can I prove the following using the definition of lenght given above?

$$S(t+h)-S(t)=L(\gamma|_{[t,t+h]}) \tag{2}$$

($\gamma |_{[t,t+h]}$ is the restriction of $\gamma$ to the interval $[t,t+h]$).

Edit: As said  the proof of $(2)$ should make use only of the definition of lenght as given above, that is $(1)$. In particular it should not use the formula 
$$L(\gamma)=\int_{a}^{b} ||\gamma'(t)|| dt$$
 A: Let
$$M(P) = \sum_{i=1}^{N}|\gamma(t_i)-\gamma(t_{i-1})|$$
where $P$ is a partition of some (connected) subset of $[a,b]$ that will be specified.
$$S(t+h) \geq M(P)$$ for any partition $P$ of $[a,t+h]$.
Choose this partition $P$ so that $t\in P$. This allows us to express $M(P)=M(Q) + M \left(P^\prime \right)$, where $Q$ is a partition of $[t,t+h]$ and $P^\prime$ is a partition of $[a,t]$. Thus,
$$ S(t+h) \geq M(Q) + M \left(P^\prime \right) $$
Taking the supremum separately over $P^\prime$ and $Q$ gives you that
$$ S(t+h) \geq L\left(\gamma|_{[t,t+h]}\right) + S(t) $$
To prove the reverse inequality, consider any partition $P$ of $[a,t+h]$. We wish to show that 
$$ S(t) + L(\gamma|_{[t,t+h]}) \geq M(P) $$
If $t\in P$, it is easy to see that the above inequality holds.
If $t \notin P$, let $P^\prime$ be the partition $P$ augmented with $t$. That is, $P^\prime = P \cup \{t \}$. Using the triangle inequality, 
$$ M(P^\prime ) \geq M (P) $$
which gives us the desired inequality. Taking the supremum over $P$ gives us the result.
A: For $a \le x \le z   \le b$ write
$$
L(x,z) = L\left(\gamma|_{[x,z]}\right)  ,
$$
the length of the curve between parameter values $x$ and $z$. Then for $x < y < z$ prove $L$ is additive: 
$$
L(x,z) = L(x,y) + L(y,z).
$$
That follows from the fact that you get a partition of the interval $[x,z]$ from any pair of partitions of each part. The suprema add up as they should.
Then
$$
S(t) + L(t, t+h) = L(a,t) + L(t, t_h) = L(a, t+h) = S(t+h).
$$
You have to fiddle a bit to check this when $h$ is negative.
A: Your definition is equivalent to:
$$L\left(\gamma|_{[a,b]}\right) = \int_{a}^b \|\gamma'(t)\} \ dt$$
Moreover, $S(t)$ is just the arc-length function which you can express as:
$$S(t) = \int_a^t \|\gamma'(u)\| \ du \Rightarrow S(t+h) - S(t) = \int_{t}^{t+h} \|\gamma'(u)\| \ du = L\left(\gamma|_{t,t+h}\right)$$
