Let $f:[a,b]\longrightarrow\mathbb{R}^n$ be a rectifiable path, $P=\{a=t_0<...<t_m=b\}$ a partition of $[a,b]$. I want to prove that $l(f)=\sum_{k=1}^m l(f_{|[t_{k-1},t_k]})$ (length of the path is equal to the length of its "parts").
I'm trying to correct my "proof", here is what I have tried.
For all $k\in\{1,\_\_,m\}$, let $Q_k=\{t_{k-1}=t_0^k<...<t_m^k=t_k\}$ be a partition of $[t_{k-1},t_k]=I_k$.
Observe that $f=\sum_{k=1}^m f_{|I_k}\chi_{I_k}$ and that $P\subset Q:=\bigcup_{k=1}^m Q_k=\{a=t_0=t_0^1<...<t_m^1=t_0^2<...<t_m^m=t_m=b\}=\{a=p_0<...<p_{m^2}=b\}$.
We have that $L(f,Q)=L(\sum_{k=1}^m f_{|I_k}\chi_{I_k},Q)=\sum_{l=1}^{m^2}||(\sum_{k=1}^m f_{|I_k}\chi_{I_k})(p_{l-1})-(\sum_{k=1}^m f_{|I_k}\chi_{I_k})(p_l)||=\sum_{i=1}^m\sum_{j=1}^m ||f_{|I_i}(t_{j-1}^i)-f_{|I_i}(t_j^i)||=\sum_{i=1}^m L(f_{|I_i},Q_i)$
(If at this point everything is right, this what I want to prove)
$l(f)=\sup_Q L(f,Q)=\sup_{\bigcup_{i=1}^m Q_i}\sum_{i=1}^m L(f_{|I_i},Q_i)=\sum_{i=1}^m \sup_{Q_i} L(f_{|I_i},Q_i)=\sum_{k=1}^m l(f_{|I_k})$.
Surely my proof has a lot of holes and stupid asumptions, please be patient. If it is too tricky to correct it, let me know a correct proof. Or If I can find it in a book, please let me know.
Up until the last line, everything looks good. At the end, you need to be a little more careful when taking supremums.
You do have : $$\sum_{i=1}^m L(f|_{I_i},Q_i) = L(f,Q) \leq l(f)$$
and therefore, by taking supremum on each $Q_i$ (which is any partition of $I_i$), we get : $$\sum_{i=1}^m l(f|_{I_i}) \leq l(f)$$
To get the other equality you need to take a supremum on any subdivision $P$ of $[a,b]$. We can do this the following way : let $P$ be any subdivision of $[a,b]$.
For $i\in \{ 1,\ldots, m\}$, let $Q_i$ be the trace of $P$ on $I_i$ i.e. the set of elements of $P$ which lie in $I_i$, to which we add the end-points of $I_i$ if they are not in $P$.
The, let $Q = \cup_{i=1}^m Q_i$ as in the OP.
Then, you have $P\subset Q$ and therefore : $$L(f,P) \leq L(f,Q) = \sum_{i=1}^m L(f|_{I_i},Q_i) \leq \sum_{i=1}^m l(f|_{I_i}) $$
By taking the supremum on $P$ we get : $$l(f) \leq \sum_{i=1}^m l(f|_{I_i})$$