Proving every rectifiable path (bounded in some sense) is integrable 
Prove that every rectifiable path $f:[a,b]\to\mathbb{R}^n$ is
  integrable

Where by rectifiable we mean a path that, for every partion $P = \{t_0,\cdots,t_n\} $ of its domain, we have that
$$l(f,P) = |f(t_1)-f(t_0)| + |f(t_2)-f(t_1)| + \cdots + |f(t_n)-f(t_{n-1})|$$
is bounded.
By integrable path we mean a path that has its coordinates as integrable functions.
It's intuitive to say that since it has a "finite graph" then it should be integrable. However, there is no assumption about continuity of the coordinate functions, so how I should prove they are integrable?
 A: You know that $\sup_P l(f,P)<\infty$.
Let $f_i$ denote the $i$th coordinate function of $f$.
We have $|f_i(t_{k+1})-f_i(t_k)|\leq|f(t_{k+1})-f(t_k)|$.
There are many ways to see this: Writing the vector difference in terms of coordinate differences and using the triangle inequality, or using the fact that the coordinate projection is 1-Lipschitz.
It follows from this estimate that $l(f_i,P)\leq l(f,P)$ for any partition $P$.
Therefore $\sup_P l(f_i,P)<\infty$ and so $f_i$ has bounded variation.
A function of bounded variation on the real line is the difference of two increasing functions.
See this question for a proof and more details.
Thus $f_i=g_i-h_i$, where both $g_i$ and $h_i$ are increasing.
Using the construction given in the linked question (or rather the answer to it), you can see that the increasing functions can be chosen to be bounded since the interval is compact.
If both $g_i$ and $h_i$ are integrable, then $f_i$ is, too.
It therefore suffices to know that a bounded increasing function on a compact interval is integrable.
I assume this is covered in whatever material you are using to study.
If not, let me know.
I believe it has already been covered on this site, but I couldn't find it yet.
