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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?

Guerlando OCs
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    a single number, like $l(f,P)$, is certainly bounded. A path is rectifiable if the set of these $l(f,P)$ is bounded when $P$ ranges over all partitions (and $n$, respectively, ranges over all positive integers). I do not get the "finite graph" part. Think of the following examples, where $0\le x \le 1$: (a) $x^2 \sin\frac1x$ and (b) $\sqrt{x} \sin\frac1x$ (I didn't verify the details but I think one of them ought to be rectifiable, though none has a "finite graph") – Mirko Apr 11 '17 at 02:13
  • I meant 'finite plot' – Guerlando OCs Apr 11 '17 at 04:13
  • Each coordinate of $f$ has bounded variation, so it is the difference of two nondecreasing functions. – G.Kós Apr 16 '17 at 14:06

1 Answers1

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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.

Joonas Ilmavirta
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