# $f[a,b]$ has measure zero

Let $f:[a,b]\to \mathbb R^n$ ($n>1$) be a rectifiable continuous curve. I want to prove $f([a,b])$ has measure zero, i.e., for every $\epsilon>0$ there are blocks $\{C_i\}_{i=1}^{\infty}$ covering $f[a,b]$ such that $\sum_{i=1}^{\infty} \text{vol}\ C_i<\epsilon$.

Following the comments to this question below, it's false when $n=1$.

I've already tried to use the function is continuous and rectifiable without any success.

• It is clearly false if $n = 1$, since the identity function on $[a,b]$ is a rectifiable continuous curve. Jan 20 '17 at 19:45
• @Vik78 I edit the question, thank you Jan 20 '17 at 19:57

Intuitively in $\Bbb R^2$ the curve has a given arc length $L$. You can surround the curve with a shape of width $\epsilon$ and the area will be no more than $L\epsilon$ (plus some small ends). Now let $\epsilon \to 0$ It works the same in higher dimensions. This shows why it works in $2$ and higher dimensions and fails in $1$. You can cover a curve of length $L$ with $L/\epsilon$ balls of diameter $\epsilon$. In dimensions higher than $1$ the volume of the balls decreases with a power of $\epsilon$, so the sum of the volumes decreases as $\epsilon$ decreases. This doesn't happen in one dimension. Use your definition of rectifiable to show that the covering by balls works.

Although Ross already gave a correct way to tackle the problem, here is another one, which relies on Lipschitz functions and their properties:

Theorem 1. A continuous path is rectifiable if and only if it is a parametrization of a Lipschitz path.

Theorem 2. If $$X \subset \mathbb{R}^n$$ has measure zero and $$f: X \rightarrow \mathbb{R}^n$$ is locally Lipschitz then $$f(X)$$ has measure zero in $$\mathbb{R}^n$$.

It suffices to assume that $$f$$ is a Lipschitz continuous path. That's because if $$f$$ is not, we can obtain a Lipschitz path $$g$$ such that $$f$$ is a parametrization of $$g$$ (and thus their images will be the same).

Now let $$0 \in \mathbb{R}^{n - 1}$$. Then, defining the function: \begin{align} \phi: [a, b] \times \{ 0 \} &\rightarrow \mathbb{R}^n \\ (x, y) &\rightarrow f(x) \end{align}

we get that $$\phi$$ is also Lipschitz. Using this, combined with the fact that the set $$[a, b] \times \{ 0 \}$$ is a subset of $$\mathbb{R}^n$$ with measure zero, we obtain that $$Im(\phi) = Im(f) = f([a, b])$$ has measure zero.

It is enough to prove that every line in $R^{n}$ has measure zero. It follows that every "rectification" of f has measure zero.

• what is a "rect"? Jan 20 '17 at 20:13
• i mean "line". i am wrong anyway. sorry Jan 20 '17 at 20:20