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.
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.
