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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.

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    It is clearly false if $n = 1$, since the identity function on $[a,b]$ is a rectifiable continuous curve. – 2017-01-20
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    @Vik78 I edit the question, thank you – 2017-01-20

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

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    what is a "rect"? – 2017-01-20
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    i mean "line". i am wrong anyway. sorry – 2017-01-20
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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.