rectifiable continuous function could you please help me with this question:
if $F:[a,b] \rightarrow \mathbb{R}^{m}$ is continuous and rectifiable,
then $ F ([a, b]) $ has $ m $ - measure zero.
Any suggestions are welcome.
 A: I assume you want $m>1$. For a subdivision $a=t_0<t_1<t_2<\ldots<t_n=b$, let $x_k = F(t_k)$, and let $r_k = \operatorname{diam} F([t_{k-1},t_k])$. Then $$F([a,b]) \subseteq \bigcup_{k=1}^n \overline{B(x_k,r_k)},$$ where $\overline{B(x,r)}$ is the $m$-dimensional closed ball with center $x$ and radius $r$. So $$\lambda_m(F([a,b])) \le C_m \sum_{k=1}^n r_k^m,$$ where $\lambda_m$ is $m$-dimensional Lebesgue measure and $C_m$ is the volume of the $m$-dimensional unit ball. Now on the other hand there exist parameters $t_{k,1}$ and $t_{k,2}$ with $t_{k-1} \le t_{k,1} \le t_{k,2} \le t_k $ and $|F(t_{k,2})-F(t_{k,1})| = r_k$, so that $$\sum_{k=1}^n r_k = \sum_{k=1}^n |F(t_{k,2})-F(t_{k,1})| \le L,$$ where $L$ is the length of $F([a,b])$. By uniform continuity, for every $\epsilon > 0$ we can find a subdivision as above such that $r_k < \epsilon$ for all $k=1,\ldots,n$. Then $$ \lambda_m (F([a,b])) \le C_m L \epsilon^{m-1}.$$
Since this is true for arbitrary positive $\epsilon$, we get that $\lambda_m(F([a,b]))=0$.
