Question : $f : \mathbb{S}^1\rightarrow \mathbb{S}^2$ is a continuous map s.t. for almost all $x\in \mathbb{S}^1$ in Lebesgue measure, there is an open ball $B(x,\delta)$ s.t. $f|B(x,\delta)$ is a rectifiable path. Then $f$ is not a onto map. How can we prove this ?
Remark : (1) There is a continuous map $f: [0,1]\rightarrow [0,1]^2$ s.t. a closure of $f([0,1])$ is $[0,1]^2$.
(2) $\mathbb{R}\subset \mathbb{R}^2$ has a measure $0$ by Sard theorem.
Reference : Exercise 6 in 45p. in Differential topology - Guillemin and Pollack
Counterexample: The Cantor function maps $[0,1]$ onto $[0,1]$ and is constant on each connected component of the complement of the Cantor set $E$. Then using a space-filling curve you can map $[0,1]$ continuously onto $\mathbb S^2$. Almost every point of $[0,1]$ is in $E^c$ and thus has a neighbourhood on which this function is constant. If you don't like your "rectifiable path" to be a single point, you can replace it by a small loop in each component of $E^c$.