1.21

Let $f: \mathcal{C} \to [0,1]$ be the Cantor-Lebesgue function. It's continuous, and surjective. Certainly, for any interval in $\mathbb{R}$, $\exists \mathcal{N}$, non-measurable. (Simply scale the non-measurable set constructed on p.24.)

Since $f$ is continuous, and $\mathcal{N} \subset [0,1]$, the pre-image, $f^{-1}(\mathcal{N}) \subset \mathcal{C}$.

Notice: $$0 \leq m^*(f^{-1}(\mathcal{N})) \leq m^*(\mathcal{C}) = 0$$ ...and thus, $$m^*(f^{-1}(\mathcal{N})) = 0$$ ...and since all sets of outer-measure zero are measurable (by Property 2, in Section 1.3) $f^{-1}(\mathcal{N})$ is a measurable set. Thus, the continuous map $f$ maps a measurable set to a non-measurable set.