Stein 1.21

Show that there exists a continuous function $F$ that maps a measurable set to a non-measurable set.

$\textit{Example:}$ Consider $F$ to be the Cantor-Lebesgue function $$F: \mathcal{C} \longrightarrow [0,1]$$ which is continuous (in fact, since $\mathcal{C}$ is compact, $F$ is uniformly continuous). Exercise 32(b) establishes that given any subset of $\mathbb{R}$ with positive outer measure, we can embed a non-measurable subset in it. So, let $\mathcal{N} \subset [0,1]$ be non-measurable. Then, $F^{-1}(N) \subset \mathcal{C}$. Since $m(\mathcal{C})=0$ then $m(F^{-1}(N)) = 0$ and so $F^{-1}(N)$ is measurable. Of course, the restriction of a continuous function is still continuous. So, $$F|_{F^{-1}(N)}$$ is a continuous function that maps a measurable set to a non-measurable set. This gives an example showing that continuity is not a sufficiently strong enough condition to preserve measurability.
For the converse, consider $$\phi: \mathcal{N} \longrightarrow \{0\} \quad \text{where} \quad \phi = 0$$ Then, $\phi$ is trivially continuous and maps a non-measurable set to a measurable set.

One such condition on a function to preserve measurability is absolute continuity. Problem 3.19 establishes that if $f\in{AC(\mathbb{R})}$ then
(a) $f$ maps sets of measure 0 to sets of measure 0
(b) $f$ maps measurable sets to measurable sets
Part (a) is also known as the Lusin N property.