Pre-image of a measurable set A is always measurable?

Let $f$ be a continuous function on a set $E$. Is it always true that $f^{-1}(A)$ is always measurable if $A$ is measurable?

I say no. We know that $\bar{\psi(x)}:=\frac{\phi(x)+x}{2}$ where $\phi$ is the Cantor (or Cantor-Lebesgue) function and $\bar{\psi}:[0,1] \rightarrow [0,1]$. We know this function maps a measurable subset of $C$ (the Cantor set) into a non-measurable set $W$.

A few observations about $\bar{\psi(x)}$. It is strictly increasing and continuous, so it has a continuous inverse $\bar{\psi(x)}^{-1}$. Thus $\bar{\psi(x)}^{-1}(W)=c \subset C$; that is, $\bar{\psi(x)}^{-1}(W)$ is non-measurable, but c is.

Is this correct?

Solution 1:

The question of whether the pre-image under a continuous function of a measurable set is measurable depends on two things: the topologies on the spaces (continuity) and the $\sigma$-algebras (measurability).

For example, if $f: \mathbb R \to \mathbb R$ is continuous and $\mathbb R$ comes with the standard topology and the Borel $\sigma$-algebra then the answer is, yes, all pre-images of measurable sets will be measurable.

On the other hand, we know by a cardinality argument that the Borel $\sigma$-algebra is properly contained in the Lebesgue $\sigma$-algebra so there exists a set that is Lebesgue measurable but is not Borel measurable. If you take $f$ to be the identity map $\mathrm{id}: \mathbb R \to \mathbb R$ with the standard topology on both domain and codomain but put the Lebesgue $\sigma$-algebra on the codomain and the Borel $\sigma$-algebra on the domain then any pre-image of a set that is not Borel measurable will not be measurable -- yet the function is continuous.