Is the inverse of a measurable set under a continuous function is always measurable?

Question

From book Real Analysis by Royden & Fitzpatrick,

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

Answer 1

If you mean $A$ is Lebesgue measurable, then the conclusion is NO. A classic counterexample is constructed based on cantor function, from which we can obtain a homeomorphism (continuous bijection with continuous inverse) $f:[0,1]\to [0,2]$ and maps cantor set to a set with positive measure. We know any set with positive measure contains a nonmeasurable subset. Then consider the inverse of Cantor funciton $g=f^{-1}$. Then we know there exists a subset $A$ of cantor set (measure $|A|=0$, hence $A$ is Lesbesgue measurable) such that $g^{-1}=(f^{-1})^{-1}(A)=f(A)$ is NOT Lebesgue measurable.

However, if you mean $A$ is Borel measurable, then the conclusion is YES. You can show this by proving $\mathcal{A}=\{O\mid f^{-1}(O)\;\textrm{is Borel measurable}\}$ is a $\sigma$-algebra containing all open sets, hence contains Borel-sigma algebra, which contains all Borel measurable sets. Also note, if $f^{-1}(A)$ is Borel measurable, it's Lebesgue measurable.

Answer 2

If $A$ is a Borel set and $f$ is continuous, then $f^{-1}(A)$ is also a Borel set. This is because the Borel $\sigma$-algebra is generated by open sets, and $f^{-1}(A)$ is open whenever $A$ is open.

If $A$ is Lebesgue measurable, however, the inverse image $f^{-1}(A)$ may not be Lebesgue measurable.