The preimage of a Lebesgue measurable set under a measurable function need not be measurable

Question

I am reading measure theory from Royden, and I am stuck in some of them. I have this question:

suppose $E$ is a measurable set and let $f: E \to \mathbb{R}$. Prove that : $f$ is measurable if and only if $f^{-1}(A)$ is measurable for any $A \subseteq \mathbb{R}$.

I know this is not true if measurable means "Lebesgue measurable", can anyone give a counterexample in details ?

Answer 1

Let $C \subset [0,1]$ denote the Cantor set and let $f : [0,1] \to [0,1]$ be the Cantor function.

The function $g(x) = f(x) + x$ has the property that $g : [0,1] \to [0,2]$ is strictly increasing, continuous, and maps the Cantor set to a set of positive measure. Thus $g(C)$ contains a nonmeasurable set $Z$, and $g^{-1}(Z) \subset C$ is measurable since $C$ has measure zero.

Now define $h = g^{-1}$. Then $h : [0,2] \to [0,1]$ is continuous and thus measurable, $g^{-1}(Z) \subset [0,1]$ is measurable, but $Z \subset [0,2]$ is not. Finally observe $h^{-1}(g^{-1}(Z)) = Z$.

Thus it is possible for $h$ to be measurable, $A$ to be measurable, but $h^{-1}(A)$ to be nonmeasurable.

Answer 2

What is your definition of measurability? Mine goes like this : a function $f : (X, \mathcal F) \to (Y, \mathcal G)$ is measurable if and only if $f^{-1}(\mathcal G)\subseteq \mathcal F$. At this point there are two things to do :

1. Know the definitions of all the objects involved in your question (Measurable spaces, $\sigma$-algebras, measurable functions, inverse images). Write them down explicitly, it helps. See what are the elements of the sets involved and the properties they satisfy.

2. Think for a bit. (We're doing mathematics after all!)

Hope that helps,