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

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 ?

• Since he knows it is not true, maybe "measurable" means "Lebesgue measurable". So we need one of those examples of a measurable function $F$ and Lebesgue measurable set $A$ such that $F^{-1}(A)$ is not Lebesgue measurable. – GEdgar Nov 6 '15 at 15:17
• exactly what I need – Safeyya Malkawy Nov 6 '15 at 15:23
• @Safeyya Malkawy : Can you precise with respect to which $\sigma$-algebras the function $f$ is measurable? – Patrick Da Silva Nov 6 '15 at 15:36
• $f : E \to \mathbb R$ is "measurable" iff, for every $c \in \mathbb R$, the set $\{x : f(x)>c\}$ is Lebesgue measurable. – GEdgar Nov 6 '15 at 15:43
• I mean Lebesgue measurable functions .. – Safeyya Malkawy Nov 6 '15 at 16:10

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.

• thank you ... but c in your example is Vitali's set not cantor ,isn't it ? ? since [ 0,1] is measurable – Safeyya Malkawy Nov 6 '15 at 16:16
• No, $C$ is the Cantor set. – Umberto P. Nov 6 '15 at 16:18
• As Mr. P. has just shown, if your function maps sets of measure zero to sets of positive measure, then it cannot have this property. [Because sets of positive measure contain nonmeasurable sets.] Now for some research: find out what are the necessary and sufficient conditions on a continuous function $f$ so that it maps measurable sets to measurable sets. – B. S. Thomson Nov 11 '15 at 21:22

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,

• When the function $f$ is real-valued, the $\sigma$-algebra $\cal G$ is the Borel $\sigma$-algebra, but saying $A \subset \mathbb R$ is measurable usually means that $A$ is an element of the Lebesgue $\sigma$-algebra. There is a bit more to this problem than checking definitions. – Umberto P. Nov 6 '15 at 15:20
• I know all of these definitions I just need a counterexample I always have a problem in finding them . thank you – Safeyya Malkawy Nov 6 '15 at 15:26
• @Umberto P. : I rarely work with the Borel $\sigma$-algebra when dealing with real-valued functions... I assumed it was the Lebesgue one the whole time. But then again this was not specified in the question so I cannot know. – Patrick Da Silva Nov 6 '15 at 15:35