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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?

emka
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    The preimage of a Borel set under a continuous fnction is a Borel set. The preimage of Lebesgue measurable set under a continuous function may not be Lebesgue measurable. – Michael Greinecker Oct 04 '12 at 08:29
  • That is what I am trying to prove by using the Cantor function as a counterexample. – emka Oct 04 '12 at 08:30
  • Then I don't get what the question is. – Michael Greinecker Oct 04 '12 at 08:31
  • Im trying to verify if my proof is correct. – emka Oct 04 '12 at 08:32
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    @emka Regarding your proof: can you please clarify the last sentence? Is there a $W$ missing? If yes, you are writing $\psi^{-1}(W) = c$ and at the same time $\psi^{-1}(W)$ is non-measurable but $c$ is. – Rudy the Reindeer Oct 04 '12 at 08:34
  • In Royden's book he demonstrates that $\psi(x)$ maps a measurable set into a non-measurable set. Since $\psi(x)$ is strictly increasing on $[0,1]$ then it has continuous inverse. So when $\psi$ maps a measurable set to a non-measurable set; I was going to use the fact that $\psi^{-1}$ would pull back that non-measurable set to a measurable set. – emka Oct 04 '12 at 08:37
  • @MattN. I made a slight correction that will hopefully let this make more sense. – emka Oct 04 '12 at 09:02
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    @emka Ok, thank you. Could you also explain how the Cantor function maps a measurable set into a non-measurable set? Is this really true? – Rudy the Reindeer Oct 04 '12 at 09:09
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    Correct. Since $\psi$ maps the Cantor set $C$ to a set $E$ with positive measure, there is a non-measurable subset $W$ of $E$ and $V = \psi^{-1}(W)\subset C$ is measurable because $C$ is a null set. Consider $h = \psi^{-1}$. Then $h$ is continuous and hence measurable as a function $([0,1],\mathcal{B}) \to ([0,1],\mathcal{B})$. Since $h^{-1}(V) = \psi(V) = W$ isn't measurable, $h$ isn't measurable as a function $([0,1],\mathcal{L}) \to ([0,1], \mathcal{L})$. Here $\mathcal{B}$ and $\mathcal{L}$ are the Borel and Lebesgue $\sigma$-algebras. – commenter Oct 04 '12 at 09:11
  • Related. – Rudy the Reindeer Oct 04 '12 at 09:15
  • @MattN. It is proven in my textbook - Royden's Real Analysis. – emka Oct 04 '12 at 09:18
  • @commenter $\psi$ or $\bar{\psi}$. – emka Oct 04 '12 at 09:19
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    I wrote $\psi$ for $\bar{\psi}$, sorry. Write \bar{\psi}(x), not \bar{\psi(x)}. Compare $\bar{\psi}(x)$ with $\bar{\psi(x)}$. – commenter Oct 04 '12 at 09:25
  • Actually, do I have to even have to divide by two. $\psi(x)=\phi(x)+x$ maps a set of positive measure to a non-measurable set. – emka Oct 04 '12 at 13:19

1 Answers1

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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.

Rudy the Reindeer
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