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I've come across this problem in my studies, where the book (Real Analysis (4th Edition) by Royden) gives an example of two measurable functions whose composition is nonmeasurable. The two functions are constructed as follows:

1) Let $\Phi$ be the Cantor-Lebesgue function, and define the function $\Psi\ R \rightarrow R$ by $\Psi(x)$=$\Phi(x)+x\ \forall x \in R$ which is continuous and strictly incerasing

2) $\chi_A$ which is the characteristic function of a measurable set $A$, while $\Psi(A)$ is nonmeasurable.

First I am confused why $\Psi^{-1}$ is measurable knowing that $\Psi(A)$ is nonmeasurable. Is continuity sufficient for it to be measurable?!

Secondly, I cannot prove that $f=\chi_A \circ \Psi^{-1}$ is nonmeasurable.


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Are all occurences of $\psi$ and $\Psi$ the same function? Could you reference the book so we might find more of the details of the example? – Kevin Carlson Oct 1 '12 at 1:19
up vote 1 down vote accepted

For your first comment, since $\Psi$ is continuous and strictly increasing and its domain is an interval we know that $\Psi^{-1}$ is continuous and strictly increasing. Continuity implies Borel measurability which implies Lebesgue measurability. (The inverse image of an open set under a continuous function is open (Borel)). Hence $\Psi^{-1}$ is measurable.

For your second comment, note that $$(\chi_A \circ \Psi^{-1})^{-1}((.5,1.5)) = (\chi_A \circ \Psi^{-1})^{-1}(\{ 1 \}) = \Psi(\chi_A^{-1}(\{ 1 \})) = \Psi(A)$$ which is nonmeasurable, so $\chi_A \circ \Psi^{-1}$ cannot be measurable.

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