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Me and some classmates are stuck on a problem from our real analysis textbook (Royden/Fitzpatrick 4th Ed.), which we've given some thought. The question is:

"Suppose $f$ is a real valued function on $\Bbb R%$ such that ${f^-}^1(c)$ is measurable for each $c$. Is $f$ necessarily measurable?"

Our thoughts are that it is not true. The only example of nonmeasurable set we know is the Vitali Set $\Bbb V$, and the only nonmeasurable function we know of is the characteristic function of $\Bbb V$, denoted $X_{\Bbb V}$.

$X_{\Bbb V}(x) = \cases{ 1 & \text{if } x\in \Bbb V\cr 0 & \text{if } x\notin \Bbb V } $

I was thinking that possibly ${X_{\Bbb V}^-}^1(c)$ is measurable for all $c\in\Bbb R$, but I wasn't sure. The range of ${X_{\Bbb V}}(x)$ is just $[0] \bigcup [1]$.... Is this suitable?

Perhaps a more interesting question...on the nature of nonmeasurable sets and functions. Are there alternate constructions of nonmeasurable sets and functions?

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But $X_V^{-1}(1)=V$, right? – Gerry Myerson Nov 6 '12 at 6:17
up vote 5 down vote accepted

For the characteristic function of a non-measurable set the condition is not satisfied because $f^{-1}(\{1\})$ is not measurable. However, you can slightly modify your example by defining $f(x) = x+\chi_\mathbb{V}(x)$. Then it is easy to see that $f^{-1}(\{c\})$ has always at most two points (since $f$ is strictly increasing on $\mathbb{V}$ as well as on its complement), so the preimages of points are all measurable. However, $f$ is not measurable, because if it would be then $\chi_\mathbb{V}(x) = f(x)-x$ would be, too, as the difference of two measurable functions.

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Excellent, I knew we were getting close! Thanks – KUSH Nov 6 '12 at 6:36

With your $X_{\mathbb V}$ the function $f(x)=\frac{1}{2X_{\mathbb V}(x)-1}$ is not measurable because the preimage of the positives is $\mathbb V$. But the preimiage of a single point is at most one point.

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But $f(x)=1$ if and only if $2\chi_{\Bbb V}(x)-1=1$ if and only if $2\chi_{\Bbb V}(x)=2$ if and only if $\chi_{\Bbb V}(x)=1$ if and only if $x\in\Bbb V$. Hence the preimage of $1$ is $\Bbb V$, not a singleton or the empty set.... – Cameron Buie Nov 6 '12 at 6:37

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