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A mapping $f$ from $\mathbb R$ to $\mathbb R$ is called a choice function if, for any $x, y \ {\rm in}\ \mathbb R$, $f(x)-x \in\mathbb Q$ and $f(x)=f(y)$ whenever $x-y$ is rational.

My questions is: Is there a Lebesgue-measurable choice function?

Note: Here I use the equivalence relation for the construction of Vitali sets: x and y are in the same equivalence class iff x-y is rational. So choice functions pick up one element from each equivalence class as its representative value.

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Did you leave something out? Wouldn't a constant function be a measurable choice function under this definition? –  MJD Mar 25 '12 at 10:18
    
Yes as stated: for example $f(x)=0$. But not if you require $f(x)=f(y)$ only if $x-y$ is rational. –  Henry Mar 25 '12 at 10:20
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Shouldn't it read: $f(x) = f(y)$ if and only if $x-y$ is rational? Otherwise $f = 0$ is what you are looking for. –  martini Mar 25 '12 at 10:20
    
Sorry. I missed the condition that f(x)-x be a rational. Here I used the equivalence relation for the construction of Vitali sets: x and y are in the same equivalence class iff x-y is rational. My defined choice functions pick up one element from each equivalence class as its representative value. –  Joe Zhou Mar 25 '12 at 10:27
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There is at least no Borel measurable function with this property. Suppose there were a Borel measurable function $f$ with this property. Then it would have a Borel measurable graph $G=\{(x,y):y=f(x)\}$. Now the projection of $G$ on its second coordinate is an analytic set and hence Lebesgue measurable. But it is a Vitali set, contradicting its measurability. –  Michael Greinecker Mar 25 '12 at 12:17

2 Answers 2

up vote 9 down vote accepted

Suppose $f$ is measurable. Then $V = \{x \in \mathbb{R} \,:\,f(x) = x\}$ is a measurable representative system for the equivalence relation $x \sim y$ if and only if $x - y \in \mathbb{Q}$. That is to say, $V$ is a measurable Vitali set, but such a set doesn't exist.

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Your argument is correct, but it might not be obvious why $V$ is measurable. The argument goes like this: Intersect the Graph of $f$ with the diagonal. Now if $(A,\mathcal{A}_1)$ and $(A,\mathcal{A}_2)$ are measurable spaces, then the trace of the diagonal on $A\times A$ endowed with $\mathcal{A}_1\otimes \mathcal{A}_1$ is isomorphic to $(A,\sigma(\mathcal{A}_1\cup\mathcal{A}_2))$ under $(x,x)\mapsto x$. In the given case, the diagonal is isomorphic to the real line with the Lebesgue $\sigma$-algebra. Hence, $V$ is measurable. –  Michael Greinecker Mar 25 '12 at 13:19
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@Michael: much simpler: I look at the pre-image of $0$ under the measurable function $g(x) = f(x) - x$. –  t.b. Mar 25 '12 at 13:21
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slaps forehead –  Michael Greinecker Mar 25 '12 at 13:22

There is at least no Borel measurable function with this property. Suppose there were a Borel measurable function $f$ with this property. Then it would have a Borel measurable graph $G=\{(x,y):y=f(x)\}$. Now the projection of $G$ on its second coordinate is an analytic set and hence Lebesgue measurable. But it is a Vitali set, contradicting its measurability.

I don't know how to extend this to Lebesgue measurable functions.

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