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Define two real numbers to be rationally equivalent provided their difference is a rational number.

from Royden Real Analysis

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    $\begingroup$ Are you saying that this is a problem in Royden's book? $\endgroup$
    – Pete L. Clark
    Commented May 3, 2011 at 17:02

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If there were a choice function for this equivalence relation, there would be a set of real numbers that is not Lebesgue measurable. But the existence of such a set is not provable in ZF (i.e., without the axiom of choice). Even with the axiom of choice, i.e., in ZFC, it's consistent that no such set is definable. (On the other hand, with additional assumptions, like Gödel's axiom of constructibility, one can have a definable, in fact $\Delta^1_2$, choice function.)

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