Endow $\omega_1$ with order topology. It is easy to show that each continuous function $f\colon \omega_1\to \mathbb{R}$ is eventually constant. Is the same true for Borel functions?
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It isn't true for Borel functions. To see this, observe that the set of successor ordinals (plus zero) is an open set in the order topology of $\omega_1$, and its complement is the set of limit ordinals, so these are both Borel sets in that topology. Thus, the characteristic function of the limit ordinals, with value $1$ on the limit ordinals and $0$ on successors and zero, is a Borel function, but it is not eventually constant.
answered Mar 31, 2012 at 16:34
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$\begingroup$ Your answer arrived while I was typing mine. $\endgroup$ Commented Mar 31, 2012 at 16:36
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$\begingroup$ And we had the same counterexample! $\endgroup$ Commented Mar 31, 2012 at 16:39
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$\begingroup$ The same right down to which piece should contain 0, a decision that could equally well have gone the other way. $\endgroup$ Commented Mar 31, 2012 at 17:37
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Assuming that by "Borel function" you mean (as is customary) a function such that the inverse image of each Borel set is Borel, the answer is no. Consider the characteristic function of the set $L$ of countable limit ordinals. The inverse image of any subset of $\mathbb R$ is $L$ (which is closed), the complement of $L$ (which is open), all of $\omega_1$, or empty.
answered Mar 31, 2012 at 16:35