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Let $f : [0,1] \to \left\{ 0, 1 \right\}$ be a function that has at each point a discontinuity of the second kind. Is $f$ measurable if we equip the domain with the Borel or even Lebesgue $\sigma$-algebra and the range with the discrete $\sigma$-algebra?

The indicator of the Cantor set is not a counterexample since it is not discontinuous everywhere. The indicator of the rationals is nowhere continuous and is measurable.

This can be helpful in order to search for counterexamples: If $A \subseteq [0,1]$ is a set such that both $A$ and $[0,1] \setminus A$ are dense in $[0,1]$ then the indicator of $A$ is nowhere continuous. So, the question can be weakened: Is every such set $A$ necessarily measurable?

Wikipedia says that the set of discontinuities is an $F_\sigma$ set, thus a countable union of closed sets and therefore Borel measurable. In our case it is the whole interval $[0,1]$, so this statement doesn't help.

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    $\begingroup$ (1) How do you topologize $\{0,1\}$? (2) What's the second kind of discontinuity? $\endgroup$ – Asaf Karagila Oct 23 '14 at 11:53
  • $\begingroup$ $\{ 0, 1 \}$ as a subset of $\mathbb{R}$, so discrete topology. A point x is discontinuity of the second kind is such that both the left limit and the right limits of $f$ at $x$ do not exist. $\endgroup$ – yadaddy Oct 23 '14 at 11:56
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Using the axiom of choice we can construct a Bernstein set, which is a set $A\subseteq[0,1]$ such that no perfect set is a subset of $A$. The construction is such that $[0,1]\setminus A$ is also a Bernstein set, and $A$ meets every perfect set, but contains none. We list all the perfect sets (there are $2^{\aleph_0}$ of them, and we use the smallest possible order type) then we choose from each one two points, one which will enter $A$ and the other which will not. At each step we only chose $<2^{\aleph_0}$ points, so we can continue and choose new points at each step. This ensures that $A$ and its complement are both meeting every perfect subset of $[0,1]$ (and therefore neither contains any).

In particular $A$ is dense co-dense and not measurable. Not Lebesgue and certainly not Borel measurable.

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  • $\begingroup$ Thanks for the hint. I didn't know of Bernstein sets before. It seems to me that your construction is a typical one for such sets. So, we have a function that is nowhere continuous and not Lebesgue measurable. But does it then follow that it has everywhere discontinuities of the second kind? (It seems to me to be a stronger version of "nowhere continuous".) $\endgroup$ – yadaddy Oct 23 '14 at 14:21

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