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Does there exist an $f:\mathbb{R}\rightarrow\mathbb{R}$ that is bounded such that for any $a$ then $f^{-1}(a,+\infty)$ is open but $f$ is discontinuous everywhere?

Such a function seems too likely to exist, because it seems like you can just grab a nice continuous function, distort it slightly so that it won't ruin lower semicontinuous property but enough to make it lose continuity everywhere. Yet it seems much harder than I thought at first.

I have not been able to construct one. Though I figure out that there cannot exist an onto and monotone function from the natural number into the range of $f$. This basically ruled out construction using series of scalar multiple of characteristic functions.

I am also investigating a non-constructive method: trying to prove that discontinuous everywhere function are dense in the uniform metric space of function, and that lower semicontinuous function subspace is open in that space.

So anyone have any other ideas? Or can help me with this current ideas?

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  • $\begingroup$ If $f:\mathbb R\to\mathbb R$ is semicontinuous, then the set $\{x:f\text{ is discontinuous at }x\}$ is meager. $\endgroup$ – bof Oct 22 '13 at 8:24
  • $\begingroup$ Can you give a hint on this? I thought of the possibility, but I can only prove it with the extra condition as stated above. And are you saying that it doesn't matter if the function is bounded or not? Thanks. $\endgroup$ – Gina Oct 22 '13 at 8:36
  • $\begingroup$ A semicontinuous function $f:\mathbb R\to\mathbb R$ belongs to the first Baire class, i.e., it's a pointwise limit of continuous functions. $\endgroup$ – bof Oct 22 '13 at 9:42

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