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This is why I originally meant to ask with Are there Lebesgue-measurable functions non-continuous almost everywhere?

Does there exist a function $f\colon [0,1]\to\mathbb{R}$ such that:

  1. $f$ is Lebesgue measurable; and
  2. For every continuous $g\colon [0,1]\to\mathbb{R}$, the set of points where $f(x)\neq g(x)$ has positive measure?
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On the other hand, see Lusin's Theorem. –  Arturo Magidin Sep 23 '11 at 20:33
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The question listed as duplicate is actually a stronger statement. For this one, $f = 1_{[0,1/2]}$ is a counterexample. –  Nate Eldredge Dec 10 '11 at 2:07
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1 Answer 1

Yes. Fix any measurable set $A$ such that both $A$ and its complement have non-null intersection with each nonempty open interval. Examples are discussed here. Then the characteristic function of $A$ is as desired, since removing a null set does not change this intersection property, which rules out having a continuous extension.

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