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Lusin's theorem states that an almost-everywhere finite function is measurable if and only if it is a continuous function on nearly all its domain. However, Dirichlet function seems to be a counterexample.

The dirichlet function on the interval $[0,1]$ is defined as $D(x)=\left\{\begin{align}&0,x\in[0,1]\cap\mathbb Q\\&1,x\in\mathbb [0,1]\setminus \mathbb Q\end{align}\right.$. Since both $[0,1]\cap\mathbb Q$ and $[0,1]\setminus \mathbb Q$ are measurable, $D(x)$ is measurable.

However, it is well-known that $D(x)$ is nowhere continuous on $[0,1]$. I wonder where such a paradox comes from.

  • I think there must be some common mistakes for beginners in real analysis, though I cannot figure it out right now. – Drinzjeng Triang Mar 23 '21 at 11:13
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    The irrationals are nearly all of its domain. And restricted to the irrationals it's continuous. – Gerry Myerson Mar 23 '21 at 12:14
  • @GerryMyerson Get it, but how to define "continuous" when its domain contains no intervals? – Drinzjeng Triang Mar 23 '21 at 13:39
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    it is a continuous function on nearly all its domain --- Should be "it has a continuous restriction defined on nearly all of its domain". I don't know whether the following will help, but see this answer. – Dave L. Renfro Mar 23 '21 at 14:11
  • In my post, the definition of Lusin's theorem is taken from wikipedia. In fact the word a.e. (abbreviation of almost everywhere) is universally adapted in real analysis, e.g. "continuous a.e." means the discontinuity is zero-measured. From my perspetive, Lusin's theorem reveals that $D(x)$ is continuous on $\mathbb R\setminus\mathbb Q$, rather than $m(\mathbb R\setminus{x:w_D(x)=0})=0$. – Drinzjeng Triang Mar 23 '21 at 14:58
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    Continuity has nothing to do with intervals. The irrationals inherit a topology from the reals, and continuity means the inverse image of an open set is open. – Gerry Myerson Mar 23 '21 at 21:54
  • This question is answered here: https://math.stackexchange.com/questions/23372/lusins-theorem – Darren Ong May 13 '21 at 04:57

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