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Lusin's theorem states that for every $\varepsilon$, for every borel measure $\mu$, for every function $f:\mathbb{R}^n\to\mathbb{R}^m$, for every open set $A$ of finite measure, there exists a compact set $K$ such that $\mu(A-K)<\varepsilon$ and $f$ restricted to $K$ is continue. Now, I'm wondering about what's the form of the compact $K$ in the case of the Dirichlet's function. Can you help me?

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    @Gabrio: Could you add what the Dirichlet function is?2011-02-23
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    First restrict to the unit interval. Then consider the complement $A-K$, which just needs to be open, have a small measure, and cover the rationals.2011-02-23
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    @Jonas T: The Dirichlet function is the characteristic function on the rationals.2011-02-23
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    @Colin: I don't understand, $K$ is not given.2011-02-23
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    @Gabrio: Colin is basically saying you can do what you did in your answer. He's not saying that K is given, but rather hypothetical, used as a way of stating the complementary problem, which solves the problem upon again taking complements.2011-02-23
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    @Jonas: Right, now I see.2011-02-23

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