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The characteristic function of rationals in [0, 1] satisfies the hypothesis of Lusin's theorem. But it is no-where continuous on [0, 1]. But Lusin's theorem implies that it should be continuous on a positive measure subset of [0, 1]. What am I missing here?

Indrani Rao Mallick
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2 Answers2

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Enumerate the rationals in $[0,1]$ as $a_1,a_2,\ldots$. Remove an open interval of length $2^{-n}\epsilon$ centred at $a_n$ for each $n$. A compact set $E$ remains, of Lebesgue measure $\ge1-\epsilon$ and on this set $f$ is zero (so certainly continuous).

Angina Seng
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You are confusing continuity in the subspace topology with punctual continuity. The theorem says that for all $\varepsilon$ there is some compact subset $E$ such that $\mu(E)\ge1-\varepsilon$ and the set of continuity points of the function $\left. f\right\rvert_E:E\to \Bbb R$ is co-null. This doesn't imply that $f:[0,1]\to \Bbb R$ is continuous at any point of $E$. So to say, $\left.1_{\Bbb Q}\right\rvert_{[0,1]\setminus \Bbb Q}:[0,1]\setminus\Bbb Q\to \Bbb R$ is continuous at all points, but $1_{\Bbb Q}:[0,1]\to\Bbb R$ is discontinuous at all points.

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    +1 for precision, although a bit symbolically-cluttered for my tastes. Alternatively, as excerpted from this answer --- "Note that we're talking about the restriction of $f$ to $E$ being continuous, not $f$ itself being continuous at each point of $E$. The characteristic function of the rationals is not continuous at any point, (continued) – Dave L. Renfro Jul 12 '18 at 11:28
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    but after the removal of just countably many points (thus, "$\lambda({\mathbb R}-E) < \epsilon$" is satisfied in a very strong way), we get a constant function (thus, a function that is continuous in a very strong way)." – Dave L. Renfro Jul 12 '18 at 11:28