Counter example of Lusin's theorem

Question

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?

Answer 1

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).

Answer 2

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.