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If $f\colon E\rightarrow \mathbb{R}$ is measurable function, and $m(E)<\infty$, then by Lusin's theorem, restriction of $f$ to a large closed set is continuous.

It is in the proof, I saw, that $m(E)<\infty$ is used along with Egorove's theorem. If we drop the condition $m(E)<\infty$, then the usual proof will not work; but I am not sure, whether there is other proof, which don't use the hypothesis $m(E)<\infty$, nor I could understand whether the condition $m(E)<\infty$ can't be dropped.

Question: Is it necessary to consider $m(E)<\infty$ in Lusin's theorem?

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    If you can partition (cover) $E = \bigcup_n E_n$ with $m(E_n) < \infty$, then you can choose for each $n$ some subset $F_n \subset E_n$ with $m(E_n \setminus F_n) \leq \varepsilon \cdot 2^{-n}$ and $g_n : E_n \to \Bbb{R}$ continuous with $g_n = f$ on $F_n$. Then $m(E \setminus F) \leq \varepsilon$, where $F = \bigcup_n F_n$. You just have to find a way to "paste together" the individual pieces $(g_n)_n$ (so that the resulting $g$ is continuous, maybe by discarding some sets of small measure). To do this, it would be easiest to know something about $E$. Is $E \subset \Bbb{R}^n$? – PhoemueX Nov 21 '14 at 09:23

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