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Zymund states in his book Lusin's Theorem the following

Let $E$ be some Lebesgue measurable subset of $\mathbb{R}^d$ and let $f:E\rightarrow \mathbb{R}$ be some function. $f$ is measurable if and only if for all $\epsilon$, there exists some closed $F\subset \mathbb{R}^d$ such that $F \subset E, |E\setminus F|<\epsilon$ and $f|_{F}:F\rightarrow \mathbb{R}$ is continuous.

I wish to ask: Does the Theorem hold true for extended valued function $f: E\rightarrow \mathbb{R}\cup \{+\infty,-\infty\}$?

I searched online for this question, but there seems to be no discussion on this site yet. All I get are this post, this post, and this post, which are all discussions about the Hypothesis of Lusin's Theorem, but have not much to do with my question. Thanks in advance for help in all forms.

John Frank
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    You could deduce the theorem for $\mathbb R\cup{+\infty,-\infty}$ from the one for $\mathbb R$ using any homeomorphism between $\mathbb R\cup{+\infty,-\infty}$ and a bounded interval of $\mathbb R$, but it's more interesting to know that there is a Lusin's theorem for functions taking their values in a metrizable topological space (I'm not sure if I want to try writing the full proof though). – P. P. Tuong Nov 06 '24 at 14:42

2 Answers2

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If you equip the extended reals with the topology generated by the complements of compact sets and sets of the form $\{x < a\}$ and $\{x > a\}$, for all constants $a$ (i think this is sometimes called the two point compactification of the line) then its homeomorphic to a closed interval, using a map like $\operatorname{arctan}$. So we could just apply Lusin's to $\operatorname{arctan} \circ f$, obtain our $F$ and be done I believe.

user24142
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No, if $f$ only takes $\pm\infty$ values, no such $F$ exists. There is no notion of continuity for $\pm\infty$.

TheAlertGerbil
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