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This is the statement of Theorem (Lusin) (taken from Royden):

Let $f$ be a real-valued measurable function on $E$ a measurable set of $\mathbb R$. Then for each $\epsilon>0$, there is a continuous function $g$ on $\mathbb{R}$ and a closed set $F\subseteq E$ for which $f=g$ on $F$ and $m(E\setminus F)<\epsilon$.

Another version of Lusin's Theorem is stated as follows:

Let $f$ measurable and finite a.e. on $E$ a measurable set of $\mathbb R$. For all $\varepsilon>0$ there is a closed set $F_\varepsilon\subset E$ s.t. $f|_{F_{\varepsilon}}$ is continuous and $m(E\backslash F_\varepsilon)\leq \varepsilon$.

Question: Is it necessary to consider that $f$ is finite a.e. in Lusin's theorem?

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Yes. If $f$ is not supposed to be finite, set $f(x)=\infty$ everywhere in its domain. Would Lusin's theorem hold?

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  • $\begingroup$ The focal point is that $f$ is a real-valued measurable function. $\endgroup$
    – unicornki
    Nov 10, 2019 at 11:16

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