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Let $E = [0,1]$. Lusin's theorem provides that a measurable $\space f: E\to R$ equals to some continuous $g$ at points of a set $F \subset E$. Must $f$ itself be continuous at any point (as a function on $E$)?

I have Lusin's Theorem defined as: "Let $\space f:E\to R$ be measurable. Then for each $\epsilon > 0$, there is a continuous function $g:R\to R$ and a closed set $F$ contained in $E$ such that $f=g$ on $F$, and $\operatorname{m}(E\setminus F) < \epsilon$

I understand that $f$ needs to be continuous on the interior of $F$ since $f$ coincides with a continuous function there, but I read somewhere that the interior of $F$ could be empty. Could someone please explain that part?

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If f is 1 at rational points and 0 at irrationals then F cannot have any interior point.

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