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Lusin's theorem states that if $f:[a,b]\rightarrow\mathbb{R}$ is a Lebesgue measurable function, then for any $\epsilon>0$ there exists a compact subset $E$ of $[a,b]$ whose complement has Lebesgue measure less than $\epsilon$ and a function $g$ continuous relative to $E$ such that $f=g$ on $E$. My question is, what is an example of a measure on $\mathbb{R}$ for which this is not true?

I know that it's true for all Borel measures on $\mathbb{R}$ which assign finite measure to all bounded intervals, so a counterexample would either have to be a non-Borel measure or a Borel measure which assigns infinite measure to some bounded intervals.

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Try with the counting measure and the indicator function of the rationals.

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  • $\begingroup$ I'm maybe missing something obvious, but I don't see how the counting measure can provide the example. Lusin happens inside a set of finite measure, which in this case will be finite; and any function is continuous on a finite set. $\endgroup$ Commented Feb 26, 2020 at 15:16
  • $\begingroup$ What needs to be finite (in fact, less than $\varepsilon$) is the measure of $E^c$. If the counting measure of $E^c$ is less than $1$, then $E^c$ is actually empty. $\endgroup$
    – Federico
    Commented Mar 5, 2020 at 14:17

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