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Let $1_\mathbb{Q}$ be the indicator of the rational numbers. Is $1_\mathbb{Q}$ continuous almost everywhere?

From what I understand, this is not true. I would say that $1_\mathbb{Q}$ is continuous nowhere. But if I restrict the domain to $\mathbb{R}\setminus\mathbb{Q}$ (i.e. I remove a set of measure zero), then $1_\mathbb{Q} \equiv 0$ and is continuous. Some of the responses I've seen seem to consider this as being "continuous almost everywhere". For example: Measurability of almost everywhere continuous functions

Which definition is correct/standard?

klein4
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    A function is continuous almost everywhere if the set of discontinuity is a null set. Obviously, $\boldsymbol 1_{\mathbb Q}$ is not, because as you remarked, it's continuous nowhere, and thus the set of discontinuity has infinite measure. Nevertheless, it's has a continuous represent, i.e. there is a continuous function $f$ s.t. $f(x)=\boldsymbol 1_{\mathbb Q}$ almost everywhere. Namely, if $f(x)=0$ for all $x$, then $f(x)=\boldsymbol 1_{\mathbb Q}$ almost everywhere. – Surb Oct 30 '20 at 20:15
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    A continuous a-e function is not the same as a function that is eual a.e. to a continuous function. – Tito Eliatron Oct 30 '20 at 20:17
  • In the post that I linked to, the proof seems to imply a weaker statement, i.e. a function $f:\mathbb{R}\to\mathbb{R}$ is measurable if $f|_E$ is continuous where $\mathbb{R}\setminus E$ is a set of measure zero. Is that correct? – klein4 Oct 30 '20 at 20:18
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    @klein4 Yes. More generally for a complete measure space - i.e. a measure space $(X,\mathcal E,\mu)$ such that $A\in\mathcal E$ for all $A\subseteq X$ such that there is some $B\in\mathcal E$ such that $\mu(B)=0$ and $B\supseteq A$ - a function which is equal almost-everywhere to a measurable function is measurable. –  Oct 30 '20 at 20:44

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When you restrict to $\mathbb{R} \setminus \mathbb{Q},$ you're changing the topology. This is a subtle but important point.

The criterion you describe says that in the original topology, the function is continuous at almost every point. This is not the same as "when you remove a null set, the restricted function is continuous on that subspace topology."

A similar confusion arises in Lusin's theorem, which says that if $f : [0, 1] \rightarrow\mathbb{R}$ is measurable, then for any $\epsilon > 0$ there's $F$ so that $f$ restricted to $F$ is continuous and $m(F) > 1-\epsilon.$

Just because $f$ restricted to $F$ is continuous at a point $p \in F,$ doesn't mean that $f$ is continuous at $p$. $f$ and the restriction of $f$ onto $F$ are two different functions on two different topological spaces.