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(a) $f$ is continuous almost everywhere

(b) there exists a continuous function $g$ such that $f = g$ almost everywhere (on every set of non-zero measure)

(c) $f$ is nearly a continuous function (continuous everywhere but a set of measure epsilon, for small positive epsilon)

Here $f,g$ are functions from a measurable compact domain $D \subset \mathbb{R}$ to the whole real line, our measure is the Lebesgue measure, and we use the standard topology on $\mathbb{R}$.

Clearly (c) does not imply (a) as (a) is a stronger statement. Similarly, (a) implies (c).

It seems that if $f$ is a continuous function almost everywhere then we can set up a function $h$ that is continuous everywhere and set $h=f$ so (a) $\Rightarrow$ (b).

If there exist functions $g,f$ as in (b) then $f$ is nearly continuous so (b) implies (c).

Are these relations correct? Am I missing anything?

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  • $\begingroup$ I'm afraid I don't know what "nearly" means in (c). However $a\to b$ is wrong. Consider the Heavyside function. $\endgroup$ – Hagen von Eitzen Apr 12 '13 at 16:58
  • $\begingroup$ I took the liberty of adding a few tags to attract more potential responses. $\endgroup$ – Sammy Black Apr 12 '13 at 16:59
  • $\begingroup$ Good idea, I should have thought of that. And to clarify: a function is nearly continuous if it is continuous everywhere but a set of epsilon measure, for epsilon positive but small. $\endgroup$ – user72279 Apr 12 '13 at 17:06
  • $\begingroup$ Since the formulation is "nearly a continuous function" and not "a nearly continuous function", I'd like to ask for clarification again - this rather sounds like "For all $\epsilon>0$ there exists a continuous function $g_\epsilon$ such that $f$ and $g_\epsilon$ differ on a set of measure $<\epsilon$" instead of "There exists some $\epsilon>0$ such that the measure of the set of points where $f$ is not continuous is less than $\epsilon$" $\endgroup$ – Hagen von Eitzen Apr 13 '13 at 16:05
  • $\begingroup$ I stand corrected and the definition of nearly continuous seems to be yet another concept (according to math.stackexchange.com/questions/38989/…), which I will use in my answer $\endgroup$ – Hagen von Eitzen Apr 13 '13 at 16:32
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On the set $D=[-1,1]$ consider the function $$f(x)=\begin{cases}0&\text{if }x\le0\\1&\text{if }x> 0\end{cases}$$ has property (a) because it is continuous everywhere except at $x=0$, but not (b) because such continuous $g$ must coincide with $f$ on a dense set, hence $g^{-1}(\{0\})$ and $g^{-1}(\{1\})$ are nonempty, hence the open set $g^{-1}(]0,1[)$ is nonempty, contradiction. Thus (a) does not imply (b).

The function $$f(x)=\begin{cases}0&\text{if }x\notin\mathbb Q\\1&\text{if }x\in\mathbb Q\end{cases}$$ has property (b) (with $g(x)=0$), but not (a) as it is nowhere continuous. Thus (b) does not imply (a).

The answer to Nearly, but not almost, continuous shows that (c) implies neither (a) nor (b).

Note that if $f$ is continuous at $x$ then so is any restriction $f|_S$ provided $x\in S$. Therefore, if we have (b) with some continuous $g$, then the set $S=\{x\in D\mid f(x)\ne g(x)\}$ has measure $0<\epsilon$ and $f|_{D\setminus S}$ is continuous, i.e. $f$ has property (c). we conclude that $(b)\Rightarrow (c)$.

In summary, the only implication among the three properties is $$(b)\Rightarrow(c).$$

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