# Logical Relations Between Three Statements about Continuous Functions

(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?

• I'm afraid I don't know what "nearly" means in (c). However $a\to b$ is wrong. Consider the Heavyside function. – Hagen von Eitzen Apr 12 '13 at 16:58
• I took the liberty of adding a few tags to attract more potential responses. – Sammy Black Apr 12 '13 at 16:59
• 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. – user72279 Apr 12 '13 at 17:06
• 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 $\epsilon$" – Hagen von Eitzen Apr 13 '13 at 16:05
• 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 – Hagen von Eitzen Apr 13 '13 at 16:32

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).
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 $$(b)\Rightarrow(c).$$