Discontinuous functions without removable discontinuities A function $f:\mathbb{R}\rightarrow \mathbb{R}$ has a removable discontinuity at a given real $x$ in case the left and right limits are equal but not to the function value, i.e. $f(x+)=f(x-)$ but $f(x)\ne f(x+)$.
Which function classes consist of (possibly discontinuous) functions that do not have removable discontinuities?
I have two fairly nice examples (see below), but would be interested in more (that are not variations of the below).

*

*The cadlag functions (continuous on the right, limit on the left).


*'Regularised' functions, i.e. such that $f(x)=\frac{f(x+) + f(x-)}{2}$ everywhere.
The second kind is relevant in the study of Fourier series.  I welcome function classes from logic, i.e. they need not be mainstream mathematics.  I would also welcome references on early uses of the above or new classes.
 A: The indicator function of any set with no isolated points is a simple example (eg: intervals, open sets).
One more would be the class of precise representatives in the sense of geometric measure theory, defined for locally integrable functions $f$ by
$$f^*(x) := \lim_{y \to x} \frac{1}{|B_r (x)|} \int_{B_r(x)} f(y) \, dy$$
wherever the limit exists, and $f^*(x) = f(x)$ otherwise.
A: I claim that one can always get rid of all removable discontinuities of a function to obtain a function without removable discontinuities. For generality, we shall work in the framework of general topology and ideals. In this post, we shall not automatically assume that a mapping $f:X\rightarrow Y$ between two topological spaces is continuous.
If $f:X\rightarrow Y$ is a function with removable discontinuities, then would like to replace $f$ with a function $g$ without any removable discontinuities by defining $g(x_0)=\lim_{x\rightarrow x_0,x\neq x_0}f(x)$ whenever this limit exists and $g(x_0)=f(x_0)$ otherwise. We are not done yet since the function $g$ may have new removable discontinuities, but if we transfinitely iterate this procedure of removing points of discontinuity, then we will eventually end up with a function without any removable discontinuities. We shall show that this transfinite procedure always converges.
Let $\mathcal{I}\subseteq P(X)$ be an ideal. We say that $\mathcal{I}$ is nowhere full if each $R^\circ=\emptyset$ for each $R\in\mathcal{I}$
Let $X,Y$ be topological spaces where $Y$ is Hausdorff. Suppose that $\mathcal{I}\subseteq P(X)$ is a nowhere full ideal. Let $f:X\rightarrow Y$ be a function. Define a mapping $\hat{f}:X\rightarrow Y$ by setting $\hat{f}(x_0)=y_0$ if there exists some $E\in\mathcal{I}$ where $\lim_{x\rightarrow x_0,x\not\in E}f(x)=y$ and where $\hat{f}(x_0)=f(x_0)$ otherwise. We need for $Y$ to be Hausdorff in order for the limit to be unique.
If $f:X\rightarrow Y$, then define a function $f^{(\alpha)}:X\rightarrow Y$ for each ordinal $\alpha$ by setting $f^{(0)}=f$, $f^{(\alpha+1)}=\hat{f^{(\alpha)}}$, and
$f^{(\lambda)}=\lim_{\alpha<\lambda}f^{(\alpha)}$ where the limit is taken pointwise. We are using the same notation as we do for differentiation since $f^{(\alpha)}$ is similar to the Cantor-Bendixson derivative.
Theorem: Suppose that $X,Y$ are topological spaces with $Y$ regular. Let $f:X\rightarrow Y$. Then

*

*$f^{(\alpha)}$ is well-defined for all $\alpha$.


*If $f^{(\gamma)}(x_0)=f^{(\gamma+1)}(x_0)$ and $\gamma<\alpha$, then $f^{(\gamma+1)}(x_0)=f^{(\alpha)}(x_0)$.


*If $U$ is open and $E\in\mathcal{I}$, then
$f^{(\alpha)}[U\setminus E]\subseteq \overline{f^{(\beta)}[U\setminus E]}$ whenever $\beta\leq\alpha$.


*There is an ordinal $\beta<|X|^+$ where for all $\gamma\geq\beta$, we have
$f^{(\gamma)}=f^{(\beta)}$.
Proof: We observe that $4$ follows from $1$ and $2$. If we set $A_\alpha=\{x\in X\mid f^{(\alpha)}(x)\neq f^{(\alpha+1)}(x)\}$, then the sets $(A_\alpha)_{\alpha}$ are disjoint, so since there are more ordinals than elements in $X$, there must be an ordinal $\beta<|X|^+$ with $A_\beta=\emptyset$. But this means that if $\gamma\geq\beta$, then $f^{(\gamma)}=f^{(\beta)}$.
We shall now prove $1,2,3$ using simulaneous transfinite induction.

*

*The only way in which $f^{(\alpha)}$ can be undefined is where $\alpha$ is a limit ordinal and the pointwise limit $\lim_{\gamma\rightarrow\alpha}f^{(\gamma)}$ does not exist. But the pointwise limit $\lim_{\gamma\rightarrow\alpha}f^{(\gamma)}$ always exists by $2$.


*If $\alpha$ is a limit ordinal, then $f^{(\alpha)}(x_0)=\lim_{\beta\rightarrow\alpha}f^{(\beta)}(x_0)=f^{(\gamma+1)}(x_0)$. Suppose now that $\alpha=\beta+1$. Then for $x_0\in X$, there is some $E\in\mathcal{I}$ where $\lim_{x\rightarrow x_0,x\not\in E}f^{(\gamma)}(x)\rightarrow f^{(\gamma+1)}(x_0)$. If $\alpha=\gamma+1$, then the proof is complete. Therefore, assume that $\alpha>\gamma+1$. Then $f^{(\beta)}(x_0)=f^{(\gamma+1)}(x_0)$. I now claim that
$\lim_{x\rightarrow x_0,x\not\in E}f^{(\beta)}(x)=f^{(\gamma+1)}(x_0)$ from which we can conclude that $f^{(\alpha)}(x_0)=f^{(\gamma+1)}(x_0)$ as well.
Suppose now that $V$ is an open neighborhood of $f^{(\gamma+1)}(x_0)$. Then by regularity, there is some open neighborhood $V_1$ of $f^{(\gamma+1)}(x_0)$ with $\overline{V_1}\subseteq V$. Therefore, there is some open set $U$ where $f^{(\gamma)}[U\setminus E]\subseteq\overline{V_1}$. However, by $3$, we know that
$f^{(\beta)}[U\setminus E]\subseteq\overline{f^{(\gamma)}[U\setminus E]}\subseteq \overline{V_1}\subseteq V$. This means that $f^{(\alpha)}(x_0)=f^{(\gamma+1)}(x_0)$.


*Suppose that $\beta\leq\alpha$. The result is clearly true when $\beta=\alpha$, so we can assume that $\beta<\alpha$. If $\alpha$ is a limit ordinal, then
$f^{(\alpha)}(x_0)=\lim_{\gamma\rightarrow\alpha}f^{(\gamma)}(x_0)\in\overline{f^{(\beta)}[U\setminus E]}.$ Suppose now that $\alpha=\delta+1>\beta$. Then $\delta\geq\beta$. By the induction hypothesis, we know that $f^{(\delta)}[U\setminus E]\subseteq\overline{f^{(\beta)}[U\setminus E]}$. Suppose now that $x_0\in U\setminus E$. If $f^{(\alpha)}(x_0)=f^{(\delta)}(x_0)$, then we know that $f^{(\alpha)}(x_0)\in\overline{f^{(\beta)}[U\setminus E]}$. Suppose now that $f^{(\alpha)}(x_0)\neq f^{(\delta)}(x_0)$. Then there is some $F\in\mathcal{I}$ where $\lim_{x\rightarrow x_0,x\not\in F}f^{(\delta)}(x)=f^{(\alpha)}(x_0)$. In this case, we have $\lim_{x\rightarrow x_0,x\not\in 
 E\cup F}f^{(\delta)}(x)=f^{(\alpha)}(x_0)$ as well. However, since
$f^{(\delta)}[U\setminus(E\cup F)]\subseteq\overline{f^{(\beta)}[U\setminus E]}$, we know that $f^{(\alpha)}(x_0)\in\overline{f^{(\beta)}[U\setminus E]}$ as well.

Q.E.D.
Therefore, define $f^{(\infty)}=f^{(\alpha)}$ where $\alpha$ is an ordinal where $f^{(\alpha)}=f^{(\alpha+1)}$. If $\mathcal{I}$ is contains all singletons, then $f^{(\infty)}$ has no removable discontinuities. If $\mathcal{I}$ is the collection of all finite subsets of $X$, then the functions of the form $f^{(\infty)}$ are precisely the functions $f:X\rightarrow Y$ with no removable discontinuities.
A: The class of functions whose restriction to any interval is onto has only nonremovable discontinuities. An example of such a function is the Conway base 13 function.
If a function remains onto when restricted to any interval, then it will be discontinuous at every point, and none of these discontinuities will be removable.
It is interesting to notice that despite their bad behavior, these functions all fulfill the intermediate value theorem — every intermediate value strictly between $f(a)$ and $f(b)$ is realized at some point between $a$ and $b$.
Another class: the functions $f$ whose graph is dense in the plane.
Another class: the functions $f$ whose graph has no isolated points. Every removable discontinuity comes from an isolated point.
