Does the "continuous locus" of a function have any nice properties?

Question

Suppose $f:\mathbf{R}\to\mathbf{R}$ is a function. Let $S=\{x\in \mathbf{R}|f\text{ is continuous at }x\}$. Does $S$ have any nice properties?

Here are some observations about what $S$ could be:

• $S$ can be any closed set. For a closed set $S$, let $g$ be a continuous function whose vanishing locus is $S$ (for example, you could take $g(x)$ to be the distance of $x$ from $S$ if $S$ is non-empty). Then define $$ f(x) = \begin{cases} g(x) &\text{if }x\in \mathbf{Q}\\ 0 &\text{otherwise}. \end{cases} $$ Then the continuous locus of $f$ is exactly $S$.
• $S$ can be an open interval. For an open interval $S$, define $$ f(x) = \begin{cases} 0 &\text{if }x\in S \wedge x\in\mathbf{Q}\\ 1 &\text{otherwise}. \end{cases} $$ Then the continuous locus of $f$ is exactly $S$.
• $S$ can be the complement of any countable set. Let $T = \{t_1,t_2,t_3,\ldots\}=\{t_i\}_{i\in\mathbf{N}}$ be a countable set, and let $\sum_i a_i$ be some absolutely convergent series all of whose terms are non-zero (like $a_i=\frac{1}{2^i}$). Define $$ f(x) = \sum_{i\text{ s.t. }t_i < x} a_i. $$ Then the continuous locus of $f$ is exactly the complement of $T$.

Here are some questions I'd like to know the answers to:

• Can $S$ be any open set?
• Can $S$ be non-measurable? (if $$ f(x) = \begin{cases} 0 &\text{if }x\in S\\ 1 &\text{otherwise}. \end{cases} $$ $f(x)=0$ if $x\in S$ and $f(x)=1$ otherwise, what will the continuous locus be?)

Answer 1

Yes, here's a quick proof that any given $G_\delta$ (in $\mathbb{R}$) can be realized as the set of continuity points of some real-valued function.

Let $G$ be a given $G_\delta$ set in $\mathbb{R}$, meaning $G = \cap_{i=1}^\infty G_i$, each $G_i$ an open set. Define a function $f:\mathbb{R} \to \mathbb{R}$ as follows: $f(x)=0$ if $x$ is in $G$. If x is not in $G$, there is some $k$ such that $x$ is not in $G_k$; let $k$ be minimal with that property. Define $f(x)=1/k$ if $x$ is rational and $f(x)=-1/k$ if $x$ is irrational.

If I'm not very much mistaken, $G$ is precisely the set of continuity points of this $f$. I'm happy to leave this as an exercise for now :-) Let me know if you're not sure how to do it, or - worse - if I'm just wrong about the construction.

Answer 2

It's a standard result that the continuous locus is always $G_\delta$. For each $r>0$, let $U(r)$ be the set of points $x$ such that some neighborhood of $x$ maps into some ball of radius $r$. Then each $U(r)$ is open, and the continuous locus is their intersection. Conversely, given a $G_\delta$ set, I'm pretty sure it's not hard to construct a function with that continuous locus, though I don't remember how off the top of my head.

Answer 3

I hope nobody would mind if I try to do the exercise.

Clearly $f$ is continuous on $G$. Suppose $f$ is continuous on $x$ and $f\left(x\right)=1/k$. Take $\epsilon=1/k$. Let $U$ be any neighborhood of $x$. $U\cap G_1\cap .. \cap G_{k-1}$ contains an irrational number $y$. Hence $\left|f\left(x\right)-f\left(y\right)\right|=2/k>\epsilon$. (If $f\left(x\right)=-1/k$, take $y$ to be a rational number.)