Suppose f:R→R$f:\mathbf{R}\to\mathbf{R}$ is a function. Let S={x∈R|f is continuous at x}$S=\{x\in \mathbf{R}|f\text{ is continuous at }x\}$. Does S$S$ have any nice properties?
Here are some observations about what S$S$ could be:
- S$S$ can be any closed set. For a closed set S$S$, let g$g$ be a continuous function whose vanishing locus is S$S$ (for example, you could take g(x)$g(x)$ to be the distance of x$x$ from S$S$ if S$S$ is non-empty). Then define f(x)=g(x) if x∈Q and f(x)=0 otherwise. Then $$ f(x) = \begin{cases} g(x) &\text{if }x\in \mathbf{Q}\\ 0 &\text{otherwise}. \end{cases} $$ Then the continuous locus of f$f$ is exactly S$S$.
- S$S$ can be an open interval. For an open interval S$S$, define f(x)=0 if x∈S or x∈Q and f(x)=1 otherwise. $$ f(x) = \begin{cases} 0 &\text{if }x\in S \wedge x\in\mathbf{Q}\\ 1 &\text{otherwise}. \end{cases} $$ Then the continuous locus of f$f$ is exactly S$S$.
- S$S$ can be the complement of any countable set. Let T={t1,t2,t3,...}$T =
\{t_1,t_2,t_3,\ldots\}=\{t_i\}_{i\in\mathbf{N}}$ be a countable set, and let ∑ai$\sum_i a_i$ be some absolutely convergent series all of whose terms isare non-zero (like ai=1/2i$a_i=\frac{1}{2^i}$). Define
f(x) = ∑i such that ti < x a_i.
$$ f(x) = \sum_{i\text{ s.t. }t_i < x} a_i. $$ Then the continuous locus of f$f$ is exactly the complement of T$T$.
Here are some questions I'd like to know the answers to:
- Can S$S$ be any open set?
- Can S$S$ be non-measurable? (if f(x)=0$$ f(x) = \begin{cases} 0 &\text{if }x\in S\\ 1 &\text{otherwise}. \end{cases} $$ $f(x)=0$ if x∈S$x\in S$ and f(x)=1$f(x)=1$ otherwise, what will the continuous locus be?)