Suppose $f$: $[0, 1]\longrightarrow\mathbb{R}$ satisfies $$\lim\limits_{x\longrightarrow a^-} f(x)=\lim\limits_{x\longrightarrow a^+} f(x)$$ everywhere, and that this value is finite.

The set of discontinuities of $f$ must be countable, but how large (in any other sense) can it get?

In particular, does any prescribed countable $D\subset [0, 1]$ work?

  • $\begingroup$ Thomae's function gives you $[0,1] \cap \mathbb{Q}$. $\endgroup$ – Daniel Fischer Jan 18 '16 at 15:31
  • $\begingroup$ @DanielFischer I don't see how that satisfies the original property. $\endgroup$ – Ian Jan 18 '16 at 15:33
  • $\begingroup$ The one-sided limits at all points are $0$, since the denominators of a sequence in $\mathbb{Q}\setminus \{a\}$ converging to $a$ must converge to $+\infty$. $\endgroup$ – Daniel Fischer Jan 18 '16 at 15:34

Yes, every countable set $D\subset [0,1]$ works. If $D$ is finite, we can take the characteristic function $\chi_D$, so in the following assume $D$ infinite. The construction is analogous to Thomae's function.

Let $(d_n)_{n \in \mathbb{N}\setminus \{0\}}$ be an enumeration of $D$. Define

$$f(x) = \begin{cases} \frac{1}{n} &, x = d_n \\ 0 &, x\notin D.\end{cases}$$

Then $f$ has one-sided limits $0$ everywhere, and is discontinuous at all $x\in D$ (and nowhere else).


Here is an answer to a related question, based off of this Is there a monotonic function discontinuous over some dense set?, with similar (but not quite the same as yours) discontinuities. Let $D=\{a_n:n\in\mathbb{N}\}$ be a countable subset of $\mathbb{R}$. Let $f:\mathbb{R}\to\mathbb{R}$ be defined by $$ f(x)=\sum_{a_n\leq x}\frac{1}{2^n}. $$ The above sum makes sense since the series converges absolutely. Moreover, at each $a_n$, we have $$ \lim_{x\to a_n^-}=\sum_{b_n<a_n}\frac{1}{2^n}<\sum_{b_n\leq a_n}\frac{1}{2^n}=f(a_n) $$


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