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Assume that $f:\mathbb{R}\to\mathbb{R}$ is such a function that $\lim_{x\to a^-}f(x)$ exists for every $a\in\mathbb{R}$.

Then it is known that $f$ is continuous except for a countable set. Especially $\lim_{x\to a^+}f(x)$ exists as a finite real number for all $a\in\mathbb{R}\setminus C$ where $C$ is countable.

Is it possible that $C=\{a\in\mathbb{R}:\lim_{x\to a^+} f(x)\text{ does not exist}\}$ is dense in $\mathbb{R}$, that is, $C$ intersects all nonempty open intervals?

happyEddie
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  • Look at the solution to this question. – lulu Oct 12 '24 at 11:29
  • @lulu That function is nondecreasing so it has limits from the left and limits from the right at each point (although they are not equal at the rationals). I'm looking for a function that does not have limits from the right at all in a dense set. – happyEddie Oct 12 '24 at 13:45

1 Answers1

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For any countable $C \subset \Bbb R$ there is a function $f: \Bbb R \to \Bbb R$ such that $\lim_{x\to a^-}f(x)$ exists for all $a \in \Bbb R$, and $\lim_{x\to a^+}f(x)$ exists if and only if $a \notin C$. In particular such a function exists for $C = \Bbb Q$, which is dense in $\Bbb R$.

One can mimick the proofs of Function with limits only at irrational points and Construct a monotone function which has countably many discontinuities .

Define $f: \Bbb R \to \Bbb R$ as $$ f(x) = \sum_n 2^{-n} h(x-r_n) $$ where $(r_n)$ is an enumeration of $C$ and $h: \Bbb R \to \Bbb R$ is $$ h(x) = \begin{cases} 0 & \text{ for } \le 0 \, ,\\ \sin(1/x) & \text{ for } x > 0 \, . \end{cases} $$ $h$ is continuous everywhere except at $x=0$, where the left limit exists, but not the right limit.

If $a \notin \Bbb C$ then $f$ is continuous at $a$ because each term in the sum is continuous at $a$ and the series is uniformly convergent.

If $a = r_m \in \Bbb C$ then $$ f(x) = 2^{-m} h(x-r_m) + \sum_{n \ne m} 2^{-n} h(x-r_n) $$ where the sum on the right is continuous at $a$, so that $f$ has a left limit but no right limit at $a$.

Martin R
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