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Suppose a real-valued function f, whose domain is an interval, has the property that at every point in its domain it has a left-limit and a right-limit, and it equals at least one of these. Is it true that

1: f has at most a countable number of discontinuities. (Young's theorem would appear to say "yes".)

  1. f can be called piecewise continuous. (Some say "piecewise continuity" requires a finite number of discontinuities.)
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    $\begingroup$ Here's an easy example: enumerate the rationals in $[0,1]$ as $(a_n)$. Then for $t \in [0,1]$ define $f(t) = \sum \{2^{-n}: a_n \leq t\}$. This is an increasing function which is discontinuous at every rational. $\endgroup$
    – Nik Weaver
    Commented Dec 19, 2020 at 15:58
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    $\begingroup$ The answer to 2. and 3. is no. In fact it is no even if the two one-sided limits agree. See for instance: math.stackexchange.com/q/698448/787383 $\endgroup$
    – Alessandro Della Corte
    Commented Dec 19, 2020 at 15:59
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    $\begingroup$ On question 1. this can help: mathoverflow.net/a/231462/167834 $\endgroup$
    – Alessandro Della Corte
    Commented Dec 19, 2020 at 16:10
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    $\begingroup$ Are you a student? Here is an exercise for you. Prove that a function on a closed interval has the property you want if and only if it is a uniform limit of a sequence of step functions. $\endgroup$
    – Bill Johnson
    Commented Dec 19, 2020 at 16:23
  • $\begingroup$ Thanks. I was not aware of Thomae's function or Young's theorem. Please comment on my current understanding. If a real-valued function, whose domain is an interval, has left- and right-limits and equals at least one of them at each point in the domain, then it is continuous, except possibly at a countable number of points. $\endgroup$
    – immeasurable
    Commented Dec 20, 2020 at 17:41

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Perhaps you know this example. (Found in the comment by DellaCorte.) It is likely to be seen in a calculus course ….

$$ f(x) = \begin{cases} \frac{1}{b}, & \text{if $x = \frac{a}{b}$ is rational in lowest terms,} \\ 0, & \text{if $x$ is irrational.} \end{cases} $$ This function is continuous at every irrational, discontinuous at every rational, and has left and right limit $0$ everywhere.

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