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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.)
immeasurable
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    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. – Nik Weaver Dec 19 '20 at 15:58
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    The answer to 2. and 3. is no. In fact it is no even if the two one-sided limits agree. See for instance: https://math.stackexchange.com/q/698448/787383 – Alessandro Della Corte Dec 19 '20 at 15:59
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    On question 1. this can help: https://mathoverflow.net/a/231462/167834 – Alessandro Della Corte Dec 19 '20 at 16:10
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    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. – Bill Johnson Dec 19 '20 at 16:23
  • 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. – immeasurable Dec 20 '20 at 17:41
  • Addendum: I had intended to add the following to my comment: "And one can say the function is piecewise continuous." – immeasurable Dec 20 '20 at 17:50

1 Answers1

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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.

LSpice
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Gerald Edgar
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