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Motivation:

Let $X\subseteq \mathbb{R}$ and $Y\subseteq\mathbb{R}$ be arbitrary sets, where we define a function $f:X\to Y$.

I want a measure of discontinuity which ranges from zero to positive infinity, where

  • When the limit points of the graph of $f$ is continuous on the closure of $X$, the measure is zero
  • When the limit points of graph of $f$ can be split into functions, where $n$ of those functions are continuous on the closure of $X$, the measure is $n-1$
  • When $f$ is discrete, the measure is $+\infty$
  • When $f$ is "hyper-discontinuous", the measure is $+\infty$
  • When the graph of $f$ is dense in the limit point set of $X\times Y$, the measure is $+\infty$
  • When the measure of discontinuity is between zero and infinity, the more "disconnected" the graph of $f$, the higher the measure of discontinuity.

Question: Is there a measure of discontinuity which gives what I want?

Attempt: I tried to answer this using a previous question, but according to users it's needlessly complicated and likely is incorrect. I'm also struggling to explain why the answer has potential.

Arbuja
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    You can think about how well approximated it can be by a pointwise limit of continuous functions (since as a corollary of Luzin’s theorem, pointwise limits of continuous functions are continuous a.e), I’m not sure if this is helpful though. – Malady Jan 27 '25 at 17:11
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    I'm not entirely sure these conditions are consistent with each other. For instance, the function mapping $p/q\mapsto 1/q$ for $p/q$ rational and in reduced form, and sending irrationals to zero certainly has a graph whose closure cannot be split into the graph of a single continuous function, yet this function is continuous almost everywhere. – Eli Seamans Jan 28 '25 at 04:20
  • @EliSeamans Made edits. I meant continuous “almost everywhere”. – Arbuja Jan 28 '25 at 14:17
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    The problem is not with that. For instance, the graph of the function $f$ I described in my comment has both $(1/2,1/2)$ and $(1/2,0)$ in its closure, and so it's impossible to cover the closure with the graph of one function, let alone a continuous one. – Eli Seamans Jan 28 '25 at 16:37

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