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While reading this comment and thinking about how you could change the functions without convergence (because the Lebesgue integral doesn't care about changes at countably many places), I just arrived at a concept which could be called "liar functions", for reasons which should be obvious after reading this.

The concept is based on the known fact that the set of all real numbers you can name (no matter in which way) is countable, because the number of names is countable. Therefore you can change a function in all "named" positions, without changing its integral (or any other property which is only dependent on values "almost everywhere).

Let $N\subset\mathbb R$ be the set of named values (that is, the set of values you can uniquely specify). Then I call a "liar function" a function $f:\mathbb R\to \mathbb R$ which has the following properties:

  • The restriction of $f$ to $N$ has an obvious extension $f_1$ to all of $\mathbb R$.
  • The restriction of $f$ to $\mathbb R\setminus N$ also has an obvious extension $f_2$ to $\mathbb R$.
  • $f_1$ and $f_2$ are significantly different.

As a simple example, consider the function $$f(x)=\cases{1 & for $x$ in $N$\\ 0 & otherwise}$$ This function simply lies about its value: Whenever you evaluate it at a position you can specify, it gives $1$. However, it is actually $0$ almost everywhere, except where you can look.

Another function, which lies more subtly, would be the function $$g(x)=\cases{1/q & for $x=p/q\in \mathbb{Q}$, $\operatorname{lcd}(p,q)=1$\\ 0 & for $x\in N\setminus \mathbb Q$\\\frac{1}{1+x^2} & otherwise}$$

This on named values looks like the well known function which is continuous on all irrational numbers but discontinuous at all rational numbers. However actually it equals the Lorentz function almost everywhere (which e.g. means its integral over $\mathbb R$ exists, but is not $0$ as one might infer if one onle knows it on named values), and it is actually nowhere continuous (but almost everywhere equal to a continuous function).

Now my question: Is this concept of "liar functions" (probably under another name) already known?

Also, does there exist a liar function which is discontinuous in any $x\in N$, but continuous for any $x\notin N$?

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  • $\begingroup$ I also know little on this subject, but your 'known number' seems to mean the concept of computable number. According to Wikipedia, there has been some attempts to develop analysis based on the field of computable numbers, so it may be helpful to search in these literature. Finally, I guess that the function $f(x)$ that assigns the reciprocal of the minimal Gödel number representing $x$ for computable $x$, and $0$ otherwise seems work for you last question. $\endgroup$
    – Sangchul Lee
    Aug 7, 2012 at 16:38
  • $\begingroup$ This isn't quite the same as the computable numbers. For instance, Chaitin's constant $\Omega$ is nameable but not computable. I think you'd need to fix a language in which you're doing your naming before starting to ask questions like the last one. Once you've done that, as long as your language has at most countably many symbols, you might try defining a function that's $1/n$ at numbers with $n$-symbol names and 0 at unnameables. But the continuity argument seems tricky. $\endgroup$
    – Kevin Arlin
    Aug 7, 2012 at 16:48
  • $\begingroup$ While all comutable numbers are "named" (you can just use the algorithm which computes it as "name"), I'm not sure the reverse is also true. That is, there might be a number for which you can write down an unique identifying definition (i.e. some property which exactly one real number has), but you are not able to actually compute the number (that is, there is no algorithm which can give you an arbitrary good approximation). For example, assume you've got a specific numbering of all algorithms (e.g. as program in Jot), and ... $\endgroup$
    – celtschk
    Aug 7, 2012 at 16:50
  • $\begingroup$ ... define your number as "0." + the concatenation of all programs that always terminate, in lexicographic order. That's a well-defined number, but you cannot compute it (because that would, by design, involve solving the halting problem). $\endgroup$
    – celtschk
    Aug 7, 2012 at 16:52
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    $\begingroup$ The concept of "nameable numbers" is problematic on its face. Let $x$ be the smallest positive integer not nameable in $100$ characters or less. (Uh-oh.) $\endgroup$
    – mjqxxxx
    Aug 7, 2012 at 16:53

2 Answers 2

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For any sequence of reals $r_1,r_2,\dots$, there is a function that is discontinuous precisely at elements of this sequence. For example, define $f(x)$ to be the sum of $2^{-n}$ over all $n$ such that $r_n<x$.

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  • $\begingroup$ Nice construction. That definitely answers my second question, thank you. $\endgroup$
    – celtschk
    Aug 7, 2012 at 17:05
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None of these constructions work, for a non-obvious reason: it is not actually possible to define the set $N$, using the usual axioms of set theory. In fact there are models of ZFC in which every real number is definable (even though this would appear to violate the countability of the definable numbers; but it's not possible to define a bijection between the definable numbers and the natural numbers either). This is thoroughly explained by Joel David Hamkins here on MathOverflow.

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  • $\begingroup$ Is this related to the Lowenheim-Skolem theorem? That there is a countable model of ZFC that contains uncountable sets $\endgroup$
    – David Lui
    Sep 2 at 23:52
  • $\begingroup$ @David: it's related in the sense that the same "can't define the bijection" issue occurs there. $\endgroup$
    – Qiaochu Yuan
    Sep 3 at 0:48

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