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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 $x\notin N$?

Community
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celtschk
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  • 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. – Sangchul Lee Aug 07 '12 at 16:38
  • 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. – Kevin Carlson Aug 07 '12 at 16:48
  • 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 ... – celtschk Aug 07 '12 at 16:50
  • ... 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). – celtschk Aug 07 '12 at 16:52
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    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.) – mjqxxxx Aug 07 '12 at 16:53
  • I've just seen @KevinCarlson's post, and looking up Chaitin's constant reveals that it is defined in a spirit very close to what I just described, only instead of concatenating the halting programs, it just tells with each bit whether the corresponding program halts. – celtschk Aug 07 '12 at 16:54
  • @mjqxxxx: The concept "nameable in less than 100 characters" is not the same as the concet "nameable at all". Without size restriction, the descriptions in any language are equivalent, because you can just prefix the description in one language by a description of that language in the other language to get a complete description in the other language. – celtschk Aug 07 '12 at 16:57
  • The notion you seem to be looking here is "definable number". – Harry Altman Aug 07 '12 at 17:33
  • Given a language $L$ with finitely many symbols, an ordering of those symbols defines a lexicographic enumeration of the strings in language $L$: $S_1, S_2, \ldots$. Let $x$ be the real number with decimal digits $.x_1 x_2 x_3 \ldots$ such that $x_i = 1$ if $S_i$ defines a number $y$ whose i'th digit $y_i = 7$, $x_i = 7$ otherwise. Then $x$ is not nameable in language $L$ --- but I just gave it a definition. – Robert Israel Aug 07 '12 at 18:02
  • A lot of these paradoxes about nameability are avoided with a moment of precise thought. The predicate "smallest number not nameable in 100 characters or less" includes an implicit quantifier over all languages. If we just interpret a language by a function from names to $\mathbb{R}$, though, then this predicate is obviously never satified: for all $x$ in $\mathbb{R}$, the singleton language $a \mapsto x$ names $x$. Besides, without a more exact definition I may as well allow uncountable languages, and have each real number name itself! – Kevin Carlson Aug 07 '12 at 19:52
  • @RobertIsrael: You didn't prove that this number exists. What if the string $i$ in corresponds to exactly this definition? Then $x_i=7$ iff $x_i\ne 7$, which is a contradiction. Note that this doesn't invalidate the language $L$ because there's no requirement that each sequence of symbols describes a number. – celtschk Aug 24 '12 at 15:55
  • @KevinCarlson: Of course a language is defined as the set of finite sequences of a finite set of symbols, together with rules of what sequences are valid, and what valid sequences mean. – celtschk Aug 24 '12 at 16:03
  • @celtschk: Yes, you get a contradiction, which shows that for such a language there is no satisfactory interpretation of "defining a number". – Robert Israel Aug 24 '12 at 16:35
  • No, it just says that this is not the definition of a number. Just as "the prime between 24 and 26" is not the definition of a number, despite it superficially looking like one. – celtschk Aug 24 '12 at 16:38
  • The basic question is how to interpret "string $S$ in language $L$ defines number $x$". Let's suppose we fix a certain function $F$ from the strings of language $L$ into ${\mathbb R} \cup {\text{"not-a-definition"}}$. We interpret "string $S$ defines number $x$" as $F(S) = x$ where $x \in \mathbb R$. Then I insist that my definition is perfectly good: the $i$'th digit of $x$ is $1$ if $F(S_i) = y \in \mathbb R$ with $y_i = 7$, and $7$ otherwise. This is a real number. It's just that there is no $S_i$ with $x = F(S_i)$. – Robert Israel Aug 24 '12 at 18:29
  • If $L$ contains a String $S_i$ which formulates your definition in the language $L$, then your definition does not define a proper number, because if it did, then $S_i$ would also (remember, $S_i$ is your definition, just expressed in the language $L$), in contradiction to the fact that it can't. – celtschk Aug 24 '12 at 19:01

2 Answers2

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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 $r_n<x$.

Colin McQuillan
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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.

Qiaochu Yuan
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