10
$\begingroup$

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 $x\notin N$?

  • 0
    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.2012-08-24

1 Answers 1

4

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.

  • 0
    Nice construction. That definitely answers my second question, thank you.2012-08-07