Context: I was solving an integral which had a typo that made it impossible to solve and when I wrote it on wolfram it gave me "no result found in terms of standard mathematical functions" this made me wonder how many functions that we need to define so that we can solve every integral or express anything with.

Let $S$ be the set that contains almost everywhere continuous functions on $\mathbb{R}$ such that if $f,g \in S$ then $f+g, \, f\times g, \ f^g, \ f(g) \in S$ , $af, \ f^a \in S \ \forall a \in \mathbb{R}$ provided they are defined for all $x$ .

My question is How many elements should $S$ have such that any almost everywhere continuous real function could be expressed as a finite combinations of elements of $S$.

By combinations here I mean that $f+g, \, f\times g, \ f^g, \ f(g), af, \ f^a \ \forall a \in \mathbb{R}$

Of course $S$ should be an infinite set but could it be countable set and still could express any continuous function with its elements?

I think it is impossible for such set to exist but I don't know how to prove or disprove it.

  • $\begingroup$ What does $f^g$, $\frac{1}{f}$, $f^a$ mean when the quantities such as $f(x)^{g(x)}$ are not defined $\endgroup$
    – Jakobian
    Commented Jun 14 at 16:18
  • $\begingroup$ @Jakobian I forgot to write provided they are defined for all x $\endgroup$
    – pie
    Commented Jun 14 at 16:21
  • $\begingroup$ I think one should go one step further and instead things like $f^g$ consider $\exp, \ln\in S$. Nonetheless I doubt there exists such countable set of generators $\endgroup$
    – Jakobian
    Commented Jun 14 at 16:24
  • $\begingroup$ Given the complexity of the space of a.e.-continuous functions, my guess would be that no countable set of generators would get everything. By way of analogy, any continuous periodic function on an interval $[-L,L]$ can be represented by its Fourier series, which is a countable sum of orthogonal functions. But this is not enough to capture every "nice enough" periodic function (though it is dense in that set). Yes, you are permitting more operations than just scaling and addition, but I doubt it is going to make any difference to the general character of the problem. $\endgroup$
    – Xander Henderson
    Commented Jun 14 at 16:40
  • $\begingroup$ Also, this may (or may not) be of relevance / interest: en.wikipedia.org/wiki/Risch_algorithm . $\endgroup$
    – Xander Henderson
    Commented Jun 14 at 16:40

1 Answer 1


The set of finite combinations of $S$ has cardinality $\mathfrak{c}$ (the continuum). But the set of almost-everywhere continuous functions has cardinality $\mathfrak{c^c}$, which is greater. In fact, let $E$ be the Cantor set, and consider real-valued functions which are $0$ on the complement of $E$ and arbitrary on $\mathbb E$. These are continuous on the complement of $E$ (and thus almost everywhere), and their cardinality is $\mathfrak{c^c}$. Thus not all of these can be expressed as finite combinations of $S$.

  • $\begingroup$ What is $S$? ( ) $\endgroup$ Commented Jun 14 at 17:11
  • $\begingroup$ I'm assuming you start with a set $S$ of cardinality at most $\mathfrak c$ and take all finite combinations. The resulting set then still has cardinality at most $\mathfrak c$. $\endgroup$ Commented Jun 17 at 1:10

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