# Is $\operatorname{card}(I)=\operatorname{card}(D)$

When I was answering number of integrable functions is greater than number of differentiable functions I got to wonder if the inequality was strict.

So with $\mathcal I$ being the set of integrable functions $\mathbb R\to\mathbb R$ and $\mathcal D$ being the set of differentiable functions $\mathbb R\to\mathbb R$: Is $\operatorname{card}(\mathcal I)=\operatorname{card}(\mathcal D)$? Does it matter whether we're talking Riemann- or Lebesgue integrability (or one of the variants I'm not very familiar with, and don't remember the name of)? Would the answer be different if either $\mathbb R$ was replaced with another set?

$\geq$ is obvious, and integrable functions that aren't differentiable are well-known too, but I imagine that both sets have cardinality $2^{2^{\aleph_0}}$?

This alone gives you $2^{2^{\aleph_0}}$ integrable functions whose integral is $0$. (Note that this works for Riemann integrable functions as well, since on $[0,1]$ the function is integrable, since the points of discontinuity are measure $0$, and outside $[0,1]$ it is a constant function.)
On the other hand, a differentiable function is continuous, and there are only $2^{\aleph_0}$ of those.
• Do you mean "only non-zero on ..."? Then I get that. Now that you wrote it, I do believe I've seen a proof that there's only $2^{\aleph_0}$ continuous functions, but I don't remember the argument (that might just be because it's getting late), could you give a hint on that? – Henrik Apr 14 '15 at 22:57
• Yes, that is what I meant. I'll edit that in, thanks. As for only $2^{\aleph_0}$ continuous functions, note that a continuous function is fully decided by its values on the rational numbers, so by knowing all functions from the rationals to the reals we know all the continuous functions (and more, of course). How many of those do we have? – Asaf Karagila Apr 14 '15 at 23:05