The cardinality of continuous real functions is $|\mathbb{R}|$ but I was wondering wether allowing functions to be almost everywhere continuous would increase the cardinality or not.

On the one hand, there are infinite sets with measure 0 (e.g. $\mathbb{Q}$, Cantor set) and allowing functions to be discontinuous at each of these points will blow up the cardinality of the set. However, it is impossible to have a function continuous everywhere but on $\mathbb{Q}$ because of its density (Edit: Actually it is possible). By defining $K=\{\frac{1}{n},n\in \mathbb{N}\}$, the function $\mathbb{1}_K$ is still almost everywhere continuous so even if a set $A$ of measure 0 has a limit point, it's still possible to find a function continuous everywhere but on $A$. I introduced by myself the following definition of a set: A set $A$ is locally dense if $m(A)=0$ and $m(\bar{A})>0$. I haven't proven yet that a function that is continuous everywhere but on a locally dense set is not almost everywhere continuous but I'm pretty sure it's true and I think I have a sketch of proof in mind. So we would be left with functions that are continuous everywhere but on non-locally dense sets.

On the other hand, if we want to prove that the cardinality of almost everywhere continuous functions is at most $|\mathbb{R}|$, we cannot proceed the same way we did for continuous functions since an almost everywhere continuous function has to be defined on all rationals and we have to know in which point the function is non-continuous. Yet, if all non-locally dense sets are countables and if my previous guess is true (functions that are continuous everywhere but on a locally dense set are not almost everywhere continuous), that means an almost everywhere continuous function is just defined by its value on the rationals and by a countable set of discontinuity points, namely $D$. So the cardinality would be $|\mathbb{R}^{\mathbb{Q}}\times D|=|\mathbb{R}|$ Moreover, finding an injection from "non-continuous" functions to almost everywhere continuous functions seems to be totally impossible.

If I had to take a guess, I would say that this set has cardinality $|\mathbb{R}|$. Any idea on how to conclude anything ?


1 Answer 1


You can have a function continuous except on $\Bbb Q$. If $x$ is irrational, $f(x)=0.$ if $x$ is rational, wrote it in lowest terms as $a/b$ and $f(x)=1/b$. Each irrational is far enough from rationals with small denominators to be OK .

Now you have to show the set of discontinuity points is countable rather than a measure $0$ uncountable set


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