Measurable functions in a countable co-countable $\sigma$-algebra

I found a interesting problem that says: Let $(X,S)$ be a measurable space where $X=\mathbb R$ and $S$ is the countable co-countable $\sigma$-algebra in $\mathbb R$, i.e. $S=\{A\subset\mathbb R: A\ \lor\ \mathbb R-A\ \ is\ finite\ or\ countable\}$. So the point is to describe the S-measurable functions $\ f:X\to \Bbb R^*$, where $\Bbb R^*$ are the extended real numbers.

Solution 1:

Let $I$ denote the image of $f$. If $I$ is not countable then some $r\in\mathbb R$ will exist such that $I\cap(-\infty,r)$ and $I\cap[r,\infty)$ are both not countable. Then the disjoint preimages of these sets cannot be countable, hence $f$ cannot be meausurable.

We conclude that $I$ must be countable. Then for $x\in I$ fibres $f^{-1}(\{x\})$ form a countable partition of $X=\mathbb R$, so at least one of these fibres is not countable. If for distinct $x,y\in I$ the fibres $f^{-1}(\{x\})$ and $f^{-1}(\{y\})$ are both uncountable then we can choose some $r$ in between $x$ and $y$ such that the disjoint preimages of $(-\infty,r)$ and $(r,\infty)$ are both uncountable, and again $f$ cannot be measurable.

We conclude that there is exactly one $x\in I$ with an uncountable fibre. The complement of this fibre is covered by other fibers. Each of them is a countable set and also the number of these fibers is countable. Conclusion: the complement of the mentioned uncountable fiber is countable, wich means that the uncountable fiber is cocountable.

Our final conclusion: $$f\text{ is measurable if and only if }f\text{ is constant on a cocountable set}$$ In that case the preimages of $(-\infty,r)$ with $r$ ranging over $\mathbb R$ will all be cocountable or countable.

edit (concerning question of @dan in a comment on this answer)

Let it be that $I$ is an uncountable subset of $\mathbb R$.

Let $A:=\left\{ x\in\mathbb{R}\mid\left(-\infty,x\right)\cap I\text{ is countable}\right\} $ and let $B:=\left\{ x\in\mathbb{R}\mid\left(x,\infty\right)\cap I\text{ is countable}\right\} $.

Note that the fact that $I$ is uncountable implies that $A\cap B=\varnothing$.

It is our aim to prove that $A\cup B\neq\mathbb{R}$.

So we assume that $A\cup B=\mathbb{R}$ and from here it is enough to find a contradiction.

At first hand for the shape of $A$ we see three possibilities: $A=\varnothing$, $A=\left(-\infty,s\right]$ for some $s\in\mathbb{R}$ or $A=\mathbb{R}$.

But if $A=\mathbb{R}$ then $I=\bigcup_{n=1}^{\infty}\left(I\cap\left(-\infty,n\right)\right)$ is countable as well, so the third possibility falls off.

Then similarly for the shape of $B$ we find two possibilities: $B=\varnothing$ or $B=\left[i,\infty\right)$ for some $i\in\mathbb{R}$.

Then based on $A\cup B=\mathbb{R}$ we find that also the possibilities $A=\varnothing$ and $B=\varnothing$ fall off.

So $\mathbb{R}=A\cup B=\left(-\infty,s\right]\cup\left[i,\infty\right)$ implying that $i\leq s$.

But then $A\cap B=\left[i,s\right]\neq\varnothing$ and a contradiction is found.

Solution 2:

Let $f$ be a measurable function. Divide the line into a countable number of half-closed intervals and look at their preimages under $f$. Exactly one of them is uncountable. Break that interval into two and look at their preimages. Exactly one of them is uncountable. Continue like this. By the nested interval theorem there is exactly one point whose preimage is uncountable and cocountable. So the function is constant except on a countable set.