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.

  • $\begingroup$ Is codomain $\mathbb R^*$ equipped with the Borel $\sigma$-algebra? $\endgroup$ – drhab Oct 13 '15 at 18:04
  • $\begingroup$ Yes, it's the Borel $\sigma$-algebra for $\Bbb R^*$ $\endgroup$ – Arnulf Oct 13 '15 at 18:06
  • $\begingroup$ I'm trying to see how this functions should be analyzing the generator class $B_1 =\{A\subset \Bbb R^* : A=(a,\infty )\}$. First, I'm trying to discard all the continuos functions, but I'm a little stuck. Assuming $f$ is continuos we get that: $$\forall\ A\in B_1,\ f^{-1}(A)\ is\ open,\ so\ f^{-1}(A)\ is\ not\ finite\ neither\ countable\ and,\ thus,\ f^{-1}(A)\notin S $$ And so it's only left to prove that also $$\Bbb R - f^{-1}(A)\ is\ not\ finite\ neither\ countable$$ But I got stuck here since the only thing I know is that $\Bbb R - f^{-1}(A)$ is closed. $\endgroup$ – Arnulf Oct 13 '15 at 18:52

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.

  • $\begingroup$ In the second part, where you suppose theres another not countable fiber, I don't get why choose an r ($\in \Bbb R$) and also I don't get it why f cannot be measurable. $\endgroup$ – Arnulf Oct 14 '15 at 2:53
  • $\begingroup$ The preimages of $(-\infty,r)$ and $(r,\infty)$ both contain an uncountable fiber, hence are both uncountable. Also they are disjoint. Consequently also their comlements are uncountable, wich means that neither of them is cocountable. So these preimages are not measurable. $\endgroup$ – drhab Oct 14 '15 at 6:59
  • $\begingroup$ The same conclusion holds for measurable functions $ \ f: X \to Y \, $, where $X$ is an arbitrary uncountable set with the same $\sigma$-algebra of countable and co-countable sets and $Y$ is a Hausdorff topological space with some $\sigma$-algebra that contains the Borel sets. $\endgroup$ – Gustavo Jan 14 '17 at 22:55
  • $\begingroup$ @drhab May you please explain why there is $r\in \mathbb{R}$ such that $I \cap (-\infty , r)$ and $I \cap [r,\infty)$ are both uncountable? Thanks!! $\endgroup$ – dan Dec 2 '18 at 13:32
  • $\begingroup$ @dan More or less I said that on intuition. It appeared that a proof (or at least my proof) of it is too much for a comment. That's why I edited the answer to add a proof of the statement. I is quite well possible that a more elegant proof can be given, but I am content that I could find one. $\endgroup$ – drhab Dec 2 '18 at 20:01

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