Measurable function remaining constant This is a problem which appeared in one of my tests, which i wasn't able to solve.
Let $\Omega$ be a uncountable set. Let $S$ be the collection of subsets of $\Omega$ given by: $A \in S$ if and only if $A$ is countable or $A^{c}$ is countable. Suppose $f: \Omega \to \mathbb{R}$ is a real measurable function. Prove that there exists a $y \in \mathbb{R}$ and a countable set $B$ such that the $f(x)=y$ is on $B^{c}$.
 A: Here's a fun way to write the solution.
Define on $(\Omega, \mathcal{S})$ the probability measure $P(A) = 0$ if $A$ is countable, $P(A)=1$ if $A^c$ is countable.  Then $f$ can be seen as a random variable $X$.  Since all events have probability $0$ or $1$, all events, and hence all random variables, are independent.  Now there must be some $N$ with $P(|X| \le N) > 0$ (since $\Omega = \bigcup_{N=1}^\infty \{|X| \le N\}$ and $P$ is countably additive), hence $P(|X| \le N) = 1$.  So $X$ is a.s. bounded and in particular is $L^2$.  But $Var(X) = Cov(X,X) = 0$ since $X$ is independent of itself!  So $X = EX$ a.s., i.e. except on a countable set.
A: Note, that the countable intersection of co-countable sets (i.e., sets whose complement is countable) is co-countable ( since its complement is a countable union of countable sets).
Now, the union of the inverse image of the sets $[n,n+1]$ under as $n$ varies over all integers is all of $\Omega$. Since the inverse image of each $[n,n+1]$ is either countable or co-countable, so at least one of them must be co-countable (since $\Omega$ is uncountable).
Say $[n_1,n_1+1] = [a_1,b_1] $ is a set with co-countable (and hence uncountable) inverse image. Clearly, one of $[n_1, n_1 + 1/2]$ and $[n_1+1/2,n_1+1]$ must have a co-countable inverse image, call it $[a_2,b_2]$, ... proceeding in this manner we get a sequence of nested intervals $[a_n,b_n]$ each of whose inverse image is co-countable and $\lim_{n\to\infty} b_n - a_n = 0$, their intersection consists of a single point, say $y$, and  $f^{-1}(\{y\}) = \cap f^{-1}( [a_n,b_n] )$ being an intersection of co-countable sets is co-countable. Call $B^{c} = f^{-1}(\{y\})$, we are done.
