# 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 $B^{c}$.

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You mean to say that $S$ is a collection of subsets of $\Omega$ and not all of the subsets. – Asaf Karagila Oct 1 '10 at 16:16
I guess you mean $S$ is the collection of all subsets $A$ such that $A$ or $A^c$ is infinite countable (many authors say that a set is countable if it is finite or infinite countable). – AD. Oct 1 '10 at 18:14
What about $f:\mathbb{R}\ni x\mapsto x\in\mathbb{R}$ (the identity) - $f$ is not constant on any subset except singleton sets. – AD. Oct 1 '10 at 18:17
@AD $f^{-1}(0,1) = (0,1)$ which is neither countable not co-countable, so the identity function is not measurable. – ACARCHAU Oct 1 '10 at 18:22
@A.D. Given a set of infinite cardinality $\kappa$ and an infinite cardinal $\lambda\lt\kappa$, the collection of all subsets that have cardinalty at most $\lambda$ or are complements of subsets of cardinality at most $\lambda$ is a $\sigma$-algebra. So here, $S$ is a $\sigma$-algebra. – Arturo Magidin Oct 1 '10 at 22:26

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.

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Nice :) – AD. Oct 2 '10 at 5:39
Why can you say that R is the union of $[n,n+1]$ with $n$ an integer? – user54297 Mar 23 '14 at 21:01

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.

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