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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.2010-10-01
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    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).2010-10-01
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    What about $f:\mathbb{R}\ni x\mapsto x\in\mathbb{R}$ (the identity) - $f$ is not constant on any subset except singleton sets.2010-10-01
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    @AD $f^{-1}(0,1) = (0,1)$ which is neither countable not co-countable, so the identity function is not measurable.2010-10-01
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    @acarchau: You mean $S$ is the $\sigma$-algebra on which $S$ is defined?2010-10-01
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    The inverse image of a measurable set in $R$ has to belong to $S$.2010-10-01
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    @acarchau 1) That definition of measurable function is a bit too strong. It is enough to claim that $f^{-1}(U)$ is measurable. 2) Ok if this is the problem then $S$ should be the $\sigma$-algebra generated by all finite sets -- $S=\{A: A$ or $A^c$ is at most countable$\}$ (which is a $\sigma$-algebra).2010-10-01
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    @Chandru1 Do you agree to the last comments of mine above? (I will think about the problem...)2010-10-01
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    @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.2010-10-01

2 Answers 2

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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 :)2010-10-02
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    Why can you say that R is the union of $[n,n+1]$ with $n$ an integer?2014-03-23
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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.