Measurable functions on an uncountable set

Question

If X is an uncountable set and A is sigma algebra on X that is generated by all singletons in X and f is a function from X to real numbers and the sigma algebra on real numbers is set of the Borel sets.then f is a measurable function if an only if....?

Answer 1

Hint:

$\mathcal{A}:=\left\{ B\in\wp\left(X\right)\mid B\text{ is countable}\vee X-B\text{ is countable}\right\} $ is a $\sigma$-algebra.

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addendum:

function $f:X\rightarrow\mathbb R$ is a measurable function if and only if its image is countable and exactly one of its fibres is uncountable.

Fibres of $f$ are preimages of singletons, i.e. sets of the form $\left\{ x\in X\mid f\left(x\right)=r\right\} $ where $r\in\mathbb{R}$. My hint was meant as an indication the $\mathcal A$ is here the $\sigma$-algebra on $X$ generated by the singletons in $X$.

Proof:

Suppose that the image of $f$, i.e. the set $f\left(X\right):=\left\{ f\left(x\right)\mid x\in X\right\} \subseteq\mathbb{R}$ is uncountable. Then a Borelset $B\subset\mathbb{R}$ can be found such that $f\left(X\right)\cap B$ and $f\left(X\right)\backslash B$ are both uncountable. For a proof of that look here. Consequently the preimages under $f$ of the sets $B$ and its complement $B^{c}=\mathbb{R}\backslash B$ are both uncountable hence do not belong to $\mathcal{A}$. Proved is now that $f$ can only be measurable if its image is countable. If its image is indeed countable then $f$ must have fibres that are uncountable. This because the union of these fibres form the uncountable set $X$. If $r,s\in\mathbb{R}$ are distinct then fibres $f^{-1}\left(r\right)$ and $f^{-1}\left(s\right)$ are disjoint so if they are both uncountable then none of them can belong to $\mathcal{A}$. Proved is now that $f$ can only be measurable if exactly one of its fibres is uncountable.

If conversely both conditions are satisfied then indeed $f$ is measurable. If $f^{-1}\left(r\right)$ is its unique uncountable fibre then $X\backslash f^{-1}\left(r\right)=\bigcup_{s\in f\left(X\right)\backslash\left\{ r\right\} }f^{-1}\left(s\right)$ is a countable union of countable sets, hence is countable itself. For any preimage $f^{-1}\left(U\right)$ we have $f^{-1}\left(U\right)\subset X\backslash f^{-1}\left(r\right)$ or $X\backslash f^{-1}\left(U\right)\subset X\backslash f^{-1}\left(r\right)$ and in both cases $f^{-1}\left(U\right)\in\mathcal{A}$.