Take $X \neq \emptyset$. Let $\mathbb{A}$ be the following $\sigma-$algebra: $$\mathbb{A}=\{ A \in P(X) : A \text{ is countable or } A^c \text{ is countable} \}$$
I want to know if it's true or false:
"if $f:R \to R $ is $\mathbb{A}$ -measurable then for all $\delta>0$ exist $a \in \mathbb{R} $ such that $(f^{-1}(a,a+\delta))^c $ is countable "
My suspicion is that it is false, I tried to do it as absurd, but I did not get anything.
As $f$ is $\mathbb A$-measurable, either $f^{-1}(a, a + \delta)$ or $(f^{-1}(a, a + \delta))^c$ is countable. As $\mathbb{R} = \bigcup\limits_{n={-\infty}}^\infty (\frac{n\delta}{2}, \frac{n\delta}{2} + 1)$, we have $\mathbb{R} = f^{-1}(\mathbb{R}) = f^{-1}(\bigcup\limits_{n={-\infty}}^\infty (\frac{n\delta}{2}, \frac{n\delta}{2} + \delta)) = \bigcup\limits_{n=-\infty}^\infty f^{-1}(\frac{n\delta}{2}, \frac{n\delta}{2} + \delta)$
As $\mathbb R$ is uncountable and union at the right side is countable, at least one of terms (corresponding to say $m$) is uncountable - thus for $a = \frac{n\delta}{2}$, $f^{-1}(a, a + \delta)$ is uncountable and $(f^{-1}(a, a + \delta))^c$ is countable.