$f^{-1}(\{ a \})$ measurable for $a \in \mathbb{K}$ but $f$ not measurable

Question

Let $(X, \mathcal{A})$ a measurable space and $f:X \to \overline{\mathbb{R}}$, we know that the following conditions are equivalent.

(i) $f$ is measurable.

(ii) $\forall a \in \mathbb{Q}, \hspace{1ex} \{ x \in X \hspace{1ex} | \hspace{1ex} f(x)>a \} \in \mathcal{A}$.

(iii) $\forall a \in \mathbb{Q}, \hspace{1ex} \{ x \in X \hspace{1ex} | \hspace{1ex} f(x) \geq a \} \in \mathcal{A}$.

(iv) $\forall a \in \mathbb{Q}, \hspace{1ex} \{ x \in X \hspace{1ex} | \hspace{1ex} f(x)<a \} \in \mathcal{A}$.

(v) $\forall a \in \mathbb{Q}, \hspace{1ex} \{ x \in X \hspace{1ex} | \hspace{1ex} f(x) \leq a \} \in \mathcal{A}$.

The question is:

Do we have (i) $f$ is measurable $\Longleftrightarrow$ (vi) $\forall a \in \mathbb{K}, \{ x \in X \mid f(x) = a \} \in \mathcal{A},$ where $\mathbb{K}= \mathbb{R}$ or $\mathbb{Q}$? We do have (i) $\Longrightarrow$ (vi), but do we have (vi) $\Longrightarrow$ (i) ?

Answer 1

NO.

Let $B$ be a non-Lebesgue-measurable subset of $[0,1]$ and let $C=B\cup [1,2].$ This is to ensure that $C$ is non-measurable with cardinal $|C|=|\overline {\Bbb R}\setminus C|$ so that there is a $bijection$ $f:\overline {\Bbb R}\to \overline {\Bbb R}$ such that $f(x)>0$ if $x\in C$ and $f(x)\le 0$ if $x\in \overline {\Bbb R}\setminus C.$

If $a\in \overline {\Bbb R}$ then $f^{-1}\{a\}$ contains just one point, so $f^{-1}\{a\}$ is Lebesgue-measurable. But the set $S=\{a\in \overline {\Bbb R}: a>0\}$ is open and $f^{-1}S=C,$ which is not a measurable set, so $f$ is not a measurable function.