Let $A \notin \mathcal{L}$. Let $f: (\Bbb{R}, \mathcal{L}) \to (\Bbb{R}, \mathcal{B})$ be given by $$ f(x) = \begin{cases} x&\text{if}\, x\in A\\ -x&\text{if}\, x\notin A \end{cases}. $$ We want to show that $f$ fails to be $(\mathcal{L}, \mathcal{B})$-measurable, yet $f^{-1}(\{x\}) \in \mathcal{L}$.
To prove that $f$ is not $(\mathcal{L}, \mathcal{B})$-measurable, I want to find a Borel set $B$ such that $f^{-1}(B) = A \notin \mathcal{L}$. But I cannot find such a $B$ using the above definition. What I found is if I restrict the domain of $f$ to $[0,1]$ and let $A \subset [0,1]$ with $A \notin \mathcal{L}$, then $f^{-1}((0,\infty)) = \{x \in \Bbb{R} \mid f(x) > 0\} = A \notin \mathcal{L}$, so $f$ is not $(\mathcal{L}, \mathcal{B})$-measurable.
So is the original definition valid or not? If yes, how can I find such a Borel set $B$?
Original definition: "To prove that $f$ is not $(\mathcal{L}, \mathcal{B})$-measurable, I want to find a Borel set $B$ such that $f^{-1}(B) = A \notin \mathcal{L}$." This statement is true.
Solution. Suppose that $f$ is measurable. Hence $f^{-1}((0, \infty))$ is measurable. Hence $(0, \infty) \cap f^{-1}((0, \infty)) $ is measurable. $$x \in (0, \infty) \cap f^{-1}((0, \infty)) \Leftrightarrow (x > 0) \cap (x \in A) \Rightarrow $$ $$(0, \infty) \cap f^{-1}((0, \infty)) = A \cap (0,\infty)$$ Hence $ A \cap (0,\infty)$ is measurable. Similarly, we prove that $ A \cap (-\infty, 0)$ is measurable. Hence $A = (A \cap (0,\infty)) \sqcup ( A \cap (-\infty, 0)) \sqcup (A \cap \{ 0\})$ is measurable since $(A \cap \{ 0\}) \subset \{ 0\}$ is measurable. So we got that $A$ is measurable and got a contradiction.