Vestrup Measure and Integration Exercise 5.1.6

Solution 1:

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.