Given a sample space $\Omega = [0, 1] \times [0, 1]$, on which a $\sigma$-algebra $\mathcal{A}$ is generated by $A_1, A_2$, and $A_3$, where $A_1 := \{(x, y) \in \Omega: y < \frac{x}{2}\}$, $A_2 := \{(x, y) \in \Omega: y > 1 - \frac{x}{2}\}$, and $A_3 := \Omega \setminus(A_1 \cup A_2)$, it is quite easy to find an $\mathcal{A}-\mathcal{B}$-measurable function such as $f(\omega)= \mathbf{1}_{A_1}(\omega)$.
But how to find a function, $f: \Omega \rightarrow \mathbb{R}$, that is not $\mathcal{A}-\mathcal{B}$-measurable?
This is a question from an old test that will not let me sleep tonight, unless a kind helper offers some useful insight.
A thousand thanks :))
This $\sigma$-algebra is really simple since $A_i\cap A_j = \emptyset$. Convince yourself that $C = \{(0,\frac{1}{2})\}\ \notin \mathcal{A}$ and define a function $$f(x)=\begin{cases} 0, \quad x\in C\\ 1, \quad \text{otherwise}.\end{cases}$$ Then $f^{-1}(\{0\}) \notin \mathcal{A}$, so $f$ is not measurable (why?).