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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 :))

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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?).

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  • $\begingroup$ Doesn't $C$ fall into $A_3$? $A_3$ is literally all $\Omega$ except $A_1$ or $A_2$? $\endgroup$ – Mitch Baker Nov 2 '17 at 23:50
  • $\begingroup$ Ah, I get it now. You made it more sublime. Now this event is no longer measurable because our $\sigma$-algebra is too coarse or primitive. $\endgroup$ – Mitch Baker Nov 2 '17 at 23:53
  • $\begingroup$ Many thanks, good sir! $\endgroup$ – Mitch Baker Nov 2 '17 at 23:56
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    $\begingroup$ Yeah, $C$ is a subset of $A_3$ but it is not an element of $\sigma$-algebra $\mathcal{A}$. Hope you will sleep well ;) $\endgroup$ – ElChorro Nov 2 '17 at 23:57

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