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Consider $\pi$ that is measurable variable with respect to the $\sigma-\text{algebra}$ $\mathcal{F}$. Is this equivalent to write $\pi\in\mathcal{F}$? By using the concept of measurability it means that $\pi$ as measurable is known and we can find it or simply we can use is as a constant am I right?

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A function $\pi:(\Omega_1,\mathcal{F}_1)\to(\Omega_2,\mathcal{F}_2)$ is said to be measurable if given $A\in\mathcal{F}_2$ ($A$ is a set of that belongs to the $\mathcal{F}_2$ $\sigma$-algebra), then $\pi^{-1}(A)\in\mathcal{F}_1$.

Some books use the notation $\pi\in\mathcal{F_1}$ as a shorthand for "$\pi$ is measurable with respect to $\mathcal{F}_1$". But, obviously, $\pi$ is not a subset of $\Omega_1$, so it can't be an element of $\mathcal{F}_1$. It's just a kind of abuse of notation that should be unambiguous and harmless.

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    $\begingroup$ well thank you very much. It is not some books, it is in the vast majority of every graduate and undergraduate book... $\endgroup$ – Hunger Learn Jun 7 '20 at 9:41

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