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In "Probability Theory, A Comprehensive Course", by Achim Klenke, Theorem 1.92 :

Let $X_1, X_2, \dots$ be measurable maps $(\Omega, \mathcal A) \to (\overline {\mathbb R}, \mathcal B(\overline {\mathbb R}))$. Then the following maps are also measurable: $\inf_{n \in \mathbb N} X_n$, $\sup_{n \in \mathbb N} X_n$, ${\lim \inf}_{n \to \infty} X_n$, ${\lim \sup}_{n \to \infty} X_n$.

For the proof of the measurability of $\inf_{n \in \mathbb N} X_n$, the following equality is considered:

\begin{equation} \left(\inf_{n \in \mathbb N}X_n\right)^{-1}([-\infty, a]) = \bigcup_{n=1}^{\infty} X_n^{-1} ([-\infty, a])\end{equation}

Could someone provide intermediate steps in the proof of this equality ?

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    $\begingroup$ Did you try to show that? (You want to show equality of two sets, so you pick one element in one of it, and show that this is also in the another set... ) $\endgroup$ Aug 22, 2021 at 15:42

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Klenke says $$\bigg(\inf_{n \in \mathbb{N}}X_n\bigg)^{-1}([-\infty,a\color{red}{)})=\bigcup_{n \in \mathbb{N}}X_n^{-1}([-\infty,a\color{red}{)})\in \mathcal{A}$$ or equivalenty $$\bigg\{\inf_{n \in \mathbb{N}}X_n<a\bigg\}=\bigcup_{n \in \mathbb{N}}\{X_n< a\}\in \mathcal{A}$$ which is true: $(\subset)$. Let $\omega \in \{\inf_{n \in \mathbb{N}}X_n<a\}$. Assume $\omega \notin \cup_{n \in \mathbb{N}}\{X_n< a\} $. This would mean that $X_n(\omega)\geq a, \, \forall n \in \mathbb{N}$ which would imply $\inf_{n \in \mathbb{N}}X_n(\omega)\geq a$, and that is a contradiction. $(\supset)$. Let $\omega \in \cup_{n \in \mathbb{N}}\{X_n< a\} $, then there exists $n \in \mathbb{N}$ such that $X_n(\omega)<a$ so $\inf_{n \in \mathbb{N}}X_n(\omega)<a$.

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