Why $\displaystyle \left(\inf_{n\in \mathbb{N}}X_n\right)^{-1}([-\infty,a))=\bigcup^{\infty}_{n=1}\left(X_n\right)^{-1}([-\infty,a))$? Why is $\displaystyle \left(\inf_{n\in \mathbb{N}}X_n\right)^{-1}([-\infty,a))=\bigcup^{\infty}_{n=1}\left(X_n\right)^{-1}([-\infty,a))$ true?
Isn't it supposed to be an intersection?
 A: The union is correct. Recall that $\inf_nX_n$ is the greatest lower bound of the $X_n$. Therefore if $X_{m}<a$ for some $m$, then $\inf X_n\leq X_{m}<a$.
Conversely, if $\inf X_n<a$ then $X_m<a$ for some $m$, for otherwise $a$ would be a lower bound for $\{X_n\}$.
A: I'm assuming by the notation you're using that the $(X_{n})_{n \in \mathbb{N}}$ is a sequence of functions or random variables all defined on the same domain $\Omega$.
Given $\omega \in \Omega$, $\inf \{X_{n}(\omega) \, \mid \, n \in \mathbb{N}\} < a$ if and only if $X_{m}(\omega) < a$ for some $m \in \mathbb{N}$.  Hence $(\inf X_{n})^{-1}((-\infty,a)) = \bigcup_{n \in \mathbb{N}} X_{n}^{-1}((-\infty,a))$.
It would be harder if we were looking at sets of the form $(-\infty,a]$ since then we would have to consider the possibility that $\inf X_{n} = a$, but even then we are first interested in unions.  In fact,
$$(\inf X_{n})^{-1}((-\infty,a]) = \bigcap_{K \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} X_{n}^{-1}((\infty,a + K^{-1})).$$
