The measurability of $f(x)$ given that $\lim f_n(x) = f(x)$. 


I have difficulty in interpreting the equality "$\{x \in X : \sup_{n\in N} f_n(x) > \alpha\} = \cup_{n=1}^\infty \{x \in X : f_n(x) > \alpha\} $". Is this right to say that $\sup_{n\in N}f_n(x)$ is the largest $f_n(x)$, so it contains $x \in X$ the most on the condition given, so it is equal to the union of sets?
Furthermore, it is quite difficult for me to understand the proof for $f^*, F^*$. Could you elaborate on either of one?
Thank you in advance.
 A: Let $z \in \cup_{n=1}^\infty \{ x \in X : f_n(x) > \alpha \}$. This means that for some $m \in \mathbb N$, $f_m(z) > \alpha$, which implies that $\sup_n f_n(z) > \alpha$. In other words, $$ \bigcup_{n=1}^\infty \{ x \in X : f_n(x) > \alpha \} \subseteq \{ x \in X : \sup_n f_n(x) > \alpha \}.$$
To see the reverse inclusion, let $z \in \{ x \in X : \sup_n f_n(x) > \alpha \}$. This implies that there is a sequence $\{ m_i \}_{i=1}^\infty \subseteq \mathbb N$ such that
$$ \lim_{i \to \infty} f_{m_i} (z) > \alpha. $$
In particular, this means that $f_{m_i}(z) > \alpha$ for all large enough $i$. Thus, there exists an $n \in \mathbb N$ such that $f_n(z) > \alpha$. This proves the reverse inclusion.
Now let $g_n(x) = \inf_{m \ge n} f_m (x)$. The previous argument shows that $g_n(x)$ is measurable. The same reasoning shows that $f^*(x) = \sup_n g_n(x)$ is measurable. The exact same argument applies to $F^*$.
Note that the displayed equalities for $f^*$ and $F^*$ are equivalent definitions for $\limsup$ and $\liminf$.
