Difficulty in understanding the equality I am having difficulty to understand one equality which came in proof of showing " If $f_{n}: X \to [ -\infty, \infty] $ are measurable, then $\sup f_{n}=g$ is measurable".
At the start of proof, the author claims that
  $$g^{-1}((\alpha, \infty])= \bigcup_n f_{n}^{-1}( (\alpha, \infty]).$$ 
 I am not getting why this is true. Thanks.
 A: You first verify that $g^{-1}(\alpha,\infty]\subset \cup f_n^{-1}(\alpha,\infty]$:
\begin{align*}
x\in g^{-1}(\alpha,\infty]\ 
\Rightarrow \alpha<g(x)
&\Rightarrow \alpha<f_n(x) \text{ for at least one $n$ (since otherwise } g(x)=\sup f_n(x)\leq \alpha)\\
&\Rightarrow x\in\cup f_n^{-1}(\alpha,\infty].
\end{align*}
And then verify that $\cup f_n^{-1}(\alpha,\infty]\subset g^{-1}(\alpha,\infty]$:\begin{align*}
x\in\cup f_n^{-1}(\alpha,\infty]
\ &\Rightarrow \alpha<f_n(x)\ \text{ for at least one $n$}\\
&\Rightarrow \alpha< \sup f_n(x)\\
&\Rightarrow \alpha< g(x)\\
&\Rightarrow x\in g^{-1}(\alpha,\infty].
\end{align*}
A: Consider some point $x\in g^{-1}((a,\infty])$. This means that $\sup_n f_n(x)\in (a,\infty]$ and so there must exist some $m$ for which $f_m(x)\in (a,\infty]$ and hence $x\in f_m^{-1}((a,\infty])\subseteq \bigcup_n f_n^{-1}((a,\infty])$.
For the converse inclusion, suppose that $x\in\bigcup_n f_n^{-1}((a,\infty])$ then there exists at least one $m$ such that $x\in f_m^{-1}((a,\infty])$ and so $f_m(x)\in (a,\infty]$. Since by definition, $g_n(x)\geq f_m(x)$ it follows that $g_n(x)\in (a,\infty]$ and the result follows. 
As intuition think: if the supremum of the sequence of functions at a point $x$ is greater than $a$ then this must mean at least one of the individual functions evaluated at $x$ is also greater than $a$. This is reflected in the union since the union is equivalent to 'or', so $x$ being in the union means at least one of these functions evaluates $x$ at a value greater than $a$. 
A: 

Lemma Suppose $S\subset[-\infty,\infty]$ and $\alpha\in[-\infty,\infty]$. Then $\sup S>\alpha$ if and only if $\alpha>s$ for some $s\in S$.


Proof: Suppose $s\in S$ and $s>\alpha$. If $U$ is an upper bound for $S$ then $U\ge s>\alpha$. But $\sup S$ is in particular an upper bound for $S$; hence $\sup S>\alpha$.
Suppose on the other hand that  $s\le\alpha$ for every $s\in S$. That says precisely that $\alpha$ is an upper bound for $S$; since $\sup S$ is the least upper bound it follows that $\sup S\le\alpha$.
Now fix $x\in X$, and let $S=\{f_n(x)\}$. The following are equivalent:
$$x\in g^{-1}(\alpha,\infty]).$$
$$\sup S=g(x)>\alpha$$
$$\exists n(f_n(x)>\alpha)$$
$$\exists n(x\in f_n^{-1}((\alpha,\infty])$$
$$x\in\bigcup_n f_n^{-1}((\alpha,\infty]).$$
Similar things come up all the time in proofs that this or that is measurable:


Exercise (warning) $\sup S\ge\alpha$ is not equivalent to $\exists s\in S(s\ge\alpha)$. One implication does hold; which one?
Exercise $\inf S<\alpha$ if and only if $s<\alpha$ for some $s\in S$.
Exercise Previous exercise with $\le$ in place of $<$ is false.


Authors are not going to mention these facts explicitly when they apply them as above, taking them as obvious..
