Inverse image of a particular measurable function, from rudin real and complex analysis theorem 1.14 Taken from Rudin - Real and complex analysis theorem 1.14
If $f_n : X \rightarrow [-\infty,+\infty], n = 1,2,\ldots$ is a sequence of measurable functions. Defining 
$$
g = \sup_{n \geq 1} f_n
$$
Why for fixed real $\alpha$ we have
$$
g^{-1}((\alpha,+\infty]) = \bigcup_{n \geq 1} f^{-1}_n ((\alpha,+\infty])
$$
?
 A: One direction: Suppose $x$ is in the union on the right. Then $x\in f_{n_0}^{-1}((\alpha,\infty])$ for some $n_0.$ This means $f_{n_0}(x) >\alpha.$ Hence $g(x) \ge f_{n_0}(x) > \alpha.$ Thus each set in the union on the right is contained in the set on the left, hence so is the union of such sets. 
A: First, to show that $g^{-1}((\alpha,\infty]) \subset \bigcup_{n=1}^\infty f_n^{-1}((\alpha,\infty])$, we can assume the left side is non-empty, for this is vacuously true when empty.   
Fix an $x\in g^{-1}((\alpha,\infty])$,  then $g(x) \in (\alpha,\infty]$.  Now, in order to show $x \in \bigcup_{n=1}^\infty f_n^{-1}((\alpha,\infty])$ it suffices to show that $f_{n_0}(x) \in (\alpha, \infty]$ for some $n_0 \geq 1$.  
Let $g(x) = \alpha_0 \in (\alpha,\infty]$. Because $\alpha_0$ is the supremum, we have $f_k(x) \le \alpha_0$ for every $k \ge 1$.  However, we can find a $f_{n_0}(x)$ close enough to $\alpha_0$ so that it is still on the interval $(\alpha, \infty].$
By properties of supremum, for any $\epsilon > 0$, there exists a $f_{k}(x)$ such that 
$$\alpha_0 -  \epsilon < f_{k}(x).$$
If $\epsilon = \alpha_0 - \alpha$ $(>0)$, then $\alpha < f_{n_0}(x)$ for some $n_0$ and we are done.
Now to show $\bigcup_{n=1}^\infty f_n^{-1}((\alpha,\infty]) \subset g^{-1}((\alpha,\infty])$, assume the left side is non-empty, similarly. Fix $x \in \bigcup_{n=1}^\infty f_n^{-1}((\alpha,\infty])$, then $f_{n_0}(x) \in (\alpha, \infty]$ for some $n_0$.  Once again, because $g(x)$ is the supremum, $$\alpha <f_{n_0} \le g(x)$$and we are done.
