Inverse image of a particular measurable function, from rudin real and complex analysis theorem 1.14

Question

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

$$ g^{-1}((\alpha,+\infty]) = \bigcup_{n \geq 1} f^{-1}_n ((\alpha,+\infty]) $$ ?

Answer 1

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.