## Supremum of a sequence of function [closed]

Let $(f_n)_{n \in \mathbb{N}}$ a sequence of function from a nonempty-set $X$ to $\bar{\mathbb{R}}=[-\infty,\infty]$, $g\colon X \to \bar{\mathbb{R}}$ defined by $g(x)=\sup(n \in \mathbb{N}) f_n (x)$. Then $g^{-1}\bigl( (a,+\infty] \bigr)=\bigcup_{n \in \mathbb{N}} f^{-1}\bigl((a,+\infty]\bigr)$. Why?

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This question does not seem appropriate for Mathoverflow. Please read the FAQ and visit instead math.stackexchange.com, – Malik Younsi Oct 28 2011 at 13:00
This is a problem that i found in the book "Real Analysy" of Gerald B. Folland. – Katy24 Oct 28 2011 at 13:22
@Katy24: the fact that you found the problem in that book does not make it suitable for MathOverflow (quite the opposite, actually). MathOverflow is for research-type questions, not for routine exercises (even high-level routine exercises). Be assured that there are lots of people on math.stackexchange.com who can answer your question. – Thierry Zell Oct 28 2011 at 15:53