# Supremum of a sequence of function

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 $$g^{-1}\bigl( (a,+\infty] \bigr)=\bigcup_{n \in \mathbb{N}} f_n^{-1}\bigl((a,+\infty]\bigr).$$ Why?

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You are doing Tao 18.5.10 ? That was my reason for searching for that :D – Epsilon Dec 30 '12 at 15:08

Note that \begin{align*} x\in g^{-1}\bigl( (a,+\infty] \bigr)&\iff g(x)\in(a,+\infty]\\\\ & \iff\sup_{n\in\mathbb{N}}\;f_n(x)\in(a,+\infty]\\\\ *&\iff\exists\,n\in\mathbb{N}\text{ such that }f_n(x)\in(a,+\infty]\\\\ &\iff\exists\,n\in\mathbb{N}\text{ such that }x\in f_n^{-1}\bigl((a,+\infty]\bigr)\\\\ &\iff x\in\bigcup_{n\in\mathbb{N}}f_n^{-1}\bigl((a,+\infty]\bigr) \end{align*} The step labeled $\ast$ is the key step; if we had chosen to look at $[a,+\infty]$, instead of $(a,+\infty]$, it would be false. We need that $$\sup\; a_n>b\iff \lnot(\sup\; a_n\leq b)\iff\lnot(a_n\leq b\text{ for all }n\in\mathbb{N})\iff a_n>b\text{ for some }n\in\mathbb{N}.$$
Thus, we have shown that $$g^{-1}\bigl( (a,+\infty] \bigr)=\bigcup_{n \in \mathbb{N}} f_n^{-1}\bigl((a,+\infty]\bigr).$$