Proving a property of $L^\infty$ spaces Let $f$ be continuous and bounded on $\mathbb{R}^d$. Show that $\|f\|_\infty=$ sup$\{|f(\vec{x})|: \vec{x}\in\mathbb{R}^d\}$ relative to Lebesgue measure.
Basically I need to prove that $\|f\|_\infty$ is just the largest magnitude that $f$ takes on $\mathbb{R}^d$. I already have from the definition that:
$\|f\|_\infty=$ inf$\{t>0:|f|\le t $ $ a.e.\}$. Since $f$ is continuous and bounded shouldn't I be able to trivially argue that sup$\{|f(\vec{x})|: \vec{x}\in\mathbb{R}^d\}=$ inf$\{t>0:|f|\le t \: a.e.\}$?
 A: We have $\|f\|_{\infty}=\inf\{M|m(\{x\in [a,b]:|f(x)|>M\})=0\}$ for all $f\in L^{\infty}[a,b]$. Thus for all $n\in \mathbb{N}$, there exists $M_n$ such that $M_n<\|f\|_{\infty}+\frac{1}{n}$ and $m(A_n)=0$, where $A_n=\{x\in [a,b]:|f(x)|>M_n\}$. Let $A=\bigcup\limits_{n=1}^{\infty}A_n$. Thus $M_n\to \|f\|_{\infty}$. We claim that $A=\{x\in [a,b]:|f(x)|>\|f\|_{\infty}\}$. Let $x\in A$. Thus there exists $n\in \mathbb{N}$ such that $x\in A_n$ and so $|f(x)|>M_n\geq \|f\|_{\infty}$. Again let $x\in [a,b]$ such that $|f(x)|>\|f\|_{\infty}$. Thus there exists $n\in \mathbb{N}$ such that $|f(x)|>M_n\implies x\in A_n$. Therefore $x\in A$. Hence $A=\{x\in [a,b]:|f(x)|>\|f\|_{\infty}\}$. Since $m(A)\leq \sum\limits_{n=1}^{\infty}m(A_n)=0$, therefore $\|f\|_{\infty}\in \{M|m(\{x\in [a,b]:|f(x)|>M\})=0\}$ and so $\|f\|_{\infty}=\min\{M|m(\{x\in [a,b]:|f(x)|>M\})=0\}$.
Let $f$ be continuous. We claim that $\|f\|_{\infty}=\max\{|f(x)|: x\in [a,b]\}$. Let $A=\{x\in [a,b]:|f(x)|\leq \|f\|_{\infty}\}$. We shall show that $A=[a,b]$. If possible, let $x\in [a,b]\setminus A$. Since $m([a,b]\setminus A)=0$, there exists a sequence $(x_n)\subset A$ such that $x_n\to x$. By continuity of $f$ at $x$, we have $f(x_n)\to f(x)$. Now $x_n\in A\implies |f(x_n)|\leq \|f\|_{\infty}$ for all $n\in \mathbb{N}$ and so $|f(x)|\leq \|f\|_{\infty}\implies x\in A$ which is a contradiction. Hence $A=[a,b]$ and so $\|f\|_{\max}=\max\{|f(x)|:x\in [a,b]\}\leq \|f\|_{\infty}$. Again $\{x\in [a,b]: |f(x)|>\|f\|_{\max}\}=\emptyset$ and so is of measure zero. Thus by definition $\|f\|_{\infty}\leq \|f\|_{\max}$. Hence $\|f\|_{\infty}=\|f\|_{\max}$.
$[a,b]$ can be replaced by $\mathbb R^d$. Then $\max$ will be replaced by $\sup$.
