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If $f \in L_\infty(X, \mathcal{X}, \mu)$, then $|f(x)| \leq ||f||_\infty$ for almost all $x$. Moreover, if $A < ||f||_\infty$, then there exists a set $E \in \mathcal{X}$ with $\mu(E) > 0$ such that $|f(x)| > A$ for all $x \in E$.

I'm studying for my finals and working some problems in my book. I stumbled upon this one and I'm having difficulty approaching it. I was trying to look for a contradiction, but I seem to be going in circles.

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I'll take the definition of $\Vert f\Vert_\infty=\Vert f\Vert$ to be $ \Vert f\Vert= \inf\bigl\{ \alpha>0\mid \mu\bigl( \{ x\ \bigl|\ |f(x)|>\alpha\}\bigr)=0\bigr\}. $ Note that

$\tag{1} \beta>\Vert f\Vert \quad\Longrightarrow\quad\mu\bigl(\{ x \mid |f(x)|>\beta\}\bigr)=0. $


To show $|f(x)|\le\Vert f\Vert$ for almost all $x$:

Let $M=\bigl\{ x \mid |f(x)|>\Vert f\Vert\bigr\}$. For each positive integer $n$, set $M_n=\bigl\{ x \mid |f(x)|>\Vert f\Vert+{1\over n}\bigr\}$. Then by $(1)$ we have $\mu(M_n)=0$ for all $n$. Note that $M=\bigcup_{n=1}^\infty M_n$. Now note that, since a countable union of sets of measure zero has measure zero:
$\mu\bigl(\{x\mid |f(x)|> \Vert f\Vert \bigr) =\mu\Bigl(\bigcup_{n=1}^\infty M_n\Bigr) =0. $ Consequently, $|f(x)|\le \Vert f\Vert $ almost everywhere.


To show that if $A < ||f||_\infty$, then there exists a set $E \in \mathcal{X}$ with $\mu(E) > 0$ such that $|f(x)| > A$ for all $x \in E$:

Suppose that $A<\Vert f\Vert$. By the definition of $\Vert f\Vert$ as the infimum of all $\alpha>0$ such that $\mu\bigl(\{ x \mid |f(x)|>\alpha\}\bigr)=0$, it follows that $\mu\bigl(\{ x \mid |f(x)|>A\}\bigr)\ne0$. So simply take $E=\{ x\mid |f(x)|>A\}$.


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By definition, for all sets $S$ with $\mu(S) > 0$ you have $|f(s)| \leq \|f\|_\infty$ for almost all $s \in S$.

Assume there is a constant $K$ such that $K < \|f\|_\infty$ and there is a set $S$ such that $\mu(S) = 0$ and $|f(s)| \leq K$ on all of $S$. Then this is a contradiction to the definition of $\|f\|_\infty = \inf \{ K > 0 \mid |f(x)| \leq K \text{ for all } x \text{ except on a set of measure } 0\}$.

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    The first line's part of what katari wants to prove, so I suspect the definition of the $\sup$-norm in use is $\|f\|_\infty = \inf{\{M \gt 0\,:\,\mu(\{|f| \gt M\} = 0\}}$, so you should maybe add a word more.2012-05-06