$|f(x)| > A$ for all $x \in E$. 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.
 A: 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 $E=\{ x\mid |f(x)|>A\}$.

A: 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 $\|f\|_\infty = \inf \{ K > 0 \mid |f(x)| \leq K \text{ for all } x \text{ except on a set of measure } 0\}$.
