If $\overline{M}:=\inf\left\{M:\mu(\left\{x:|f(x)|>M\right\})=0\right\}$ and $\overline{a}=\sup\left\{a:\mu(\left\{x:|f(x)|>a\right\})>0\right\}$
I want proves $\overline{a}=\overline{M}$
I haves this!
$\overline{a}$ is upper bound of $\left\{a:\mu(\left\{x:|f(x)|>a\right\})>0\right\}$. Then $\forall a>\overline{a},\ \mu(\left\{x:|f(x)|>a\right\})=0$ (if $\mu(\left\{x:|f(x)|>a\right\})>0,\ \overline{a}$ would not be the supreme)
Therefore, for all $a>\overline{a},\ a\in \left\{M:\mu(\left\{x: |f(x)|>M\right\})=0\right\}$ Then $\inf\left\{M:\mu(\left\{x: |f(x)|>M\right\})=0\right\}\leq a$
And I do not know how to continue.
You have proved that $a>\overline {a}$ implies $\overline {M}\leq a$. Take limit as $a \to \overline {a}$ to get $\overline {M} \leq \overline {a}$. Now let $\mu \{x:|f(x)| >a\}>0$. Then, for any $M \leq a$ we have $\mu \{x:|f(x)| >M\}>0$. Hence $\{M:\mu \{x:|f(x)| >M\}=0\}$ is disjoint from $(-\infty, a]$ which shows that the infimum of the set is $\geq a$. Thus $\overline {M} \geq a$. Take sup over $a$ to get $\overline {M} \geq \overline {a}$