To have a clearer idea I will make the following definition.
Let $ (X,\mathscr{A},\mu) $ a measure space. We define \begin{equation} \mathscr{L}_{\infty}=\{f\in (M,\mathscr{A}): f \mbox{ is bounded } (a.e.) \} \end{equation} and for $ f\in\mathscr{L}_{\infty} $ we have: \begin{equation} \|f\|_{\infty}=\inf\{a>0:\mu(\{x\in X:|f(x)|>a\})=0\}. \end{equation}
What I want is to prove that proposition.
Let $ f\in\|f\|_{\infty}$, then \begin{equation} \|f\|_{\infty}=\sup\{c\geq 0:\mu(\{x\in X:|f(x)|\geq c\})>0\} \mbox{ (alternative definition)}. \end{equation}
I tried to do this:
Let $b=\sup\{c\geq0:\mu(\{x\in X:|f(x)|\geq c\})>0\}$. If $b=0\Rightarrow \mbox{for all } c>0$ we have $\mu(\{x\in X:|f(x)|> c\})=0$, then $\|f\|_{\infty}=0$. Conversely, if $\|f\|_{\infty}=0$, we obtain $|f(x)|\leq\|f\|_{\infty}=0, \ a.e.$ then $f=0, \ a.e. \Rightarrow \ b=0$. Thus $\|f\|_{\infty}=b$.
For the case $\|f\|_{\infty}>0$, I don't know how to do it. I would appreciate your help.