The proposition and the first part of its proof are given below:
My questions are:
1- why in case $p=\infty,$ we are sure that there is a subset $E_{0}$ of $E$ of measure zero for which $g$ is bounded on $E \setminus E_{0}$?
2- How is the simple approximation lemma leads to that there is a sequence of simple functions on $E \setminus E_{0}$ that converges uniformly on $E \setminus E_{0}$ to $g$ and why this leads to that it converges with respect to the $L^{\infty}(E)$ norm.
Here is the picture of the Simple Approximation Lemma.
The first is because of the fact that the essential supremum of $g$ is finite (as it is in $L^\infty(E)$). Then on $E\setminus E_0$ the Simple Approximation Lemma applies. We find such $\phi_n$ and $\psi_n$ for $\varepsilon=\frac{1}{n}$ (if you like we can extend them to $E$ by setting them $0$ or $+\infty$ or whatever you like on $E_0$ to have them to be simple functions on $E$, which are still essentially bounded), and quite clearly on $E\setminus E_0$, $\|\phi_n - f\|_\infty \le \frac{1}{n}$ (everything is finite here), so this holds as essential sup on $E$ too. Basically we can ignore what happens on $E_0$.