I only have a question about the proof of this theorem.
In the proof , we took any cauchy sequence $(f_n)$ in $L^\infty$ and we want to show that this sequence converges. We also considered the set $A_k$ of measure zero and its countable union A to arrive to $\left | f(x)-f_n(x) \right |\leq \frac{1}{k}$ for all $x\in X-A,n\geq N_k$
Till here everything is clear, but then we set $f=0$ on $A$ . Why ? Is it because we want $f$ to be measurable ? And couldn't we choose something other than $0$ ?
I would say it's just a convention. You want a function $f$, you have found it's value outside of $A$, the easiest $f$ you can think about on the whole space is found setting it to be $0$ in $A$. Note that it is not really important since it will just be a representative of the class of functions in $L^\infty$.
So yes, there's nothing in the value $0$ itself in this case, you may very well choose $1/17$ or $f\equiv g$ on $A$ for some function $g$.