If $(\Omega,F,\mu)$ is the completion of a measure space $(\Omega,F_{0},\mu)$ and $f$ is a Borel measurable function on $(\Omega,F)$, then show that there exists a Borel measurable function g on $(\Omega,F_{0})$ such that $f=g$ except on a subset of a set in $F_{0}$ of measure 0.
The author proposes to start with indicator functions, but I'm not even sure how to start tackling this problem.
No topology is mentioned, so I will assume that the word "Borel" is not there.
If $X\in F$, then $X=A\cup E$, with $A\in F_0$ and $\mu(E)=0$. So $1_X=1_A$ with the possible exception of a subset of $E$; by definition, $E\subset E_0$ for some nullset $E_0\in F_0$, so $1_X=1_A$ outside of $E_0$. That is, $1_X=1_A$ almost everywhere in $F_0$ and $1_A$ is measurable in $F_0$. Using this idea, we get that for any simple $f$ measurable on $F$, there exists a simple function $g$ measurable on $F_0$ and $f=g$ a.e.
Now for any function $f$ measurable on $F$, there exists a sequence of simple functions $f_n$ with $f_n\to f$ pointwise. Let $g_n$ be simple, measurable on $F_0$, with $f_n=g_n$ up to a nullset $E_n\in F_0$. Put $E=\bigcup_nE_n\in F_0$, still a nullset. Define $g=f\,1_{E^c}$. As $g=\lim_ng_n\,1_{E^c}$, it is a limit of functions which are measurable in $F_0$, so it is also measurable in $F_0$. And $g=f$ outside of $E$.