What is this set? I have the following set,
${F}' = \{ A \subset \Omega :^\exists B,N \in {F}$ s.t. $\mu(N) = 0$ and $A \Delta B \subset N\}$
Here, $F$ is a collection of the subsets of $\Omega$, precisely $(\Omega,F,\mu)$ is a measure space. And in measure theory $F'$ is called as the completion of the $\sigma$-algebra $F$.
I would like to understand this set precisely. How can we construct this set from scratch? (I know it is an abstract thing but I am asking just to understand it).
My trial: Take $A \subset\Omega$. Check if there exists any $B,N \in F$ s.t. $\mu(N) = 0$ and $A \Delta B \subset N$. If so, take $A \in F'$, if not get rid of it. The set of all such $A$'s is $F'$. Is it true?
If there is a more-easy-to-understand explanation, I would be very happy to see that. Thanks!
 A: In the measure space $(\Omega,F,\mu)$, it is possible to have sets $E\subset D$ such that $D\in F$, $\mu(D)=0$, and yet, $E\notin F$. For a variety of reasons, we may want to avoid this situation, so that if $D$ has measure zero, so does any subset of it (in particular, any subset of $D$ is in the $\sigma$-algebra of measurable sets). 
The collection $F'$ is what results from attempting to remedy this situation: We add to $F$ all subsets of measure zero sets. Of course, doing that results into a collection of sets that may no longer be a $\sigma$-algebra. So we consider the $\sigma$-algebra this collection generates. And $F'$ is precisely what we obtain. 
So, the short description is: $F'$ is the smallest $\sigma$-algebra that contains $F$ as a subset, and has all subsets of $\mu$-measure zero sets. 
The next step to have a complete measure space is to see that we can extend $\mu$ to a measure $\mu'$ defined on all of $F'$. There is a natural candidate: If $A\in F'$ and $B,N$ are as in the definition, so $B,N\in F$, $B\Delta A\subset N$, and $\mu(N)=0$, what we are saying is that $A$ and $B$ differ from each other on a negligible set, so it makes sense to define $\mu'(A)=\mu(B)$. Of course, one needs to verify that $\mu'$ is well defined (there may be $B',N'$ different from $B,N$ with $B',N'\in F$, $A\Delta B'\subset N'$, and $\mu(N')=0$, and in that case we need to verify that $\mu(B)=\mu(B')$). We also need to verify that $\mu'$ is a measure on $F'$, of course.
