Completeness of Measure spaces A metric space X is called complete if every Cauchy sequence of points in X has a limit that is also in X. It's perfectly clear to me.
A measure space $(X, \chi, \mu)$ is complete if the $\sigma$-algebra contains all subsets of sets of measure zero. That is, $(X, \chi, \mu)$ is complete if $N \in \chi$, $\mu (N) = 0$ and $A \subseteq N$ imply $A \in \chi$. Technically, I could understand the definition, but can't get the logic behind it.
Questions:
1) Why do we care only about subsets of sets of measure zero to determine completeness?
2) How does the completeness of measure spaces relate to a completeness of metric spaces?
3) Could you suggest a concrete elementary example of a measure space (preferably, with simple sets) that isn't initially complete and then is completed?
 A: In some proofs in measure theory, one often encounters the following: we have a set $E$ of measure $0$ and a subset $A\subseteq E$, and we want to show that $A$ has measure $0$. However, if the measure space is not complete, this will not always be the case, since $A$ might not be measurable. This is fixed by considering the completion of the measure space.
Moreover, if a measure space is complete, then the following property holds: if $A,B$ are measurable with $A\subseteq B$, $\mu(A)=\mu(B)$ and $A\subseteq C\subseteq B$, then $C$ is measurable, with $\mu(C)=\mu(A)=\mu(B)$. That's why we only care for sets of measure zero, since it implies that "completeness" extends to sets of measure different than zero.
Edit: Suppose that $A\subseteq B$, $\mu(A)=\mu(B)<\infty$ and the space is complete. Suppose now that $A\subseteq C\subseteq B$. Then $C\setminus A\subseteq B\setminus A$, and the last set has measure zero, so by completeness $C\setminus A$ is measurable, with $\mu(C\setminus A)=0$. From this, we obtain that $C$ is measurable, with measure $$\mu(C)\leq\mu(C\setminus A)+\mu(A)=\mu(A).$$ But, $A\subseteq C$, so $\mu(A)\leq \mu(C)$, therefore $\mu(C)=\mu(A)$.
This fact explains my last comment before the edit, and could be a definition for "completeness for sets of positive measure". Therefore, this shows that if we complete the measure to include all sets that should have measure zero, then we include all sets that should have any given positive measure.
A: 3) Take the sample space to be $\Omega=\{1,2,3\}$, the $\sigma$-algebra to be $\mathcal F=\{\emptyset,\Omega,\{1,2\},\{3\}\}$, and let $P$ be the probability measure on $(\Omega,\mathcal F)$ such that $P(\{3\})=1$. Then $P(\{1,2\})=0$ and $\{1\}$ is a non-$\mathcal F$-measurable subset of $\{1,2\}$. The probability space $(\Omega,\mathcal F,P)$ is not complete.
A: 1) There's only one possible way to assign subsets of a set of measure zero a measure: Its measure is zero. This is not possible for other sets.
2) Unrelated (as far as I know at least)
3) Either the Borel sigma-algebra with Lebesgue measure or - if you want a very trivial example - the trivial measure on a sigma-algebra which is not the power set.
