Definition of complete in the context of Lebesgue measurable sets I came across this statement on Lebesgue measurable sets. 

The Lebesgue measurable sets are said to be complete because every subset of a null set is again measurable and the lebesgue measurable sets are a completion of the Borel sets. 

What does complete and completion mean in this context?
Any help with this doubt is appreciated.
 A: A $\sigma$-algebra is complete with respect to a measure $\mu$ if every subset of a set of $\mu$-measure zero is measurable.   The completion of a $\sigma$-algebra with respect to the measure $\mu$ is just the $\sigma$-algebra generated by throwing in all of the subsets of $\mu$-measure zero. 
A: A measure $\mu$ defined on a $\sigma$-algebra $\mathcal F$ of subsets of a set $X$ is complete if and only if any subset of a set of measure $0$ is measurable, that is, it belongs to $\mathcal F$ (and has measure $0$). For example, the Borel sets of reals with the standard measure do not form a complete measure because one can show that there are only $\mathfrak c$ many Borel sets, but the Cantor set has measure $0$ and $2^{\mathfrak c}>\mathfrak c$ subsets, so some of them are not measurable.
Any measure admits a completion. The completion of the measure $\mu$ (defined on $\mathcal F$) is defined by considering the smallest $\sigma$-algebra $\mathcal G$ of subsets of $X$ that contains all subsets of any $Y\in\mathcal F$ of measure $0$. For example, Lebesgue measure is obtained that way starting with the Borel sets. It turns out a set $A$ is in $\mathcal G$ iff there are some $B,C\in\mathcal F$ and a set $D\subset C$ with $\mu(C)=0$ and $A=B\cup D$.
For Lebesgue measure one can show that a set $A$ is measurable iff there is some Borel set $B$ such that the symmetric difference $A\triangle B=(A\setminus B)\cup(B\setminus A)$ is contained in some Borel set of measure $0$. 
One can check that the measure $\mu$ admits a unique extension to $\mathcal G$, that is, there is a unique measure $\hat\mu$ defined on $\mathcal G$ with the property that $\hat\mu(B)=\mu(B)$ for all $B\in\mathcal F$. With $A,B,C,D$ as above, we have that $\hat\mu(A)=\mu(B)$. Of course, one cannot quite define $\hat\mu$ this way, without first checking that this definition makes sense, that is, if $E,F\in\mathcal F$, and  $\mu(F)=0$, $G\subset F$, and also $A=E\cup G$, then we have $\mu(B)=\mu(E)$. This measure $\hat\mu$ is complete in the sense of the first paragraph.
The Wikipedia article has a few more details.
