Theorem (1.36)

Let $(X,\mathscr{M},\mu)$ be a Measure Space. Let $$\mathfrak{N}=\{ E\subseteq X: \exists A,B\in \mathscr{M}:A\subseteq E\subseteq B\text{ and }\mu(B\setminus A)=0 \}$$then,

1. $\mathfrak{N}$ is a σ-algebra and $\mathscr{M}\subseteq \mathfrak{N}$. We call $\mathfrak{N}$ the $\mu$-completion of $\mathscr{M}$.
2. Let $$\nu(E):=\begin{cases} \mu(E)&E\in\mathscr{M} \\ \mu(A)&E\in \mathfrak{N} \setminus \mathscr{M} \end{cases}$$whenever $A,B\in\mathscr{M}$ with $A\subseteq E\subseteq B$ and $\mu(B\setminus A)=0$. Then $\nu$ is well-defined on $\mathfrak{N}$ and $\nu$ is a Measure on $\mathfrak{N}$.
3. The Measure Space $(X,\mathfrak{N},\nu)$ is complete.