Exercise 1.4.26 (Completion of a measure space)

Let $N:=\{N\in B:\mu (N)=0\}$ be the set of all null sets on $(X,B,\mu )$, and let $\overline{N}:=\{F\subseteq N:N\in N\}$ be a “completion” of $N$. We argue that we can construct a complete measure space by defininig a new $\sigma$-algebra $\overline{B}$ by

and a new measure $\overline{\mu }$ by

We quickly verify that the properties from the exercise.

• $\overline{B}$ is a $\sigma$-algebra.

  • em We have $\varnothing \in B\subseteq \overline{B}$.
  • For any $\overline{E}\in \overline{B}$ we have $\overline{E}=E\cup N$ for some $E\in B,N\subseteq N^{\prime }\in N\subseteq B$. Then ${\overline{E}}^c=E^c\cap N^c$. We have $N^c={N{}^{\prime }}^c\cup (N^{\prime }∕N)$. Thus, ${\overline{E}}^c=E^c\cap ({N{}^{\prime }}^c\cup (N^{\prime }∕N)=[E^c\cap {N{}^{\prime }}^c]\cup [E^c\cap (N^{\prime }∕N)]$. The former term is an intersection of two elements in $B$, and is thus itself contained in $B$. The latter term is a subset of $N^{\prime }\in N$. Thus, ${\overline{E}}^c\in \overline{B}$ by definition.
  • Let ${\left({\overline{E}}_n\right)}_{n\in \mathbb{N}}$ be a sequence of sets in $\overline{B}$, let ${\left(E\right)}_{n\in \mathbb{N}},{\left({\overline{N}}_n\right)}_{n\in \mathbb{N}}$ denote elements of $B$ and $\overline{N}$ such that ${\overline{E}}_n=E_n\cup {\overline{N}}_n$ for each $n\in \mathbb{N}$. Then ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\overline{E}}_n={\bigcup }_{n=1}^{\infty }{E}_n\cup {\overline{N}}_n={\bigcup }_{n=1}^{\infty }{E}_n\cup {\bigcup }_{n=1}^{\infty }\overline{N}$. The former set is contained in $B$ by $\sigma$-algebra property, and the latter is a subset of a null set ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{N}_n$ where ${\left(N_n\right)}_{n\in \mathbb{N}}$ in $N$ is such that $\forall <!--\; FUNCTION\; APPLICATION\; -->n\in \mathbb{N}:\overline{N}\subseteq N$.

• $\overline{\mu }$ is a measure.

  • We have $\overline{\mu }(\varnothing )=\mu (\varnothing )=0$.
  • Let ${\left({\overline{E}}_n\right)}_{n\in \mathbb{N}}$ be a sequence of sets in $B$, we represent them as ${(E_n+{\overline{N}}_n)}_{n\in \mathbb{N}}$. We then have $\overline{\mu }({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{\overline{E}}_n)=\overline{\mu }({\bigcup }_{n=1}^{\infty }{E}_n\cup {\bigcup }_{n=1}^{\infty }{\overline{N}}_n)$. The former union is an element of $B$, the latter of $\overline{N}$. By definition of $\overline{\mu }$ we thus have $\overline{\mu }({\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{E}_n\cup {\bigcup }_{n=1}^{\infty }{\overline{N}}_n)=\mu ({\bigcup }_{n=1}^{\infty }E)={\sum }_{n=1}^{\infty }\mu (E_n)={\sum }_{n=1}^{\infty }\overline{\mu }(E_n\cup {\overline{N}}_n)={\sum }_{n=1}^{\infty }\overline{\mu }({\overline{E}}_n)$.
• $(X,\overline{B},\overline{\mu })$ is the coarsest refinement of $(X,B,\mu )$ that is a complete measure space.
  This space is complete, since any sub-null is contained in , and the latter is the subset of $\overline{B}$ by construction. Let $(X,{\overline{B}}^{\prime },{\overline{\mu }}^{\prime })$ be another refinement of $(X,B,\mu )$ which is complete. Then $\overline{B}\subseteq {\overline{B}}^{\prime }$ (Any $\overline{E}\in \overline{B}$ is $\overline{E}=E\cup \overline{N}$. Since ${\overline{B}}^{\prime }$ is a refinement of $B$, we have $E\in B\subseteq {\overline{B}}^{\prime }$. Since ${\overline{B}}^{\prime }$ is complete, $\overline{N}\in \overline{N}\subseteq {\overline{B}}^{\prime }$. Thus $\overline{E}\in {\overline{B}}^{\prime }$.)