> [!definition]
>
> Let $(X, \cm, \mu)$ be a [[Measure Space|measure space]], then $\mu$ is **complete** if it is defined for all $\pow{E}$ where $\mu(E) = 0$ ($E$ is a [[Null Set|null set]]).
> [!theoremb] Completion of Measures
>
> Let $(X, \cm, \mu)$ be a measure space. Let $\cn = \bracs{N \in \cm: \mu(N) = 0}$ be the collection of all null sets in $\cm$, and
> $
> \overline{\cm} = \bracs{E \cup F: E \in \cm, F \subset N \text{ where }N \in \cn}
> $
> Then $\overline{\cm}$ is a [[Sigma Algebra|sigma algebra]], and there is a *unique* extension of $\overline{\mu}$ of $\mu$ to a **complete** measure on $\overline{\cm}$.
>
> *Proof*.
>
> ![[completion_representation.png]]
>
> Let $E \cup F \in \overline\cm$ where $E \in \cm$ is a measurable section, and $F \subseteq N \in \cn$ is a "small" section. Rewrite $E \cup F = \pblue{E^\prime} \cup \poran{F^\prime}$ where
> $
> E^\prime = E \setminus N \quad F^\prime = F \cup (E \cap N)
> $
> Then $E \cup F = \pblue{E^\prime} \cup \poran{F^\prime}$ is a disjoint union. We can now decompose any element $S \in \overline{\cm}$ of into a measurable part and a small part that are disjoint, and have the measurable $\pblue{E}$ represent $S$.
>
> For countable union, we can simply do the union by component
> $
> \bigcup_{i \in I}(E_i \cup F_i) = \paren{\bigcup_{i \in I}E_i} \cup \paren{\bigcup_{i \in I}F_i} \in \overline{\cm}
> $
> where $\bigcup_{i \in I}E_i \in \cm$ and each $F_i \subseteq N_i \in \cn$, then
> $
> \bigcup_{i \in I}F_i \subseteq \bigcup_{i \in I}N_i \in \cn
> $
> since a countable union of null sets is a null set, the small component remains small.
>
> ![[completion_complement.png]]
>
> For complement, take any $E \cup F \in \overline\cm$ and assume that they are disjoint. Split $X = N \cup (X \setminus N)$, then
> $
> \begin{align*}
> (E \cup F)^c &= [N \cup (X \setminus N)] \setminus (E \cup F)\\
> &= [N \setminus (E \cup F)] \cup [(X \setminus N) \setminus (E \cup F)] \\
> &= \poran{[N \setminus F]} \cup \pblue{[(X \setminus N) \setminus E]}
> \end{align*}
> $
> where $N \setminus F \subseteq N$ is still a small set, and $(X \setminus N) \setminus E=N^c \cap E^c$ is a measurable set.
>
> So we have $\overline{\cm}$ being a $\sigma$-algebra.
>
> Let $\overline\mu: \overline{\cm} \to [0, \infty]$, $\overline\mu(\textcolor{#0070C0}{E} \cup \textcolor{orange}{F}) = \mu(\textcolor{#0070C0}{E})$, and assume that $\pblue{E}$ and $\poran{F}$ are disjoint. While $\pblue{E}$ may be different based on the choice of representative, $\mu(\pblue E)$ is always the same. To see this, let $S \in \cm$,
> $
> S = \pblue{E_1} \cup \poran {F_1} = \pblue{E_2} \cup \poran {F_2} \quad \poran{F_1} \subseteq \poran{N_1}, \poran{F_2} \subseteq \poran{N_2}
> $
> be two different decompositions.
>
> ![[completion_well_defined.png]]
>
> Then the difference between $\pblue{E_1'}$ and $\pblue{E_2'}$ must lie within $\poran{N_1 \cup N_2}$ (so it's also a null set). Let $x \in \pblue{E_1'}$ but $x \not\in \pblue{E_2'}$, then since $\pblue{E_1'} \cup \poran{F_1'} = \pblue{E_2'} \cup \poran{F_2'}$, $x \in \poran{F_2'} \subseteq \poran{N_2}$. Let $x \in \pblue{E_2'}$ but $x \not\in \pblue{E_1'}$, then since $\pblue{E_1'} \cup \poran{F_1'} = \pblue{E_2'} \cup \poran{F_2'}$, $x \in \poran{F_1'} \subseteq \poran{N_1}$. So
> $
> \begin{align*}
> \pblue{E_1'} \setminus \pblue{E_2'} \subseteq \poran{N_2} \quad \pblue{E_2'} \setminus \pblue{E_1'} \subseteq \poran{N_1}
> \end{align*}
> $
> So the choice of $\pblue{E}$ shouldn't change the value of the measure.
> $
> \begin{align*}
> \pblue{E_1} \cup \poran{F_1} &= [(\pblue{E_1 \cap E_2}) \cup (\poran{E_1 \setminus E_2})] \cup \poran{F_1} \\
> &= (\pblue{E_1 \cap E_2}) \cup (\poran{E_1 \setminus E_2}) \cup \poran{F_1} \\
> \pblue{E_2} \cup \poran{F_2} &= [(\pblue{E_2 \cap E_1}) \cup (\poran{E_2 \setminus E_1})] \cup \poran{F_2} \\
> &= (\pblue{E_1 \cap E_2}) \cup (\poran{E_2 \setminus E_1}) \cup \poran{F_2} \\
> \end{align*}
> $
> where $(\poran{E_1 \setminus E_2}) \cup \poran{F_1}, (\poran{E_2 \setminus E_1}) \cup \poran{F_2} \subseteq \poran{N_1 \cup N_2}$, and
> $
> \overline\mu(\pblue{E_1} \cup \poran{F_1}) = \mu(\pblue{E_1 \cap E_2}) = \overline\mu(\pblue{E_2} \cup \poran{F_2})
> $
> therefore $\overline\mu$ is well-defined.
>
> Since $\overline{\mu}$ inherits properties from $\mu$ on its values on the measurable sets $\pblue{E} \in \cm$, and that its behaviour is uniquely determined by the measurable component of each $\pblue{E} \cup \poran{F}$ (it just ignores the small sets), it remains a measure by """projecting""" everything in $\overline{\cm}$ down to its $\pblue{E}$ component.
>
> Now, $\overline\mu(\emptyset) = \mu(\emptyset) = 0$, and let $\pblue{E_i} \cup \poran{F_i}$ be a countable family of sets in $\overline\cm$, then
> $
> \begin{align*}
> \overline\mu\paren{\bigcup_{i \in I}\pblue{E_i} \cup \bigcup_{i \in I}\poran{F_i}} &= \overline\mu\paren{\pblue{\bigcup_{i \in I}E_i} \cup \poran{\bigcup_{i \in I}F_i}} \\
> &= \mu\paren{\pblue{\bigcup_{i \in I}E_i}} \\
> &= \sum_{i \in I}\mu(\pblue{E_i}) \\
> &= \sum_{i \in I}\overline\mu(\pblue{E_i} \cup \poran{F_i})
> \end{align*}
> $
> and $\overline\mu$ is a measure.
>
> Let $\pblue{E} \cup \poran{F} \in \overline\cm$ be a null set and $\poran{F} \subseteq \poran{N} \in \cn$, then $\overline\mu(\pblue{E} \cup \poran{F}) = \mu(\pblue{E}) = 0$, and $\poran{E} \in \cn$ is also a null set. This means that $\pow{\poran{E \cup N}} \subseteq \overline{\cm}$ and $\overline{\mu}