Completion of a measure space using null sets

Question

Let $(X,\mathcal{A},\mu)$ be a measure space, let $\mathcal{N}$ be the collection of null sets with respect to $\mathcal{A}$ and let $\mathcal{B}=\sigma(\mathcal{A}\cup\mathcal{N})$. Show that $(X,\mathcal{B},\mu)$ is complete and is the completion of $(X,\mathcal{A},\mu)$.

What I know:

$\mathcal{N}=\{A\in\mathcal{A} : \exists B\in\mathcal{A} \text{ s.t. } A\subset B \text{ and } \mu(B)=0\}$

$\mathcal{B}=\bigcap\left\{\mathcal{A}_{\alpha} : \mathcal{A}_{\alpha} \text{ is a $\sigma$-algebra, } \mathcal{A}\cup\mathcal{N}\subset\mathcal{A}_{\alpha}\right\}$

In order to show $(X,\mathcal{B},\mu)$ is complete, I must show that it contains all of the null sets with respect to $\mathcal{B}$.

Take any $C\in X$ such that there is $D\in\mathcal{B}$ with $C\subset D$ and $\mu(D)=0$. I must show $C\in\mathcal{B}$; that is, for any $\sigma$-algebra $\mathcal{A}_{\alpha}$ containing $\mathcal{A}\cup\mathcal{N}$, $C\in\mathcal{A}_{\alpha}$.

If we are able to write $D=A\sqcup N$, where $A\in\mathcal{A}$ and $N\in\mathcal{N}$, then $\mu(D)=\mu(A)+\mu(N)=0$, so $\mu(A)=0=\mu(N)$, which implies $A$ is null, so $A\in\mathcal{N}$. Thus $D$ is null with respect to $\mathcal{A}$, so $C$ is null with respect to $\mathcal{A}$, so $C\in\mathcal{N}\subset\mathcal{B}$. Therefore $(X,\mathcal{B},\mu)$ is complete.

Does this work? Are we allowed to write $D=A\sqcup N$? This doesn't seem obvious from the definition of $\mathcal{B}$.

Answer 1

Using Michael Greinecker's comment:

$\mathcal{A}\cup\mathcal{N}$ is a $\sigma$-algebra since $\varnothing=\varnothing\cup\varnothing\in\mathcal{A}\cup\mathcal{N}$, and for any countable collection $\{A_i\cup N_i : A_i\in\mathcal{A}, N_i\in\mathcal{N}\}_{ i=1,2,\ldots}$, $$\cup_{i=1}^{\infty}(A_i\cup N_i)=\cup_{i=1}^{\infty}A_i\cup\cup_{i=1}^{\infty}N_i\in\mathcal{A}\cup\mathcal{N},$$ since $\mathcal{A}$ and $\mathcal{N}$ are $\sigma$-algebras. Therefore $\mathcal{A}\cup\mathcal{N}$ is a $\sigma$-algebra. This must coincide with $\mathcal{B}$ since $\mathcal{B}$ is the smallest $\sigma$-algebra containing $\mathcal{A}\cup\mathcal{N}$.

Answer 2

I don't believe that the accepted answer is thorough enough. It is enough to show that $\mathcal{A}\cup\mathcal{N}$ is a $\sigma$-algebra.

First, $\varnothing=\varnothing\cup\varnothing\in \mathcal{A}\cup\mathcal{N}$ and $X=X\cup\varnothing\in\mathcal{A}\cup\mathcal{N}$.

Next, we need to show closed under complements. Let $A\cup N\in \mathcal{A}\cup\mathcal{N}$. We then have $(A\cup N)^C=A^C\cap N^C$. To continue further, recall that $N\subseteq M$ where $M\in\mathcal{A}$. This gives us $$(A\cup N)^C=A^C\cap N^C=A^C\cap (M^C\cup (M-N))=(A^C\cap M^C)\cup (A^C\cap (M-N))$$ Then, since $A\in\mathcal{A}$ and $N\in\mathcal{A}$, $\mathcal{A}$ being a $\sigma$-algebra is enough to guarantee that $A^C\cap N^C\in \mathcal{A}$. Moreover, $M-N\subseteq M$, so $A^C\cap (M-N)\subseteq M$ as well. This gives us $A^C\cap (M-N)\in \mathcal{N}$ by definition as $M$ is a measure zero set contained in $\mathcal{A}$.

Finally, we show closure under countable union. We have $$\bigcup_{i=1}^\infty (A_i\cup N_i) = \left(\bigcup_{i=1}^\infty A_i\right)\cup\left(\bigcup_{i=1}^\infty N_i\right)$$ Then note that $\bigcup_{i=1}^\infty N_i\subseteq \bigcup_{i=1}^\infty M_i$ by definition of $\mathcal{N}$ and $\mu\left(\bigcup_{i=1}^\infty M_i\right)=0$ via countable subadditivity. This gives us that $\bigcup_{i=1}^\infty N_i\in \mathcal{N}$. We also have $\bigcup_{i=1}^\infty A_i\in \mathcal{A}$ since $\mathcal{A}$ is a $\sigma$-algebra, completing the proof.