Completion of a $\sigma$-algebra is a $\sigma$-algebra

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space. Let \begin{align*} \mathcal{N}&= \left\{N\subseteq\Omega:\exists F\in\mathcal{F}, N\subseteq F,\mathbb{P}(F)=0\right\} \\ \mathcal{G}&=\left\{A\cup N:A\in\mathcal{F},N\in\mathcal{N}\right\} \end{align*} Prove that $\mathcal{G}$ is a $\sigma$-algebra.

I've shown that $\Omega\in\mathcal{G}$ and that if $(A_n)_n\subseteq\mathcal{G}$ then $\cup_n A_n\in\mathcal{G}$. How can I show that if $A\in\mathcal{G}$ then $A^c\in\mathcal{G}$?

I also noticed $\mathcal{F}\subseteq\mathcal{G}$ if it helps somehow.

Answer 1

Let $G=A \cup N$ for $A \in \mathcal{F}$ and $N \in \mathcal{N}$. By definition, there exists $F \in \mathcal{F}$ such that $N \subseteq F$ and $\mathbb{P}(F)=0$. Clearly,

$$G^c = A^c \cap N^c = \underbrace{A^c}_W \backslash \underbrace{(N \cap A^c)}_{U}$$

As $$\underbrace{N \cap A^c}_{U} \subseteq \underbrace{F \cap A^c}_{V} \subseteq \underbrace{A^c}_{W}$$ it follows that

$$G^c = \underbrace{\big(A^c \backslash (F \cap A^c)\big)}_{W \backslash V} \cup \underbrace{\big( (F \cap A^c) \backslash (N \cap A^c) \big)}_{V \backslash U}; \tag{1}$$

here we have used the general fact that

$$U \subseteq V \subseteq W \implies W \backslash U = (W \backslash V) \cup (V \backslash U).$$ Consequently, we have shown that

$$G^c = \tilde{A} \cup \tilde{N} \tag{2}$$

for

$$\tilde{A} := A^c \backslash (F \cap A^c)\quad \text{and} \quad \tilde{N}:= (F \cap A^c) \backslash (N \cap A^c).$$

As $\tilde{N} \subseteq F \in \mathcal{F}$ and $\mathbb{P}(F)=0$, we have $\tilde{N} \in \mathcal{N}$. Morover, $A \in \mathcal{F}$ and $F \in \mathcal{F}$ implies $\tilde{A} \in \mathcal{F}$. Consequently, $(2)$ shows that $G^c \in \mathcal{G}$.

Answer 2

Let $G$ be an event in $\mathcal G$, so we can write it in the form $G=A\cup N$ with $A\in\mathcal F$, $N\in\mathcal N$, so $N\subseteq F$, for some suitable $F\in \mathcal F$, $\Bbb P(F)=0$.

Then $$ \begin{aligned} A &\subseteq {\color{red}{G}}=A\cup N\subseteq A\cup F\text{ leads to}\\ A^c &\supseteq {\color{red}{G^c}}=A^c\cap N^c\supseteq A^c\cap F^c\ , \end{aligned} $$ so $G^c$ lies between two events in $\mathcal G$ that differ by a null set, $$ 0\le \Bbb P(A^c-(A^c\cap F^c)) = \Bbb P(A^c\cap F^{cc}) = \Bbb P(A^c\cap F) \le \Bbb P(F) =0 \ . $$

Answer 3

Lemma 1: $\cup_{i\in \mathbb{N}}F_i$ with $\mathbb{P}(F_i)=0$ is such that $\mathbb{P}(\cup_{i\in \mathbb{N}}F_i)=0$.

Proof: Taking $F_i'=F_i\setminus(\cup_{j=1}^{i-1}F_i)$ it is easy to see these are disjoint and their union is invariant ($\cup_{i\in \mathbb{N}}F_i=\cup_{i\in \mathbb{N}}F_i'$). Hence: $$ \mathbb{P}(\cup_{i\in \mathbb{N}}F_i)=\mathbb{P}(\cup_{i\in \mathbb{N}}F_i')=\sum_{i=1}^\infty \mathbb{P}(F_i')\leq \sum_{i=1}^\infty \mathbb{P}(F_i)=0\quad \quad\square$$

Lemma 2: An element is of the form $A\cup N$ if and only if it can be writen as $(A\cup N_1 )\setminus N_2$

Proof: Indeed, $A\cup N= (A\cup N)\setminus \emptyset$. For the other direction, we have $N_i\subset \hat{N_i}\in \mathcal{F}$ with $\mathbb{P}(\hat{N_i})=0$. With this notation in mind: $$(A\cup N_1)\cap N_2^C=(A\cap N_2^C)\cup (N_1 \cap N_2^C)$$ As $N_2\subset \hat{N}_2$, we have $\hat{N}_2^C\subset N_2^C$ and $A\cap N_2^C=(A\cap \hat{N}_2^C)\cup (A\cap N_2^C\cap \hat{N}_2)$. Therefore: $$(A\cup N_1)\cap N_2^C=(A\cap \hat{N}_2^C)\cup (A\cap N_2^C\cap \hat{N}_2)\cup (N_1 \cap N_2^C)$$ $A\cap \hat{N}_2^C\in \mathcal{F}$. On the other hand, the other two terms are contained in $\hat{N}_1\cup \hat{N_2}\in \mathcal{F}$. By our first result, $\mathbb{P}(\hat{N}_1\cup \hat{N_2})=0$. $\square$

All of the hard work is done and we need only apply Lemma's 1 and 2:

1. The emptyset and the entire set are clearly of the form $A\cup N$:

Indeed $\emptyset= \emptyset \cup \emptyset$ and $X=X\cup \emptyset$

2. The complement of $(A\cup N_1)\setminus N_2$ is still of the same form

$$((A\cup N_1)\setminus N_2 )^C= ((A\cap N_2^C)\cup (N_1\cap N_2^C))^C= (A^C \cup N_2)\setminus (N_1 \cap N_2^C)$$ The result follows because $A^C\in \mathcal{F}$, $N_1\cap N_2^C\subset \hat{N_1}$.

3. Countable union of the form $A\cup N $ is still of the same form:

$$(\cup_{i\in \mathbb{N}}A_i)\cup (\cup_{i\in \mathbb{N}}N_i)$$ But $\cup_{i\in \mathbb{N}}A_i\in \mathcal{F}$, $\cup_{i\in \mathbb{N}} N_i \subset \cup_{i\in \mathbb{N}} \hat{N}_i \in \mathcal{F}$ and by Lemma 1, $\mathbb{P}(\cup_{i\in \mathbb{N}} \hat{N}_i) =0$