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

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.

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 $$G^c \in \mathcal{G}$$.

• Why is $(1)$ true? @saz – J. Doe Jan 22 at 10:26
• @J.Doe It follows from the "general fact" which I stated directly afterwards.I've just rewritten it a bit... perhaps it's clearer now. – saz Jan 23 at 7:10

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, $$0\le \Bbb P(A^c-(A^c\cap F^c)) = \Bbb P(A^c\cap F^{cc}) = \Bbb P(A^c\cap F) \le \Bbb P(F) =0 \ .$$

• I got that $A^c\cap F^c\subseteq G^c\subseteq A^c$, but how does $\mathbb{P}(A^c\cap F)=0$ leads us to $G^c\in\mathcal{G}$? @dan_fulea – J. Doe Jan 18 at 18:07
• $G^c$ is then between $A_1:=A^c\cap F^c$ and $A_1\cup N_1$ for that $N_1$ that makes $A_1\cup N_1=A^c$. – dan_fulea Jan 18 at 18:11
• Sorry for stubborning but I still don't get it. We have $A_1=A^c\cap F^c\subseteq G^c\subseteq A^c=A_1\cup N_1$, but what does $G^c$ equal to for getting into the set $\mathcal{G}$? @dan_fulea – J. Doe Jan 18 at 18:23
• $G^c=A_1\cup\color{red}{(G^c-A_1)}$ and the red entry is a subset of $N_1$. Thanks for asking, please always ask, do not even hesitate to do it. It is then my fault of not finding the argument that hits the point... – dan_fulea Jan 18 at 18:27
• We need that $G^c-A_1=(A^c\cap N^c)-(A^c\cap F^c)\in\mathcal{N}$. Why is that true? I observe that $G^c-A_1\subseteq A^c\cap N^c$ but not sure if it helps. @dan_fulea – J. Doe Jan 19 at 9:59