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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.

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

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  • $\begingroup$ Why is $(1)$ true? @saz $\endgroup$ – J. Doe Jan 22 at 10:26
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    $\begingroup$ @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. $\endgroup$ – saz Jan 23 at 7:10
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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, $$ 0\le \Bbb P(A^c-(A^c\cap F^c)) = \Bbb P(A^c\cap F^{cc}) = \Bbb P(A^c\cap F) \le \Bbb P(F) =0 \ . $$

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  • $\begingroup$ 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 $\endgroup$ – J. Doe Jan 18 at 18:07
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    $\begingroup$ $G^c$ is then between $A_1:=A^c\cap F^c$ and $A_1\cup N_1$ for that $N_1$ that makes $A_1\cup N_1=A^c$. $\endgroup$ – dan_fulea Jan 18 at 18:11
  • $\begingroup$ 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 $\endgroup$ – J. Doe Jan 18 at 18:23
  • $\begingroup$ $G^c=A_1\cup\color{red}{(G^c-A_1)}$ 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... $\endgroup$ – dan_fulea Jan 18 at 18:27
  • $\begingroup$ 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 $\endgroup$ – J. Doe Jan 19 at 9:59

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