2
$\begingroup$

Let $X$ be a set. $\mathcal{M}$ be a $\sigma$-algebra and $A \subset X$. Prove that $\mathcal{N}_A = \{E \in \mathcal{M}: A \subset E \text{ or } A \cap E = \emptyset \}$ is a $\sigma$-algebra.

It is easy to show that $\mathcal{N}_A$ is closed under complement. I can show that it is close under finite union by induction but not sure how to extend it to the countable union case.

Any help would be appreciated. Thank you in advance.

$\endgroup$
1
  • $\begingroup$ There is no difference between countable case and finite case. $\endgroup$
    – Landau
    May 5, 2021 at 2:41

1 Answer 1

1
$\begingroup$

Assuming that $A\subseteq\bigcup_{n}E_{n}$ is not valid where $E_{n}\in\mathcal{N}_{A}$, then there exists some $x\in A$ such that $x\notin\bigcup_{n}E_{n}$. In particular, $x\notin E_{n}$ for all $n$. In such a case it cannot be true that $A\subseteq E_{n}$ for any $n$ since $x\in A$ but $x\notin E_{n}$. On the other hand, as $E_{n}\in\mathcal{N}_{A}$, since $A\subseteq E_{n}$ has been ruled out, this implies that $A\cap E_{n}=\emptyset$, but this is true for any $n$, so $\bigcup_{n}(A\cap E_{n})=\emptyset$, which is equivalent to $A\cap(\bigcup_{n}E_{n})=\emptyset$, so $\bigcup_{n}E_{n}\in\mathcal{N}_{A}$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.