Let be $\Omega$ a set ,$(A_n)_{n \in \mathbb{N}}$ ring on $\Omega$ so that $A_n \subset A_{n+1}, \forall n \in \mathbb{Z}.$ Show that $\cup_{n \in \mathbb{N}}$ is a ring.
If $A_n$ are $\sigma$ algebras does it conclude that $\cup_{n \in \mathbb{N}} A_n$ is a $\sigma$ algebra?
I'm pretty much sure that if $A_n$ is a $\sigma$ algebra it will conclude that $\cup_{n \in \mathbb{N}} A_n$ is a $\sigma$ algebra as well.
Thank you in advance for your help.
What you write (edit: wrote in a previous version) doesn't make any sense. You seem to be confusing the algebraic notion of "ring" (a set $R$ together with two binary operations, $+$ and $\cdot$, such that $(R,+)$ is an abelian group, $(R,\cdot)$ is a semigroup, and $\cdot$ distributes over $+$ on both sides), with the notion of "ring" from measure theory.
Given a set $\Omega$, a "ring on $\Omega$" means a nonempty collection $\mathcal{R}$ of subsets of $\Omega$ such that:
So what you need to check is that if each $\mathcal{A}_n$ satisfies these properties, and $\mathcal{A}_i\subseteq \mathcal{A}_{i+1}$ for each $i=1,2,\ldots$, then $\mathcal{R}=\cup_{n=1}^{\infty}\mathcal{A}_n$ also has these two properties.
As for $\sigma$-algebras, a ring on $\Omega$ is an algebra on $\Omega$ if it is closed under absolute complements (that is, if $A\in\mathcal{R}$ then $\Omega\setminus A\in\mathcal{R}$). It is a $\sigma$-algebra if it is also closed under countable unions: if $A_n\in\mathcal{R}$ for $n=1,2,\ldots$, then $\cup_{n=1}^{\infty} A_n\in\mathcal{R}$.
The increasing union of algebras is again an algebra, but the increasing union of $\sigma$-algebras need not be a $\sigma$-algebra, as shown in the several exampless at the link.
