Exercises:Measure Theory - 2016 - 1/Section B/Problem 1

Section B

Problem B1

Part i)

Suppose that $\mathcal{A}_n$ are algebras of sets satisfying $\mathcal{A}_n\subset \mathcal{A}_{n+1}$. Show that $\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ is an algebra.

• Caution:Is $\subset$ or $\subseteq$ desired?

Solution

1. Closed under complementation: $\forall A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]$
   • Let $A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ be given.
     • By definition of union: $\big[A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\big]\iff\big[\exists i\in\mathbb{N}[A\in\mathcal{A}_i]\big]$
       • As $\mathcal{A}_i$ is an algebra of sets itself:
         • $A^\complement\in\mathcal{A}_i$
       • Thus $A^\complement\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$
   • Since $A\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ was arbitrary, we have shown this for all such $A$, thus $\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ is closed under complementation.
2. Closed under union: $\forall A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]$
   • Let $A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ be given.
     • By definition of union we see:
       1. $\exists i\in\mathbb{N}[A\in\mathcal{A}_i]$ and
       2. $\exists j\in\mathbb{N}[B\in\mathcal{A}_j]$
       • Define $k:=\text{Max}(\{i,j\})$
         • Now $\mathcal{A}_i\subseteq\mathcal{A}_k$ and $\mathcal{A}_j\subseteq\mathcal{A}_k$ (at least one of these will be strict equality, it matters not which)
         • Thus $A,B\in\mathcal{A}_k$
         • As $\mathcal{A}_k$ is an algebra of sets:
           • $\forall C,D\in\mathcal{A}_k[C\cup D\in\mathcal{A}_k]$
         • Thus $A\cup B\in\mathcal{A}_k$
         • So $A\cup B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ (explicitly, $\exists h\in\mathbb{N}[A\cup B\in\mathcal{A}_h]$ - namely choosing $h$ to be $k$ as we have defined it, and we have this if and only if $A\cup B$ is in the union, by the definition of union)
   • Since $A,B\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ were arbitrary we have shown this for all such $A$ and $B$. As required.

Part ii)

Check that if the $A_n$ are all sigma-algebras that their union need not be a sigma-algebra.

Is a countable union of sigma-algebras (whether monotonic or not) an algebra?

Hint: Try considering the set of all positive integers, $\mathbb{Z}_{\ge 1}$ with its sigma-algebras $\mathcal{A}_n:=\sigma(\mathcal{C}_n)$ where $\mathcal{C}_n:=\mathcal{P}(\{1,2,\ldots,n\})$ where $\{1,2,\ldots,n\}\subset\mathbb{N}$ and $\mathcal{P}$ denotes the power set

Check that if $\mathcal{B}_1$ and $\mathcal{B}_2$ are sigma-algebras that their union need not be an algebra of sets