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

Section B

Problem B1

Part i)

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

• Caution:Is [ilmath]\subset[/ilmath] or [ilmath]\subseteq[/ilmath] desired?

Solution

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

Part ii)

Check that if the [ilmath]\mathcal{A}_n[/ilmath] 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, [ilmath]\mathbb{Z}_{\ge 1} [/ilmath] with its sigma-algebras [ilmath]\mathcal{A}_n:=\sigma(\mathcal{C}_n)[/ilmath] where [ilmath]\mathcal{C}_n:=\mathcal{P}(\{1,2,\ldots,n\})[/ilmath] where [ilmath]\{1,2,\ldots,n\}\subset\mathbb{N}[/ilmath] and [ilmath]\mathcal{P} [/ilmath] denotes the power set

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

Solution

Suppose all the [ilmath]\mathcal{A}_i[/ilmath]s are sigma-algebras now, and suppose that [ilmath]\mathcal{A}_{n}\subset\mathcal{A}_{n+1} [/ilmath] still holds. We wish to show that their union, [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] is not a sigma-algebra.

• Our first guess will be that the [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed property does not hold. That is:
• [ilmath]\neg\big[\forall(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]\big][/ilmath] which is equivalent to (in that [ilmath]\iff[/ilmath] or if and only if):
  • [ilmath]\exists(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n][/ilmath] (Caution:Things look very similar here! Read with care!)
• As before: [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\iff\neg(\exists k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}_k])[/ilmath] [ilmath]\iff[/ilmath] [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
  • So we need to find a [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \bigcup_{n\in\mathbb{N} }\mathcal{A}_n [/ilmath] such that [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
    • As [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1} [/ilmath] we know [ilmath]\exists A\in\mathcal{A}_{n+1}[A\notin\mathcal{A}_n][/ilmath], as they're proper subsets of each other.
      • Thus, define [ilmath]A_n:=X_n[/ilmath] where [ilmath]X_n\in\mathcal{A}_{n+1} [/ilmath] and [ilmath]X_n\notin\mathcal{A}_n[/ilmath] (which we have just shown to exist).
        • Now we must prove [ilmath]\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
          • Let [ilmath]k\in\mathbb{N} [/ilmath] be given.
            • Notice [ilmath]A_{k+1}\subseteq\bigcup_{n\in\mathbb{N} }A_n[/ilmath] (it is probably actually [ilmath]\subset[/ilmath] but we do not require this and I cannot be bothered to prove they're not equal) as each element participating in a union is a subset of the union
              • Saving work. Better to use instance (this is what the hint is for) See new notepad.