Exercises:Measure Theory - 2016 - 1

Problems

Section B

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]
          • Caution:Not convinced this is all "okay" - I also may have spotted why we're given the hint

It is easier to come up with an instance of [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1} [/ilmath] and prove this for that instance than do it in general.

Using hint as instance

Let [ilmath]\mathcal{A}_n:=\mathcal{P}(\{1,2,\ldots,n\}\subset\mathbb{N})[/ilmath]. Let [ilmath]\mathcal{A}:=\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] for convenience. We claim [ilmath]\mathcal{A} [/ilmath] is not a sigma-algebra and we suggest this based on the [ilmath]\sigma[/ilmath]-[ilmath]\cup[/ilmath]-closed property. That is we claim:

• [ilmath]\neg(\overbrace{\forall(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}]}^{\sigma\text{-}\cup\text{-closed} })[/ilmath] [ilmath]\iff\exists(A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}][/ilmath]
  • Notice: [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}[/ilmath] [ilmath]\iff\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] [ilmath]\iff\neg(\exists j\in\mathbb{N}[\bigcup_{n\notin\mathbb{N} }A_n\in\mathcal{A}_j])[/ilmath] [ilmath]\iff\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k][/ilmath]
• Combining these we see we want to show: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}[\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j]][/ilmath] or just: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j][/ilmath]

Proof: [ilmath]\exists (A_n)_{n\in\mathbb{N} }\subseteq\mathcal{A}\forall j\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j][/ilmath]

• Let [ilmath]\mathcal{A}_n:=\mathcal{P}(\{1,\ldots,n\}\subset\mathbb{N})[/ilmath] - where [ilmath]\mathcal{P} [/ilmath] denotes the power set of the finite set [ilmath]\{1,\ldots,n\} [/ilmath] - which is a portion of [ilmath]\mathbb{N} [/ilmath] - the naturals.
• We see that we have [ilmath]\mathcal{A}_n\subseteq\mathcal{A}_{n+1}[/ilmath] or more specifically in fact: [ilmath]\mathcal{A}_n\subset\mathcal{A}_{n+1}[/ilmath].
  • Define [ilmath] ({ A_n })_{ n = 1 }^{ \infty }\subseteq \mathcal{A} [/ilmath] as [ilmath]A_n:=\{n\}[/ilmath]
    • Clearly [ilmath]A_n\in\mathcal{A}_n[/ilmath] for each [ilmath]n[/ilmath].
    • Let [ilmath]j\in\mathbb{N} [/ilmath] be given (so [ilmath]j[/ilmath] is arbitrary and we show the following for all [ilmath]j[/ilmath])
      • We must show [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_j[/ilmath]
      • Lemma: [ilmath]\forall a\forall b\forall c[(a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c][/ilmath]
        • Let [ilmath]a,\ b[/ilmath] and [ilmath]c[/ilmath] be given.
          1. Suppose [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is false, then by the nature of logical implication we do not care about the truth or falsity of the RHS and we're done
          2. Suppose [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is true. We must show that in this case [ilmath]b\notin c[/ilmath]
             • Suppose [ilmath]b\in c[/ilmath]. Then [ilmath]\exists U\in c[a\in U][/ilmath] - namely [ilmath]U:=b[/ilmath] (as by hypothesis, [ilmath]a\in b[/ilmath])
               • But this is the negation of [ilmath]\forall U\in c[a\notin U][/ilmath]! Contradicting the hypothesis
             • Thus we cannot have [ilmath]b\in c[/ilmath] if [ilmath](a\in b\wedge \forall U\in c[a\notin U])[/ilmath] is true.
             • So we must have [ilmath]b\notin c[/ilmath] - as required.
        • This completes the proof of the lemma.
      • Notice [ilmath]j+1\in\bigcup_{n\in\mathbb{N} }A_n[/ilmath] (trivial: [ilmath]j+1\in\mathbb{N}[/ilmath] and [ilmath]A_{j+1}:=\{j+1\}[/ilmath], so [ilmath]j+1\in A_{j+1}\subseteq\bigcup_{n\in\mathbb{N} }A_n[/ilmath], thus [ilmath]j+1\in\bigcup_{n\in\mathbb{N} }A_n[/ilmath])
      • Notice also that [ilmath]\forall U\in \mathcal{A}_j[j+1\notin U][/ilmath]. Proof:
        • Let [ilmath]U\in\mathcal{A}_j [/ilmath] be given.
          • Then [ilmath]U\subseteq\{1,\ldots,j\} [/ilmath], by the implies-subset relation, [ilmath]x\in U\implies x\in \{1,\ldots,j\}[/ilmath].
          • Suppose [ilmath]j+1\in U[/ilmath], then [ilmath]j+1\in\{1,\ldots,j\} [/ilmath] which is a contradiction! Thus we cannot have [ilmath]j+1\in U[/ilmath]
          • We must have [ilmath]j+1\notin U[/ilmath]
      • By the lemma: [ilmath](a\in b\wedge \forall U\in c[a\notin U])\implies b\notin c[/ilmath], in this case:
      • [ilmath](j+1\in \bigcup_{n\in\mathbb{N} }A_n\wedge \forall U\in \mathcal{A}_j[j+1\notin U])\implies \bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j[/ilmath]
    • Thus we see [ilmath]\bigcup_{n\in\mathbb{N} }A_n\notin \mathcal{A}_j[/ilmath] is true for all [ilmath]j\in\mathbb{N} [/ilmath] (since it was arbitrary)
  • For this choice of [ilmath](A_n)_{n\in\mathbb{N} } [/ilmath]

Thus [ilmath]\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[/ilmath] cannot be a sigma-algebra as it isn't closed under countable union (of pairwise disjoint sets as it happens)

Part iii)

There isn't really a part 3 but the last part of part 2 is:

• Is a countable union of [ilmath]\sigma[/ilmath]-algebras (whether monotone or not) an algebra of sets?

Then there's

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

If we do the second one first, we have shown the first, as [ilmath](\mathcal{B}_1,\mathcal{B}_2,\{\emptyset\},\{\emptyset\},\ldots)[/ilmath] is a countable collection of sigma-algebras containing [ilmath]\mathcal{B}_1\cup\mathcal{B}_2[/ilmath] and thus its union cannot be a sigma-algebra.

Solution

Let [ilmath]\mathcal{B}_1[/ilmath] and [ilmath]\mathcal{B}_2[/ilmath] be [ilmath]\sigma[/ilmath]-algebras. Is [ilmath]\mathcal{B}_1\cup\mathcal{B}_2[/ilmath] an algebra?