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 $\mathcal{A}_n$ 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

Solution

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

• Our first guess will be that the $\sigma$-$\cup$-closed property does not hold. That is:
• $\neg\big[\forall(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\in\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]\big]$ which is equivalent to (in that $\iff$ or if and only if):
  • $\exists(A_n)_{n\in\mathbb{N} }\subseteq\bigcup_{n\in\mathbb{N} }\mathcal{A}_n[\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n]$ (Caution:Things look very similar here! Read with care!)
• As before: $\bigcup_{n\in\mathbb{N} }A_n\notin\bigcup_{n\in\mathbb{N} }\mathcal{A}_n\iff\neg(\exists k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\in\mathcal{A}_k])$ $\iff$ $\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]$
  • So we need to find a $({ A_n })_{ n = 1 }^{ \infty }\subseteq \bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ such that $\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]$
    • As $\mathcal{A}_n\subset\mathcal{A}_{n+1}$ we know $\exists A\in\mathcal{A}_{n+1}[A\notin\mathcal{A}_n]$, as they're proper subsets of each other.
      • Thus, define $A_n:=X_n$ where $X_n\in\mathcal{A}_{n+1}$ and $X_n\notin\mathcal{A}_n$ (which we have just shown to exist).
        • Now we must prove $\forall k\in\mathbb{N}[\bigcup_{n\in\mathbb{N} }A_n\notin\mathcal{A}_k]$
          • 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 $\mathcal{A}_n\subset\mathcal{A}_{n+1}$ and prove this for that instance than do it in general.

Using hint as instance

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

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

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

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

Thus $\bigcup_{n\in\mathbb{N} }\mathcal{A}_n$ 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 $\sigma$-algebras (whether monotone or not) an algebra of sets?

Then there's

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

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

Solution

Let $\mathcal{B}_1$ and $\mathcal{B}_2$ be $\sigma$-algebras. Is $\mathcal{B}_1\cup\mathcal{B}_2$ an algebra?