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I am going over a tutorial in my real analysis course and there is an exercise that I don't understand a part of his solution.

The exercise is:

Let $S$ be an infinite $\sigma$ algebra on $X$ .Prove that $S$ is not countable.

The proof given is:

Assume by negation $S=\{A_{i}\}_{i=1}^{\infty}$. For each $x\in X$ define $B_{x}:=\cap_{x\in A_{i}}A_{i}$. Note that $B_{x}\in S$ since this is a countable intersection.

Claim: If $B_{x}\cap B_{y}\neq\emptyset$ then $B_{x}=B_{y}$.


If $z\in B_{x}\cap B_{y}$ then $B_{z}\subseteq B_{x}\cap B_{y}$. If $x\not\in B_{z}$then $B_{x}\setminus B_{z}$ is a set in $S$ containing $x$ and is strictly contained in $B_{x}$, in contradiction to the definition of $B_{x}$.

Hence $B_{z}=B_{x}$and similarly $B_{z}=B_{y}$ hence $B_{x}=B_{y}$.

Now consider $\{B_{x}\}_{x\in X}$ - If there is a finite sets of the form $B_{x}$ then $S$ is a union of a finite number of disjoint sets hence it is finite. We conclude there is an infinite number of sets of the form $B_{x}$. $\cup_{i\in A}B_{x_{i}}$ where $A\subseteq\mathbb{N}$ is of at least cardinality $\aleph$.

There are couple of things I don't understand in this proof:

  1. Why the fact that we found a set ($B_{x}\setminus B_{z}$) in $S$ containing $x$ and is strictly contained in $B_{x}$ a contradiction ?

  2. Why if there are only a finite number of different sets of the form $B_{x}$ then $S$ is a union of a finite number of disjoint sets and is finite ?

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You may find it helpful to think about the case where the $\sigma$-algebra $S$ separates points; i.e. for every $x,y \in X$ there exists $A \in S$ with $x \in A$, $y \notin A$. In this case, $B_x = \{x\}$, and the problem reduces to showing that the discrete $\sigma$-algebra on an infinite set is uncountable. –  Nate Eldredge Mar 4 '13 at 2:36

3 Answers 3

up vote 7 down vote accepted
  1. Because $B_x$ is supposed to be the intersection of all measurable sets containing $x$, but you've found a measurable set containing $x$ strictly inside $B_x$.

  2. Because for any measurable set $T$, we have $T=\bigcup_{x\in T}B_x$. Thus, if there are $n$ distinct sets of the form $B_x$, then there are at most $2^n$ elements of $S$.

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Thanks Zev! ${}$ –  Belgi Mar 4 '13 at 1:44
No problem, glad to help! –  Zev Chonoles Mar 4 '13 at 1:45
Could someone explain why we must have strict containment? I see how it's possible, but what if nothing is removed ? –  NS248 May 26 at 5:49

As ncmathsadist mentioned before, there is also a direct prove. For those who want to know how to proceed:

Let be $\mathcal{S}$ an infinite $\sigma$-Algebra. Take any pairwise distinct sets $A_1,A_2,\ldots\in\mathcal{S}$. Then $$\begin{align*}G_0&=A_0^\complement\cap A_1\cap A_2\cap A_3\cap\cdots,\\ G_1&=A_0\cap A_1^\complement\cap A_2\cap A_3\cap\cdots,\\ G_2&=A_0\cap A_1\cap A_2^\complement\cap A_3\cap\cdots,\\ \vdots\end{align*}$$ all lie in $\mathcal{S}$, because the countable intersection is closed in $\sigma$-Algebras, and it is easy to see, that they are pairwise disjoint. But also the countable and finite union of any collection of the $G_k’s$ must lie in $\mathcal{S}$.

You will recognise, that for distinct collections, you’ll get different unions (because the $G_k$’s are pairwise disjoint). Wich Collection you choose, you can encode with a binary sequence $a_0,a_1,a_2,\ldots$ by setting $a_k=1$ for taking $G_k$ into the collection and $a_k=0$ otherwise. So you get an injection from $\{0,1\}^\mathbb{N}$ to $\mathcal{S}$. But you already know, that $\{0,1\}^\mathbb{N}$ is an uncountable set. And that’s why $\mathcal{S}$ must also be uncountable.

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Thanks. I like this direct proof. –  NS248 May 25 at 21:20
Joseph, can you point out, why all but finitely many Gi s are not empty? –  Somabha Mukherjee Aug 11 at 17:01

Show that if a $\sigma$-algebra is infinite, that it contains a countably infinite collection of disjoint subsets. An immediate consequence is that the $\sigma$-algebra is uncountable.

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Can you explain that a little more please? –  Twink Feb 24 at 1:40

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