Measure Theory

**galahad** · April 29th 2010, 05:09 PM

Quick question: Does the sigma-algebra generated by a countable number of sets have a countable number of elements?

**Tikoloshe** · April 29th 2010, 05:28 PM

It can be shown that any infinite sigma algebra has cardinality greater than or equal to that of the continuum.

**galahad** · April 30th 2010, 03:38 AM

Thanks, but how? I tried proving that you can't have a bijection from the natural numbers to the sigma-algebra, with no sucess...

**Tikoloshe** · April 30th 2010, 09:17 AM

Given a countably infinite $\sigma$ -algebra $\Sigma$ over an infinite set X, let $\{E_n\}_{n\in\mathbb{N}}$ be an enumeration of $\Sigma$ . Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in $\Sigma$ . Namely, consider a sequence $b=b_1b_2b_3\ldots$ , and let $F_b=\bigcup_{b_i=1}E_i$ .

**Focus** · May 1st 2010, 12:59 AM

Originally Posted by Tikoloshe

Given a countably infinite $\sigma$ -algebra $\Sigma$ over an infinite set X, let $\{E_n\}_{n\in\mathbb{N}}$ be an enumeration of $\Sigma$ . Wlog, we can take these to be disjoint. Then, for every binary sequence, we may produce another set in $\Sigma$ . Namely, consider a sequence $b=b_1b_2b_3\ldots$ , and let $F_b=\bigcup_{b_i=1}E_i$ .

This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.

You probably need to use the axiom of choice. Let X be a set and $\mathcal{F}$ an infinite sigma algebra over X. Then define the equivalence relation ~ as x~y if and only if $x \in A \in \mathcal{F} \iff y \in A$ . Then each equivalence class E(x) is measurable, as it is the countable intersection of sets in the sigma-algebra containing x.

Let I be the set of representatives from each equivalence class (Axiom of Choice here). Then $X= \bigcup_{x \in I} E(x)$ is a disjoint union. Hence if we suppose that $\mathcal{F}$ is countable then I must either be finite or countable. Thus we have an injection from $\{0,1\}^I$ to $\mathcal{F}$ as we can pick an element of $\mathcal{F}$ by taking $\bigcup_{x\in S \subset I}E(x)$ (as they are equivalence classes, you do have a guarantee of getting distinct sets).

This is impossible as $|\{0,1\}^I|> |I|$ if I is countable or finite.

**Tikoloshe** · May 1st 2010, 10:51 AM

Originally Posted by Focus

This is not immediate. I don't think it works quite that simply, there is no guarantee that two sequences will give different sets.

I don’t understand your objection. If $\{E_i\}_{i\in\mathbb{N}}$ is a countable sequence of disjoint, nonempty sets, then using different index sets, the unions over the $E_i$ will be distinct sets. I.e., if $A,B\subset\mathbb{N}$ where $A\neq B$ , then $\bigcup_{i\in A}E_i\neq\bigcup_{i\in B}E_i$ . Given an arbitrary countable sequence in a $\sigma$ -algebra, you can of course make a new sequence of pairwise disjoint, nonempty sets (without AC).

Originally Posted by Focus

You probably need to use the axiom of choice.

I don’t think AC is necessary for the proof (c.f. mine above). Also, here is another proof which takes a different approach.

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