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There are 2 things which create a lot of confusion in my mind.

1) I know that every sigma-algebra is an algebra. But not every algebra is a sigma-algebra. Put differently, it seems that sigma-algebras are subsets of algebras (?).

On the other hand, a sigma-algebra is an algebra complemented to include countably infinite operations (Wikipedia). Hence, it seems that algebras are subsets of sigma-algebras (?).

Could anybody please clarify? I just can't get my head around.

2) Also, it is true to say that:

finite algebra $\subset$ sigma-algebra (i.e. countably complete algebra) $\subset$ complete algebras?

If so, what do we mean by an arbitrary algebra?

Thank you very much.

Asaf Karagila
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  • If I understood your question, you are confusing algebras (and sigma algebras) with their elements. – Dalamar May 01 '14 at 18:32

3 Answers3

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An algebra, in this context, an algebra of sets, is a family of subsets of some $X$ which includes $X$, closed under complements and finite unions. For example consider all the finite and co-finite sets of natural numbers.

A $\sigma$-algebra is an algebra which is closed under countable unions. So we require not only that $A\cup B$ is in the algebra whenever $A$ and $B$ are, but if for every $n\in\Bbb N$ $A_n$ is in the algebra, then $\bigcup_{n\in\Bbb N} A_n$ is in the algebra. The example of finite and co-finite sets doesn't work anymore. Consider $A_n=\{2n\}$, then $\bigcup_{n\in\Bbb N}A_n$ is the set of even integers, which is neither finite nor co-finite.

We can have algebras whose "completion" is even larger, we can require unions of $\aleph_1$ elements from algebra to be in the algebra, or we can require that any union of elements is an element. The latter are known as complete Boolean algebras, and for example $\mathcal P(X)$ is a complete Boolean algebra for any $X$ (note that the union of any collection of subsets of $X$ is still a subset of $X$).

So we have that $\sf\text{algebra}\supsetneq\sigma\text{-algebra}\supsetneq\text{complete algebra}$, and not as you have written where the inclusions go the other way around.

Asaf Karagila
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    Your answer really helped me understand this point. However, there is an example I can't wrap my head around. Let $X =\mathbb{R}$. Define an algebra $\mathcal{A}$ which consists of sets in the form $$ \cup_{i = 1}^k (a_i, b_i]$$ where $-\infty \leq a_i < b_i \leq \infty$. The book says that this is an algebra of sets, but not quite a $\sigma$ algebra – Tyler Hilton Jan 18 '15 at 21:24
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    Yes, it's an algebra. To see why, note that it's clearly closed under finite unions and the complement of $(a_i,b_i]$ is $(-\infty,a_i]\cup(b_i,\infty]$. But it's not a $\sigma$-algebra since it does not contain any open interval (except $\Bbb R$ itself) but $(0,1)$ is the intersection of $(0,1+\frac1n]$. Also, in the presentation you give, you need to explicitly add $\varnothing$; you can fix that by allowing $a_i\leq b_i$ rather than $a_i<b_i$ and thus generate the empty set as $(0,0]$. – Asaf Karagila Jan 18 '15 at 21:36
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1)

a. Humans who wear hats are subset of all humans. b. To turn a human into a hat-human, put a hat on him/her.

Arbitrary algebra = arbitrary human.

2)

Humans $\supset$ Hat-humans $\supset$ The Pope

But if you look at Popes, set of Humans (as restricted to Popes) $=$ set of Hat-Humans

PA6OTA
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Let $X$ be a set. A $\sigma$-algebra of subsets of $X$ is a family $\Sigma$ of subsets of $X$ which contains the empty set and which is closed under the formation of complements and countable unions, that is:

(i) $\emptyset\in\Sigma$;

(ii) If $E\in\Sigma$ then $-E\in\Sigma$;

(iii) If $E_0,E_1,...,E_n,...\in\Sigma$ then $\bigcup_{n=0}^{+\infty} E_n\in\Sigma$.

Now, if a family of subsets of $X$ is closed under the formation of countable unions, then it is closed under the formation of finite unions.

Indeed, set $E=E_0$ and $F=E_n$ for every $n\geq 1$. Then $E\cup F=\bigcup_{n=0}^{+\infty} E_n\in\Sigma$.

Hence, if $\Sigma$ is a $\sigma$-algebra then it is an algebra, that is to say, $\sigma$-algebras $\subset$ algebras.

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