When is an algebra on a set S not a sigma algebra?

Question

I’m learning the basics of sigma algebras and measures, and my textbook says the following:

Definition 1.1 Let $S$ be a non-empty set. A collection $\Sigma_0 \subset 2^S$ is called an algebra (on $S$) if

(i) $S \in \Sigma_0$,

(ii) $E \in \Sigma_0 \Rightarrow E^\mathsf{c} \in \Sigma_0$,

(iii) $E, F \in \Sigma_0 \Rightarrow E \cup F \in \Sigma_0$.

And then:

Definition 1.2 Let $S$ be a non-empty set. A collection $\Sigma \subset 2^S$ is called a $\sigma$-algebra (on $S$) if it is an algebra and $\bigcup_{n = 1}^\infty E_n \in \Sigma$ as soon as $E_n \in \Sigma$ ($n = 1, 2, \ldots$).

My question is, when is a collection $\Sigma_0$ an algebra but not a $\sigma$-algebra? Diagrammatically, I can see that for any $\Sigma_0 \subset 2^S$ the union of all elements of $\Sigma_0$ always gives $S$, which seems to mean that every algebra on $S$ is a $\sigma$-algebra.

What am I not seeing here?

Answer 1

An algebra on a set $S$ may fail to be closed under countable unions, and therefore fail to be a $\sigma$-algebra.

Here is an example: let $S=\{1,2,3,\dots\}$, and let $\mathcal{A}$ be the set of all subsets $A\subset S$ such that either $A$ or $A^c$ is finite. One can check that $\mathcal{A}$ is an algebra. However, if we define $A_n=\{2n-1\}$ for $n=1,2,3,\dots$ then each $A_n\in \mathcal{A}$ but $$ \bigcup_{n=1}^{\infty}A_n=\{1,3,5,\dots\}\not\in\mathcal{A}$$ so $\mathcal{A}$ is not closed under countable unions.

Answer 2

This is an example from Theory of Measure and Integration, J Yeh.

$\mathcal{R}=\{\emptyset\}\cup\{\text{finite union of rectangles of the type }(a_{1},b_{1}]\times(a_{2},b_{2}],-\infty\leq a_{i}<b_{i}\leq\infty\}$, where we denote $(a_{i},\infty]=(a_{i},\infty)$.

Now consider $A_{n}=(n-1/2,n]\times(0,1]$ and $\displaystyle\bigcup_{n}A_{n}$ which is not an element of $\mathcal{R}$.