How do we prove a $\sigma$-algebra is also an algebra

measure-theoryprobability

I was reading Probability Essentials 2004 by Jacob and Protter and I found the following definition and explanations about $\sigma$-algebra and algebra on page 7:

Let $\Omega$ be an abstract space, that is with no special structure. Let $2^\Omega$
denote all subsets of $\Omega$, including the empty set denoted by $\emptyset$. With $\mathcal{A}$ being
a subset of $2^\Omega$, we consider the following properties:

  1. $\emptyset \in \mathcal{A}$ and $\Omega \in \mathcal{A}$;
  2. If $A \in \mathcal{A}$ then $A^\mathcal{c} \in \mathcal{A}$, where $A^\mathcal{c}$ denotes the complement of $A$;
  3. $\mathcal{A}$ is closed under finite unions and finite intersections: that is, if $A_1,…,
    A_n$     are all in $\mathcal{A}$, then $\cup_{i=1}^{n} A_{i}$ and $\cap_{i=1}^{n} A_{i}$ are in $\mathcal{A}$ as well (for this it is enough that $A$ be stable by the union and the intersection of any two sets);
  4. $\mathcal{A}$ is closed under countable unions and intersections: that is, if $A_1, A_2,
    A_3,…$     is a countable sequence of events in $\mathcal{A}$, then $\cup_{i=1}^{\infty} A_{i}$ and $\cap_{i=1}^{\infty} A_{i}$ are both also in $\mathcal{A}$.

Definition 2.1. $\mathcal{A}$ is an algebra if it satisfies (1), (2) and (3) above. It is a $\sigma$-algebra, (or a $\sigma$-field) if it satisfies (1), (2), and (4) above.

Note that under (2), (1) can be replaced by either (1'): $\emptyset \in \mathcal{A}$ or by (1''): $\Omega \in \mathcal{A}$. Note also that (1)+(4) implies (3), hence any $\sigma$-algebra is an algebra (but there are algebras that are not σ-algebras).

I am particularly confused about how (1)+(4) implies (3), ie how closed under countable unions and intersections with the addition of empty set and whole set would imply closed under finite unions and intersections. Can someone explain this to me? Thank you.

Best Answer

Suppose that $A_{k}\in\mathcal{A}$ for $1\leq k\leq n$ and $A_{k} = \varnothing$ when $k\geq n + 1$.

Consequently, according to (4) and (1), one has that:

\begin{align*} \mathcal{A}\ni\bigcup_{k=1}^{\infty}A_{k} = \left(\bigcup_{k=1}^{n}A_{k}\right)\cup\left(\bigcup_{k = n+1}^{\infty}\varnothing\right) = \bigcup_{k=1}^{n}A_{k} \end{align*} whence we conclude property (3) holds.

Hopefully this helps!

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