[Math] I don’t understand the definition of an algebra and $\sigma$-algebra

measure-theory

Here is a definition I have for an algebra:

A family $f$ of subsets of the set $\Omega$ is called an algebra if it satisfies:

  • $\emptyset, \Omega \in f$
  • If $A \in f$ then $A^c \in f$
  • If $A,B \in f$ then $A \cup B \in f$

What I don't understand about this definition is that if $f$ is a family of subsets of a set $\Omega$ then how can $\Omega$ be in $f$? I just can't follow that. I also do not understand the motivation of this definition.

I have taken a look at an example online and I have found this:

Let $ \Omega = [0,1]$ and $f$ the family of all subsets of $[0,1]$ w hich can be expressed as a finite union of intervals. Then $f$ is an algebra, but not a $\sigma$-algebra.

The page did not expand at all and simple just stated it was an algebra, and not a $\sigma$-algebra. I feel as though I'm missing something obvious, I cannot understand how $\Omega \in f$.

Best Answer

$f$ (or better, $F$) is a set of subsets of $\Omega$, and is not itself a subset of $\Omega$. Certainly $\Omega \subseteq \Omega$, so it's legitimate for $\Omega$ to be a member of $f$.

The online example you mention is a (Boolean) algebra:

  • $\emptyset, [0,1]$ are both members ($\emptyset$ is a finite union of intervals: the empty sequence of intervals);
  • the intersection of two intervals is an interval (or $\emptyset$), so the intersection of finite unions of intervals is a finite union of intervals (using $A \cap (B \cup C) = (A \cup B) \cap (A \cup C)$);
  • the complement of an interval $\subseteq [0,1]$ is a union of (zero, one or two) intervals, ... and so on.

It's not a $\sigma$-algebra because it doesn't contain some countable unions of its members – for example, $\bigcup_{0 < n \in \mathbb{N}} (\frac 1 {n+1}, \frac 1 n)$ is not a finite union of intervals.

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