# I don't understand the definition of an algebra and $\sigma$-algebra

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 $\Omega \in f$.

• $\Omega$, like any set, is a subset of itself.
– user169852
Commented Oct 12, 2015 at 22:27

$f$ (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.

• I know this is a little late, but could you elaborate on the countable unions of its members. I don't understand why your example contradicts the definition of $\sigma$-algebra Commented Oct 13, 2015 at 17:29

You're thinking the sigma algebra $\Sigma$ consists of only proper subsets of $\Omega$, but the definition is $\Sigma$ is a subset of $P(\Omega)$, which includes $\Omega$.

The difference between and algebra and sigma algebra is that the former only requires closure of finite unions, and the latter requires closure of countable unions (condition 3 you listed). Thus, every sigma algebra is an algebra.

A ordered double $(\Omega,\Sigma)$ is a said to be a measurable space and the elements of $\Sigma$ are called measurable sets. A function $\mu:\Sigma\to [0,\infty]$ is called a measure on $\Sigma$ if $\mu(\emptyset)=0$ and $\mu(\cup _{i=0}^{\infty}A_i)=\sum_{i=1}^{\infty}\mu(A_i)$ for disjoint sets $A_i\in\Sigma$ (this second property is called countable additivity of the measure). The ordered triple $(\Omega, \Sigma, \mu)$ is called a measure space and function $f:(\Omega, \Sigma, \mu)\to(\Omega', \Sigma', \mu')$ between two measure spaces is said to be measurable if $f^{-1}(A')\in\Sigma$ for all $A'\in\Sigma'$.

As for motivation, if you know to definition of a topological space, you'll notice they objects are formally described similarly. In this analogy, measurable sets are the measure-theoretic version of open sets and continuous functions are the measure-theoretic version of continuous maps. There is a bit of interplay between topology and measure theory. For example, if $\Omega$ is not just a set, but a topological space, then the sigma algebra generated by the open sets is called the Borel sigma algebra.

Another motivation/application is that probability theory where the measure space $(\Omega, \Sigma, \mu)$ is called a probability space if $\mu(\Omega)=1$. In this case, we usually denote $\mu=\mathbb{P}$ and we call the measurable sets (elements of $\Sigma$ "events"). Random variables are measurable functions are where the domain space is a probability space. You may remember from probability theory that for mutually exclusive events $A_i$, you have $\mathbb{P}(\cup_{i=1}^{\infty}A_i)=\Sigma_{i=1}^{\infty}\mathbb{P}(A_i)$. This is just the countable additivity property of measures described above.

Intuitively, you can think of an algebra as taking something, which in modern mathematics is always a set, and (typically, at least) breaking it up into pieces so that you can do things with them. This collection of pieces is called an algebra. One thing you might do with the pieces, for instance, is to assign each of them a probability measure, and then you can determine the probabilities of various groups of pieces.

At any rate, there are some algebras, called trivial algebras, in which there is only one "piece"—the whole set $X$—and you can either take it ($X$) or leave it ($\emptyset$). In such an algebra, $F = \{\emptyset, X\}$. In other algebras, called discrete algebras, every element of $X$ is a piece, and in such algebras, $F = {\cal P}(X)$; that is, $F$ is the power set of $X$, and contains every conceivable subset of $X$ (including, as always, $X$ itself).

In between are various intermediate algebras, which may contain some subsets of $X$, but not others. However, it's always the case that if you have two pieces, the union of those pieces is also a piece. If $X$ is infinite, you will generally want a $\sigma$-algebra.

Such is needed, for instance, to assign a probability to each of an infinite number of terms: If you flip a coin until you get a heads, then you have a $1/2$ chance of getting a heads on the first flip; a $1/4$ chance of getting a heads on the second flip; a $1/8$ chance of getting a heads on the third flip; and so on. Intuitively, we see that these add up to a $1/2+1/4+1/8+\cdots = 1$ probability, or certainty, that you will get a heads somewhere down the road. The ability to add all those up is predicated, fundamentally, on a probability measure assigned to a $\sigma$-algebra consisting of (subsets of) the natural numbers.

The fact that $\sigma$-algebras hold over countable unions means, among other things, that one cannot have a uniform probability distribution over the natural numbers, which follows our intuition, but we can have one over a real interval (which also follows our intuition), even though it has infinite cardinality, by making use of Lebesgue measure. But that is beginning to get out of the scope of this answer. :-)

By the way, algebras are good for things other than probability. That just happens to be where I first encountered them.

Another way to think about it is $f$ is a subset of power set of $\Omega$.

For any set $X$, the power set of $X$, denote $P(X)$ or $2^X$, is the collection of all possible subsets of $X$.

For example,

$X=\{a\}, P(X)=\{\emptyset, \{a\}\}$

$X=\{a, b\}, P(X)=\{\emptyset, \{a\}, \{b\}, \{a, b\}\}$

$X=R$, $P(X)=$ all subsets of real numbers

etc.