The notion of a $\sigma$-field is easy to understand given basic knowledge of elementary set theory. Unfortunately, many texts formally characterize σ-fields without motivating the characterization they give, which makes the notion harder to grasp that it needs to be.

A $\sigma$-field $\mathcal{F}$ is simply a class of sets with **three properties**. $\sigma$-fields are important for probability theory because **those properties ensure that $\mathcal{F}$ contains all of the sets to which the axioms of probability refer** when applied to elements of $\mathcal{F}$.

Let $\Omega$ be a partition of the sure event. Let $\mathcal{F}$ be a class of subsets of $\Omega$.

#### The Axioms of Probability

- For all $A$ in $\mathcal{F}$, $\mbox{Pr}(A)\geq 0$.
- $\mbox{Pr}(\Omega)=1$
- For all sequences $A_1, A_2, \ldots$ of disjoint sets in $\mathcal{F}$, $\mbox{Pr}( \cup_{k=1}^\infty A_k)=\sum_{k=1}^\infty \mbox{Pr}(A_k)$.

#### The Properties of a σ-Field

- Axiom 2 refers to $\Omega$, so $\mathcal{F}$
**must contain $\Omega$**. - For every $A$ in $\mathcal{F}$, Axioms 2 and 3 together imply $\mbox{Pr}(A^c)=1-\mbox{Pr}(A)$.
^{1}Thus, $\mathcal{F}$**must contain**$A^c$$A$.**if it contains** - Axiom 3 refers to $\cup_{k=1}^\infty A_k$, so $\mathcal{F}$
**must contain this set if it contains**$A_1, A_2, \ldots$**.**

More compactly, a $\sigma$-field is a class of sets that contains the sure event and is closed under complementation and countable union.

- This result follows from the fact that Axiom 2 says $\mbox{Pr}(\Omega)=\mbox{Pr}(A \cup A^c)=1$, and Axiom 3 says $\mbox{Pr}(A \cup A^c)=\mbox{Pr}(A)+\mbox{Pr}(A^c)$.

To **share your thoughts about this post**, comment below or send me an email. Comments support $ LaTeX$: surround your code in single dollar signs for inline mode or in double dollar signs for display mode.

Conor Mayo-Wilson says

Greg,

I could be wrong, but I don’t think the functional constraints on a probability function $P$ presuppose that $\mathcal{F}$ must be closed under complements. In other words, there exists a set $\Omega$, a collection ${\cal F}$ of subsets of $\Omega$, and a function $P$ from ${\cal F}$ to the unit interval satisfying the three axioms above and such that ${\cal F}$ is not a $\sigma$-algebra.

Let $\Omega$ be the set $\{0,1\}$, and let ${\cal F}$ be the collection of sets containing $\Omega$ and the singleton set $\{1\}$. Define a function $P$ from ${\cal F}$ to the unit interval such that $P(\Omega) = 1$ and $P({1}) = 1/2$. Then $P$ satisfies the first two probability axioms by stipulation. Because there are no disjoint sets in ${\cal F}$, it is also vacuously countably additive (by the way, as you know, the third axiom in your post ought to be edited to apply to disjoint events only).

The problem with the argument in your footnote, as I see it, is the following. You are right that *if* both $A$ and $A^c$ were members of ${\cal F}$, then countable additivity would entail that $P(A \cup A^c) = P(A) + P(A^c)$. However, if either $A$ or its complement is not a member of ${\cal F}$, then either $P(A)$ or $P(A^c)$ is not well-defined and the axiom is inapplicable.

Conor

Greg Gandenberger says

Thanks, Conor! I’ve corrected Axiom 3. Your point is compatible with mine and provides a useful contrast.

Here’s a different way to put what I’m saying. Take a class $\mathcal{C}$ of subsets of $\Omega$ and assign probabilities to them. Add to that collection any other subsets of $\Omega$ such that the axioms of probability tell you what the probability of that subset would have to be if it were in $\mathcal{C}$, given your previous assignments. Keep adding sets in this way until you can’t add any more. The resulting collection of sets will be a $\sigma$-field.

As you point out, it’s nevertheless true that the axioms of probability can be satisfied in classes of sets that are not $\sigma$-fields.

kiyarash says

why don’t we define the measure on the whole omega (set of all possible out comes) ?

Wikipedia says “In general, if one wants to associate a consistent size to each subset of a given set while satisfying the other axioms of a measure, one only finds trivial examples like the counting measure. This problem was resolved by defining measure only on a sub-collection of all subsets; the so-called measurable subsets, which are required to form a σ-algebra”

what other kinds of useful measures do we have in probability ? ( i mean non_theoretical ones ) that we cannot have if we define the measure over the whole omega?

Greg Gandenberger says

There are impossibiity theorems showing that you can’t assign a measure to all subsets; see e.g. Ch. 3 of Billingsley’s .

Fabian Mancilla says

So, a $\sigma$-field is the same that a $\sigma$-algebra?

Greg Gandenberger says

Yes