Is a sigma-algebra just a special kind of algebra or are the two concepts quite distinct?

My understanding is that a sigma-algebra must be defined on a set. Furthermore I read on Wikipedia that an algebra is a set with algebraic structure.

This makes it seem as if the two concepts have little to do with each other. An algebra being a special kind of set (a set with algebraic structure), and, a sigma-algebra being defined on another set as a set of subsets that fullfils the 3 special requirements of a sigma-algebra.

  • $\begingroup$ You misunderstand. The algebraic structure is in addition to the set, and is something that exists "over" the set in that sense. $\endgroup$
    – YiFan Tey
    Commented Oct 30, 2019 at 12:36
  • $\begingroup$ Wikipedia: "an algebraic structure on a set A is a collection of finitely operations on A. The set A with this structure is also called an algebra." $\endgroup$ Commented Oct 30, 2019 at 12:52
  • $\begingroup$ You are right. But if, in a broader sense, you consider an algebra as a set with additional and multiplicational operations on it, the two concepts have some overlap. $\endgroup$
    – amsmath
    Commented Oct 30, 2019 at 12:54
  • $\begingroup$ Okey, so if I understand correctly the word "algebra" may refer to "a set with algebraic structure" and may also refer to "the algebraic structure" itself, right? $\endgroup$ Commented Oct 30, 2019 at 13:02

3 Answers 3


The term algebra is a bit misleading... there is no natural scalar multiplication on the power set. However, you can view a $\sigma$-algebra as a certain type of ring.

Let $X$ be a set and note that the power-set $\mathcal{P}(X)$ is a ring under the following operations:

\begin{align} A+B&:=A\Delta B =(A\setminus B)\cup (B\setminus A) \\ A\cdot B&:= A\cap B \end{align} It is sort of tedious, but completely elementary, to check that this ring is commutative, associative, with $\emptyset$ as $0$ and $X$ as $1$ (the worst part is probably, surprisingly, checking that $+$ is associative).

Now, a set algebra is a subset $\mathcal{A}\subseteq \mathcal{P}(X)$ such that $X\in \mathcal{A}$, $A\in \mathcal{A}$ implies $X\setminus A\in \mathcal{A}$ and $A,B\in \mathcal{A}$ implies $A\cup B\in \mathcal{A}$. You can check that this is actually equivalent to stating that $\mathcal{A}$ is a unital sub-ring of $\mathcal{P}(X)$.

So what's a $\sigma$-algebra $\mathcal{A}\subseteq \mathcal{P}(X)$? It satisfies all of the above axioms, but also if $(A_n)_{n\in \mathbb{N}}$ is a sequence of elements in $\mathcal{A}$, then so is $\cup_{n=1}^{\infty} A_n$. However, taking complements, this is equivalent to the extra requirement being that $\cap_{n=1}^{\infty} A_n\in \mathcal{A}$. Hence, a $\sigma$-algebra is somehow a unital sub-ring of $\mathcal{P}(X)$ which also admits infinite products - it's worth noting that products over index sets of arbitrary cardinality make natural sense in $\mathcal{P}(X)$.

I think that's as far as actual similarities to abstract algebra goes, and I'm not actually sure how useful it is to think about $\sigma$-algebras this way.

  • 2
    $\begingroup$ I guess you can say that $\mathcal{P}(X)$ is actually an $\mathbb{F}_2$-algebra, since $A+A=\emptyset$. So in that sense, I suppose you can actually claim that a $\sigma$-algebra still manages to be an actual algebra. Again, not entirely convinced that this is a useful persepctive. $\endgroup$ Commented Oct 30, 2019 at 12:52

The term "algebra" has different meanings.

One of them is that an "algebra on a set $X$" is a subcollection of subsets of $X$ that is closed under the formation of complements and finite unions.

In that context a $\sigma$-algebra can indeed be looked at as a special algebra, because it is a collection of subsets that is closed under the formation of complements and of countable unions.


Throughout I will only discuss algebra (the object) and not algebra (the subject.)

In the loosest sense of the word, as in universal algebra, an "algebra" is a set with some operations defined on it. The operations can even be 'infinitary', as is the case with a $\sigma$-algebra.

But frequently the term "algebra" is meant in the sense of $F$-algebra, that is, an algebra over a field, associative or not. In that case the operations obey different conditions than those of a $\sigma$-algebra, but the basic idea is the same: a set with operations.

The axioms of $\sigma$ algebras are not sufficient to be interpreted as those of an $F$ algebra, so you cannot say it is a special case of an $F$-algebra. They are distinct notions that are distant relations, you could say.


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