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The Wikipedia page for $\sigma$-algebra says this set is called a "sigma-algebra" by some, and called a "sigma-field" by others. I'm writing a paper on measure theory, where the topic of sigma-algebra comes up, and wanted to use the correct term. So, I need to figure out which term is more appropriate: field or algebra?

I recall from abstract algebra that the definition of a field is a commutative ring (which itself is a triple $(S,+,\times)$ where $+:S\times S \to S$ and $\times: S \times S \to S$ are binary operators satisfying a number of properties), so that every nonzero element has a multiplicative inverse. The classic example of a field is the rational numbers, $\mathbb{Q}$.

The definition of an algebra, however, I do not recall learning. Wikipedia says an algebra is a vector space (which is itself a triple $(S,+,\cdot)$ where $+:S\times S \to S$ and $\cdot: \mathbb{R} \times S \to S$ are operators satisfying a number of properties) equipped with a bilinear product (what is this?).

Now, I can't immediately see how EITHER of these definitions relate to a $\sigma$-algebra. Let's look at the definition. A collect $\Sigma$ of subsets of $S$ is a $\sigma$-algebra in $S$ if $S \in \Sigma$, $\Sigma$ is closed under complementation, and $\Sigma$ is closed under countable unions.

So, we have the definitions of field, algebra and $\sigma$-algebra in front of us. I can't see how $\sigma$-algebra relates to either algebra or field. And back to the original question -- which name is more appropriate: "sigma-algebra" or "sigma-field"?

Thanks!


Edit As an additional question: When defining the $\sigma$-blank, would it be rigorous to define an algebra first, and then teach $\sigma$-algebra as an example (special case) of an algebra?

Edit 2 Thank you JHance for the awesome answer! With the new construction, let's show all the necessary properties are satisfied:

Additive properties:

  • $(A \cup B) - (A \cap B) = (B \cup A) - (B \cap A)$ (commutativity: $A+B=B+A$)
  • $(A \cup 0) - (A \cap 0) = A - 0 =A$ (additive identity: $A+0=A$)
  • $(A \cup A) - (A \cap A) = A - A =\emptyset$ (additive inverse: $A+A^{-1}=A$)

Multiplicative properties:

  • $A \cap B = B \cap A$ (commutativity: $A \times B = B \times A$)
  • $(A \cap B) \cap C) = A \cap (B \cap C)$ (associativity: $(A \times B) \times C = A \times (B \times C)$)
  • $A \cap X = A$ (multiplicative identity: $A \times 1 = A$)
  • $A \cap \emptyset = \emptyset$ (that is, $A \times 0 = 0$)

Distributive:

  • $((A \cup B) - (A \cap B)) \cap C =...$
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    $\begingroup$ Look up the Wikipedia entries on "universal algebra," "algebra of sets," "field of sets," "boolean algebra," and "boolean ring." Everything is answered in them. $\endgroup$
    – anon
    Commented Aug 18, 2014 at 2:45
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    $\begingroup$ The terms $\sigma$-algebra and $\sigma$-field are interchangeable in measure theory. They have nothing to do with the terms algebra and field one finds within topics of algebra. $\endgroup$
    – user139388
    Commented Aug 18, 2014 at 6:15
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    $\begingroup$ Well, that's disappointing. If that's the case, then neither name is appropriate. No wonder I couldn't see the connection. $\endgroup$ Commented Aug 18, 2014 at 15:39
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    $\begingroup$ @Mathemanic In the most loose sense, an algebra is a set with operations, so the usage in measure theory is not completely inappropriate. $\endgroup$
    – rschwieb
    Commented Aug 27, 2014 at 22:59
  • $\begingroup$ What would be the operations? An algebra is a vector space (or more loosely a set of elements) with two operations (addition and scalar multiplication) and equipped with a bilinear product, according to Wikipedia. However the elements of a $\sigma$-blank are subsets, not vectors. $\endgroup$ Commented Aug 28, 2014 at 0:38

1 Answer 1

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Let $X$ be a total space. If we let symmetric difference take the place of addition: $A \Delta B = (A \cup B) - (A \cap B)$, and let intersection be multiplication, then a sigma algebra of subsets of $X$ becomes a boolean ring with the empty set being $0$ and the total space being $1$. In general a family of sets closed under intersection and symmetric difference, and having the emptyset is a called a ring of sets and is actually a ring (or for some, a rng since it need not have unity), in addition it is a Boolean ring of characteristic 2 as for any $A$ we have $A+ A = \emptyset$ and $A^2 = A$.

An algebra of sets is a ring containing the total space, or in other words, it is a ring with unity, note that this is equivalent to closing the family under complements.

The notion of $\sigma$-rings/algebras corresponds to a more measure theoretic desire for countable operation, although they still behave well since all operations are Abelian.

Note that a $\sigma$-algebra with addition and multiplication given by symmetric difference and intersection is actually a (unital) algebra over the field $\mathbb{Z}/2\mathbb{Z}$, which can be imbedded in the algebra as the set $\{\emptyset,X\}$

Edit: Looking back at your question, you also expressed some confusion as to what an algebra is. For most things you'll see an algebra is a set $A$ that is at once equipped with the structures of a ring and vector space (or module) over a field ( or ring) $K$. However we want ring addition and vector addition on $A$ to coincide, and we want ring multiplication to be bilinear, which really just boils down to saying that for $v,w \in A$ and $\lambda \in K$, we want $$(\lambda v)\cdot w = v \cdot(\lambda w) = \lambda(v \cdot w)$$

If you're comfortable with rings, an elegant way of phrasing this in that context is simply that a unital algebra is a ring $A$ together with a ring of scalars $R$ and homomorphism of rings $f: R \to A$. Scalar multiplication is just $\lambda v = f(\lambda) \cdot v$

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  • $\begingroup$ Dear JHance, thank you so much for the awesome answer! I'm updating the original post to try to show all the necessary properties are satisfied! $\endgroup$ Commented Aug 28, 2014 at 2:10
  • $\begingroup$ I don't think that's really the most general sense of "an algebra"... $\endgroup$ Commented Aug 28, 2014 at 2:39
  • $\begingroup$ @MaliceVidrine: Yes, my notion of algebra here is over a field, associative, and the ring homomorphism definition assumes the algebra is unital. I'll edit my language. $\endgroup$
    – jxnh
    Commented Aug 28, 2014 at 2:44

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