# Sigma-algebra

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 Sigma algebra $\forall A\in\mathcal{A}[A^C\in\mathcal{A}]$$\forall\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}\left[\bigcup_{n=1}^\infty A_n\in\mathcal{A}\right]$For a [ilmath]\sigma[/ilmath]-algebra [ilmath](X,\mathcal{A}\subseteq\mathcal{P}(X))[/ilmath]
Note: A Sigma-algebra of sets, or [ilmath]\sigma[/ilmath]-algebra is very similar to a [ilmath]\sigma[/ilmath]-ring of sets.
A ring of sets is to an algebra of sets as a [ilmath]\sigma[/ilmath]-ring is to a [ilmath]\sigma[/ilmath]-algebra

## Definition

Given a set [ilmath]X[/ilmath] a [ilmath]\sigma[/ilmath]-algebra on [ilmath]X[/ilmath] is a family of subsets of [ilmath]X[/ilmath], [ilmath]\mathcal{A} [/ilmath][Note 1], such that:

• [ilmath]\forall A\in\mathcal{A}[A^C\in\mathcal{A}][/ilmath] - Stable under complements
• [ilmath]\forall\{A_n\}_{n=1}^\infty\subseteq\mathcal{A}\left[\bigcup_{n=1}^\infty A_n\in\mathcal{A}\right][/ilmath] - Stable under countable union

### Note on Alternative Definitions

Many books have slightly different definitions of a [ilmath]\sigma[/ilmath]-algebra, the definition above is actually equivalent to the longer definitions one might see around.
The two properties above give rise to all the others.

Usually on this project differing definitions are listed, but due to the trivial proofs that they share properties, this was omitted.

To show this yourself, do the following:

1. Show a [ilmath]\sigma[/ilmath]-algebra is closed under set-subtraction, [ilmath]\forall A,B\in\mathcal{A}[A-B\in\mathcal{A}][/ilmath]
2. Use this to show [ilmath]\emptyset\in\mathcal{A} [/ilmath], notice [ilmath]\emptyset=A-A\in\mathcal{A}[/ilmath] for any [ilmath]A\in\mathcal{A} [/ilmath]
3. [ilmath]X\in\mathcal{A} [/ilmath] as [ilmath]\emptyset^C\in\mathcal{A} [/ilmath]

This resolves most ambiguities.

## Immediate consequences

Among other things immediately we see that:

As [ilmath]A-B=(A^c\cup B)^c[/ilmath] and a [ilmath]\sigma[/ilmath]-algebra is closed under complements and unions, this shows it is closed under set subtraction too

One might argue that we only know [ilmath]\mathcal{A} [/ilmath] is closed under countable union, not finite (2) union. so we cannot know [ilmath]A^C\cup B\in\mathcal{A} [/ilmath], thus this proof is invalid. It is not invalid.

The person who mentioned this argued that we require set-subtraction in order to show [ilmath]\emptyset\in\mathcal{A} [/ilmath] - which we do not yet know, so we cannot construct the family:

• [ilmath]\{A_n\}_{n=1}^\infty[/ilmath] with [ilmath]A_1=A^C[/ilmath], [ilmath]A_2=B[/ilmath], [ilmath]A_i=\emptyset[/ilmath] for [ilmath]i> 2[/ilmath]

And they are not wrong, however:

• [ilmath]A^C\cup B=A^C\cup B\cup B\cup B\ldots[/ilmath]

So

• [ilmath]\{A_n\}_{n=1}^\infty[/ilmath] with [ilmath]A_1=A^C[/ilmath] and [ilmath]A_i=B[/ilmath] for [ilmath]i> 1[/ilmath] is something we can construct.

• [ilmath]\mathcal{A} [/ilmath] is [ilmath]\cap[/ilmath]-closed (furthermore, that [ilmath]\mathcal{A} [/ilmath] is in fact [ilmath]\sigma[/ilmath]-[ilmath]\cap[/ilmath]-closed - that is closed under countable intersections)

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See Properties of a class of sets closed under set subtraction

• [ilmath]\emptyset\in\mathcal{A} [/ilmath]

[ilmath]\forall A\in\mathcal{A} [/ilmath] we have [ilmath]A-A\in\mathcal{A} [/ilmath] (by closure under set subtraction), as [ilmath]A-A=\emptyset[/ilmath], [ilmath]\emptyset\in\mathcal{A} [/ilmath]

• [ilmath]X\in\mathcal{A} [/ilmath][Note 2]

As [ilmath]\emptyset\in\mathcal{A} [/ilmath] and it is closed under complement we see that [ilmath]\emptyset^c\in\mathcal{A} [/ilmath] (by closure under complement) and [ilmath]\emptyset^c=X[/ilmath] - the claim follows.

• [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-algebra [ilmath]\implies[/ilmath] [ilmath]\mathcal{A} [/ilmath] is a [ilmath]\sigma[/ilmath]-ring

To prove this we must check:
1. [ilmath]\mathcal{A} [/ilmath] is closed under countable union
• True by definition of [ilmath]\sigma[/ilmath]-algebra
2. [ilmath]\mathcal{A} [/ilmath] is closed under set subtraction
• We've already shown this, so this is true too.
This completes the proof.