Sigma-algebra

A Sigma-algebra of sets, or $\sigma$-algebra is very similar to a $\sigma$-ring of sets.

Like how ring of sets and algebra of sets differ, the same applies to $\sigma$-ring compared to $\sigma$-algebra

Definition

A non empty class of sets $S$ is a $\sigma$-algebra if

• if $$A\in S$$ then $$A^c\in S$$
• if $$\{A_n\}_{n=1}^\infty\subset S$$ then $$\cup^\infty_{n=1}A_n\in S$$

That is it is closed under complement and countable union.

Immediate consequences

Among other things immediately we see that:

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• $\mathcal{A}$ is closed under set subtraction

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

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• $\emptyset\in\mathcal{A}$

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

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• $X\in\mathcal{A}$

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

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• $\mathcal{A}$ is a $\sigma$-algebra $\implies$ $\mathcal{A}$ is a $\sigma$-ring

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To prove this we must check:

1. $\mathcal{A}$ is closed under countable union
   • True by definition of $\sigma$-algebra
2. $\mathcal{A}$ is closed under set subtraction
   • We've already shown this, so this is true too.

This completes the proof.

Important theorems

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The intersection of $\sigma$-algebras is a $\sigma$-algebra

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TODO: Proof - see PTACC page 5, also in Halmos AND in that other book

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Common $\sigma$-algebras

See also: Index of common $\sigma$-algebras

• $\sigma$-algebra generated by
• Trace $\sigma$-algebra
• Pre-image $\sigma$-algebra

See also

• Types of set algebras
• $\sigma$-algebra generated by
• $\sigma$-ring