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$\sigma$algebra
Introduction
When defining a measure for a set $E$ we usually cannot hope to make every subset of $E$ measurable. Instead we must usually restrict our attention to a specific collection of subsets of $E$, requiring that this collection be closed under operations that we would expect to preserve measurability. A $\sigma$algebra is such a collection.
Definition
Given a set $E$, a $\sigma$algebra in $E$ is a collection $\mathcal{F}$ of subsets of $E$ such that:

$\varnothing\in\mathcal{F}$.

Any union of countably many elements of $\mathcal{F}$ is an element of $\mathcal{F}$.

The complement of any element of $\mathcal{F}$ in $E$ is an element of $\mathcal{F}$.
Notes
It follows from the definition that any $\sigma$algebra $\mathcal{F}$ in $E$ also satisfies the properties:

$E\in\mathcal{F}$.

Any intersection of countably many elements of $\mathcal{F}$ is an element of $\mathcal{F}$.
Note that a $\sigma$algebra is a field of sets that is closed under countable unions and countable intersections (rather than just finite unions and finite intersections).
Given any collection $C$ of subsets of $E$, the $\sigma$algebra $\sigma(C)$ generated by $C$ is defined to be the smallest $\sigma$algebra in $E$ such that $C\subseteq\sigma(C)$. This is welldefined, as the intersection of any nonempty collection of $\sigma$algebras in $E$ is also a $\sigma$algebra in $E$.
Examples
For any set $E$, the power set $\mathcal{P}(E)$ is a $\sigma$algebra in $E$, as is the set $\{\varnothing,E\}$.
A more interesting example is the Borel $\sigma$algebra in $\mathbb{R}$, which is the $\sigma$algebra generated by the open subsets of $\mathbb{R}$, or, equivalently, the $\sigma$algebra generated by the compact subsets of $\mathbb{R}$.
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Comments
question about \sigmaalgebra
In relation to Djao's entry \sigmaalgebra, \emptyset \in \mathcal{B}(E) appears like a premise. As far as I know, \emptyset \in A \forall A. So, why this one is not redundant?
BTW, in my browser IE6 the entry title looks like [red cross]$\sigma$algebra, missing the greek letter sigma.
Re: question about \sigmaalgebra
You seem to be confusing membership with inclusion (being a subset of). The empty set is a subset of every set, but it is not, for example, a member of itself.
Re: question about \sigmaalgebra
Hi ratboy,
Yes! Now it is clearer for me why E \in \mathcal{B(E)}.
Thank you!
Pedro