Do Sigma-Algebras Need to Include the Empty Set and Underlying Set?

[ghotra]

I had a quick question concerning the definition of a \sigma-algebra \mathcal{F} over a set \Omega. Most sources I've seen (e.g. http://en.wikipedia.org/wiki/Sigma-algebra ) require that \Omega or the empty set be an element of \mathcal{F}.

Is this necessary? I ask because I am looking at "Probability: Theory and Examples" by Durrett, and he does not state that as a requirement. He only requires that an element's complement be in \mathcal{F} and that countable (possibly infinite) unions of elements (in the set) remain in the set. Additionally, he says that \mathcal{F} \neq \emptyset, but this does not necessarily imply that the empty set is in \mathcal{F}.

So, has Durrett just forgotten to include this? Do his later results assume this requirement? Or is it the case this is an unnecessary requirement?

 

[ghotra]

Specifically, he states:

if A_i \in \mathcal{F} is a countable sequence of sets then \cup_i A_i \in \mathcal{F}

I think this is my answer. Let the sequence consist of only the set \mathcal{F}. Then \mathcal{F} (and hence the empty set as well) is in \mathcal{F}.

Correct?

 

[Zone Ranger]

I had a quick question concerning the definition of a \sigma-algebra \mathcal{F} over a set \Omega. Most sources I've seen (e.g. http://en.wikipedia.org/wiki/Sigma-algebra ) require that \Omega or the empty set be an element of \mathcal{F}.

Is this necessary? I ask because I am looking at "Probability: Theory and Examples" by Durrett, and he does not state that as a requirement. He only requires that an element's complement be in \mathcal{F} and that countable (possibly infinite) unions of elements (in the set) remain in the set. Additionally, he says that \mathcal{F} \neq \emptyset, but this does not necessarily imply that the empty set is in \mathcal{F}.

So, has Durrett just forgotten to include this? Do his later results assume this requirement? Or is it the case this is an unnecessary requirement?

let \mathcal{F} be a sigma algebra over a set \Omega
since \mathcal{F} in noempty there exists an A\in\mathcal{F} since \mathcal{F} is a sigma algebra A^c\in\mathcal{F} and A\bigcup A^c=\Omega\in\mathcal{F}

 

[fourier jr]

a sigma algebra R on a set X is a nonempty collection of sets satisfying the following:
i) R closed under complements
ii) R closed under countable unions
& that's all

we can derive the fact that the set X on which the algebra is defined, is in R and also the empty set. the empty set is in every sigma algebra because if E is in R, then E\E (=empty set) is in R since R is closed under complementation. also E union E' = X is also in R. so no, the definition doesn't need to include anything about the empty set or the underlying set X is in the algebra.