# Thread: What is an example of a σ - algebra of subsets of R?

1. ## What is an example of a σ - algebra of subsets of R?

Here R represents the real numbers.

Let B represent a σ - algebra (sigma algebra)

Then i know that two examples are:

(i) B = {∅ , R }

(ii) B = P(R) (the set of all subsets of R)

I need another example.

Is it true to say that other examples are:

(iii) B = { all Lebesgue measurable sets in R }

(iv) B = { a collection of subsets of R which are countable}

(v) If {Σλ} is a family of σ-algebras over R indexed by λ then the intersection of all Σλ is a σ-algebra over R.

However i dont know what actual examples of (iii), (iv) and (v) in terms of the set R could be.

Or is there a much more obvious alternative example?

Thanks for any help

*Edit:
Having thought about it a little more, could B = {∅ , Q , R\Q , R } ? where Q is the rational numbers so R\Q is the irrational numbers.

2. Originally Posted by Siknature
Here R represents the real numbers.

Let B represent a σ - algebra (sigma algebra)

Then i know that two examples are:

(i) B = {∅ , R }

(ii) B = P(R) (the set of all subsets of R)

I need another example.

Is it true to say that other examples are:

(iii) B = { all Lebesgue measurable sets in R }
Well, there's always a sigma-algebra related to a measure. So all the measurable sets to a given measure will belong to that sigma-algebra. Hence This is a sigma-algebra by definition

(iv) B = { a collection of subsets of R which are countable}
The axiom of the complement is not satisfied.
But B = { subsets of R which are countable or which complement is countable } is a sigma-algebra (and not only for R, but for any whole set E)

(v) If {Σλ} is a family of σ-algebras over R indexed by λ then the intersection of all Σλ is a σ-algebra over R.
As far as I can remember, yes. But not the union.
Or it's inverted, I can't remember well...

However i dont know what actual examples of (iii), (iv) and (v) in terms of the set R could be.
Thanks for any help

*Edit:
Having thought about it a little more, could B = {∅ , Q , R\Q , R } ? where Q is the rational numbers so R\Q is the irrational numbers.
Yes, any set in the form B = { empty set, A, R\A, R } where A is a subset of R, is a sigma-algebra

3. Well, there's always a sigma-algebra related to a measure. So all the measurable sets to a given measure will belong to that sigma-algebra. Hence This is a sigma-algebra by definition
If I remember correctly, when I studied this topic, the measure of a set (and when it exists) was defined first. Then it was proved that the collection of measurable sets is closed under infinite unions and complements.

4. Originally Posted by emakarov
If I remember correctly, when I studied this topic, the measure of a set (and when it exists) was defined first. Then it was proved that the collection of measurable sets is closed under infinite unions and complements.
when you study the Lebesgue measure, then usually only in the end you show that it is a sigma algebra. but in general all measure space must have the set of all its measurable sets, which is a sigma algebra.

about the intersection of sigma algebras, most (or at least from what I have seen so far) of the algebraic structures are closed under intersection, and for most of them, it is very easy to prove it (for example, sets, groups, rings, vector spaces, algebras and so on)

anyway, another example for a sigma algebra is the set of all countable subsets and all the sets that have countable complement (Moo below me is right, it should be countable and not finite. if it is only finite, then it is an algebra and not a sigma algebra).
also, to generalize your last example, if you have $\displaystyle B_i ;\;1\leq i\leq k$ where $\displaystyle i\neq j \rightarrow B_i \cap B_j =\emptyset$ and $\displaystyle \bigcup _1 ^k B_i = \mathbb{R}$then $\displaystyle \{\bigcup _{j\in J} B_j \mid J \subseteq [k]\}$ is a sigma algebra.

5. Originally Posted by Prometheus
when you study the Lebesgue measure, then usually only in the end you show that it is a sigma algebra. but in general all measure space must have the set of all its measurable sets, which is a sigma algebra.

about the intersection of sigma algebras, most (or at least from what I have seen so far) of the algebraic structures are closed under intersection, and for most of them, it is very easy to prove it (for example, sets, groups, rings, vector spaces, algebras and so on)

anyway, another example for a sigma algebra is the set of all finite subsets and all the sets that have finite complement.
also, to generalize your last example, if you have $\displaystyle B_i ;\;1\leq i\leq k$ where $\displaystyle i\neq j \rightarrow B_i \cap B_j =\emptyset$ and $\displaystyle \bigcup _1 ^k B_i = \mathbb{R}$then $\displaystyle \{\bigcup _{j\in J} B_j \mid J \subseteq [k]\}$ is a sigma algebra.
Some courses may do it in this order (Lebesgue measure, then sigma-algebra), but it's not always this way
Everything is not centered in the Lebesgue measure ^^
Also, you can replace all the finite things into countable (at most countable)

@ emakarov : I checked on google... We first have a measurable space, which is a pair (X,A), where X is a set, and A a sigma-algebra over X. Then a measure space is a measurable space to which we associate a nonnegative measure.
When talking about a measure, I've always learnt that I had to specify the measure (measurable) space. It's not a question of first defining the measure of a set. You have to know where you've taken this set from.