# Sigma algebra?

1. Aug 3, 2007

### pivoxa15

What are some examples of sigma algebra operations?

2. Aug 3, 2007

### mathman

A sigma algebra is a collection of sets. The set operations are union and intersection. As you know, a sigma algebra is a collection where all countable unions and intersections of members are also members.

It is not clear to me me in your question what you mean by operation.

3. Aug 17, 2007

### littleHilbert

Why is the collection of finite unions of all sets of the form [a,b],[a,b),(a,b],(a,b) in the interval [0,1] not a sigma-algebra?

Clearly, it's an algebra, since
[0,1] is a finite union of disjoint subintervals,
the complement of a finite union is
$$({{\bigcup}_i}I_i)^c={{\bigcap}_i}(I_i)^c={\bigcap}_i}{{\bigcup}_j}J_j$$ is in the collection, since the complement of an interval is an interval or a union of intervals.
$${{{\bigcup}_i}I_i}\cap{{\bigcup}_j}J_j={\bigcup}_{i,j}(I_i\cap{J_j})$$ is in the collection, since the intersection of intervals is either empty or an interval.
But what's wrong with the countable intersection
$${\bigcap}_1}^{\infty}{\bigcup}_{\mathbb{N}}}I_{\mathbb{N}}$$

Last edited: Aug 17, 2007
4. Aug 17, 2007

### matt grime

It fails to satisfy the definition of a sigma algebra. That is why it is not a sigma algebra - the union of an infinite collection of sets in the 'algbra' is not in the 'algebra'.

5. Aug 17, 2007

### CompuChip

For example, consider the rationals intersected with [0, 1]. They can be written as a countable union
$$\bigcap_{q \in \mathbb{Q} \cap [0,1]} [q, q]$$
and therefore should be in the algebra, which you defined as "finite unions of all sets of the form ..." .

6. Aug 18, 2007

### pivoxa15

On the topic of sigma algebras, which is a family of sets, each being a subset of a universal set S. Can the property of openess and closedness discussed in sigma algebras? I guess it will depend on S? If S is the real numbers then the sigma algebra can.

7. Aug 18, 2007

### matt grime

Open and closed is a property of topological spaces. Topologies are in some sense completely different from sigma algebras. So, no, there is no need for open or closedness to have anything to do with sigma algebras, or anything else like it (D-algebras, etc). Of course, it is unlikely one would have been defined without the other, but that is does not stop there being no technical relation between the two, as opposed to a conceptual one. And in any 'real life' case, you will always be trying do measure theory on a topological space anyway.

Last edited: Aug 18, 2007
8. Aug 18, 2007

### mathman

In a practical sense, there is the concept of Borel field, which is the sigma algebra generated by open sets. Measure theory usually is taught by defining measure on Borel sets.

9. Aug 18, 2007

### pivoxa15

So there is a relation but this relation is not used to do measure theory, although in the definitions they may be related however nothing deeper than that after that?