# Why are sigma fields significant in probability theory?

1. May 30, 2011

### rukawakaede

As the title.

Why are sigma fields important in probability?

The only one reason I can think of is that sigma fields are used as domain, e.g. borel fields uses sigma fields instead of power set. However, are there any other significances of sigma fields in probability theory?

Last edited: May 30, 2011
2. May 30, 2011

### Lajka

Well, the probability, as any other measure, is defined over sigma-fields primarily because of the existence of sets which are unmeasurable. We need to exclude those somehow, so this is the best way to do it.
As far as I know, all those unmeasurable sets are pretty pathological in their nature, and any subset that you can think of is most likely measurable. So, from a practical POV, this is just a formality which needed to be done for consistency's sake.

I'm sure someone will correct me if I'm wrong.

3. May 30, 2011

### mathman

Probabilities can be added as long as the number of terms is countable. Sigma fields insure that if you have a countable number of events the union is also an event, so the calculation of its probability is meaningful.

4. May 30, 2011

### disregardthat

The proof of their existence actually provides the mathematical means of calculating the measure (which is necessary for a measure to be of any use) of any sigma-measurable set. In the lebesgue-measure, we are sure that by approximating with for example unions of disjoint n-dimensional boxes (such as [a_1,b_1)x[a_2,b_2) x ... x [a_n,b_n)) that converge (as sets) to our measurable set, the corresponding approximation of (easily calculated) measures will also converge. E.g: we know that closed sets (in the standard topology of R^n) are lebesgue-measurable, so we can calculate their measure by making covers of disjoint unions of such boxes that converges as sets to the closed set.

For general sigma-algebras with corresponding measures, we can approximate by disjoint unions of some generating semi-ring. We could use the boxes in the lebesgue-measure case since the set of boxes is a generating semi-ring for the set of lebesgue-measurable sets.

If we do not know that the set is contained in our sigma-algebra (that is, being measurable), we can't use this method.

In fact, sigma-algebras have no use in consistency. Not using them, and still applying a measure to some set would be meaningless as we would have no method of calculating it. Applying a measure does mean that we have such a method (and generating semi-rings provide in many cases a constructive method!). We could try by e.g. approximating with some arbitrary method-such as by disjoint union of boxes-but we don't know that this will converge uniquely. And this is exactly what the existence+uniqueness of measures proves for sigma-algebras.

Similarly for probability theory; P(A) does not makes sense unless A is contained in a set where the measure P can be meaningfully applied.

Last edited: May 30, 2011
5. May 30, 2011

### bpet

Further to the above comments, the use of sigma fields simply puts probability theory on a common foundation with measure theory, i.e. general integration.

It is possible to define versions of probability theory based on finite additivity rather than sigma-additivity, with various implications, for example densities of sets on the natural numbers such as $d(A)=\lim_{n\to\infty}\frac{1}{n}\#(A\cap\{1,...,n\})$ are finitely additive but not sigma additive because all singletons have zero density.

Last edited: May 30, 2011
6. Jun 2, 2011

### disregardthat

This is not a measure on P(N) though. An example of a set for which d(A) is ill-defined is the set A of integers starting with the digit 1 in base 10.