Why are sigma fields significant in probability theory?

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Sigma fields are significant in probability theory because they provide a structured domain for defining probabilities, ensuring that only measurable sets are considered, which excludes pathological unmeasurable sets. They allow for the meaningful calculation of probabilities by ensuring that the union of countably many events remains an event within the sigma field. The existence and uniqueness of measures within sigma-algebras enable consistent methods for approximating measures of sets, particularly through disjoint unions of generating semi-rings. This framework aligns probability theory with measure theory, facilitating general integration. Overall, sigma fields are essential for establishing a rigorous foundation for probability calculations.
rukawakaede
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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?

Thanks for your response.:smile:
 
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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.
 
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.
 
Lajka said:
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.

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.
 
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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.
 
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bpet said:
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.

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.
 
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