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Σ-Algebras#

In measure theory, we want to be able to "measure" aspects of the subsets of some set \(X\). In general, not all subsets have all aspects which we might want to measure, just like we cannot really measure the humidity of a sound. It simply does not make sense to speak of how humid a sound wave is. Similarly, some subsets might be measurable in certain aspects but not in others. Therefore, we need to know which things can be measured in the way we want, i.e. we need a notion of measurable sets in \(X\).

Definition: Σ-Algebra

Let \(X\) be a set.

A σ-algebra over \(X\) is a collection \(\Sigma\) of subsets of \(X\) with the following properties:

  • The empty set \(\varnothing\) and \(X\) itself are in \(\Sigma\).
  • If a subset \(S \subset X\) is in \(\Sigma\), then so is its complement \(X \setminus S\).
  • If \(\mathcal{S}\) is a countable subcollection of \(\Sigma\), then its union \(\bigcup \mathcal{S}\) is also in \(\Sigma\).

The elements of \(\Sigma\) are called measurable sets.

A σ-algebra is just a selection of subsets with the aspects that we want to measure. The intuition behind the requirements of the definition is the following:

  • Intuitively, \(\varnothing\) should be in \(\Sigma\) because measuring any aspect of "nothing" should simply result in zero. Otherwise, "nothing" would have "something" and so it would not really be "nothing". Furthermore, we want the "whole thing" \(X\) itself to be measurable and so \(X\) should be in \(\Sigma\).
  • If we can measure the whole thing \(X\) and we can measure a part \(S\), then we should intuitively be able to measure the remaining part \(X \setminus S\) as well.
  • The last requirement guarantees that if something is made of components which have the desired aspects, then it should itself have these aspects.

Theorem: Countable Intersections in Σ-Algebras

Let \(X\) be a set and let \(\Sigma\) be a σ-algebra on \(X\).

If \(\mathcal{S}\) is a countable subcollection of \(\Sigma\), then its intersection \(\bigcap \mathcal{S}\) is also in \(\Sigma\).

Proof

TODO