Serlo: EN: Generated sigma-algebras

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In this article we learn what the $\sigma$ algebra generated by a set system is. We prove some important properties and get to know the Borel $\sigma$-algebra.

Motivation

Let $𝒞$ be a set system over a basic set $\mathrm{\Omega }$ and $\mu :𝒞\to [0,\mathrm{\infty }]$ a function on sets. Our goal is to find out how and under what conditions $\mu$ can be continued to a measure on a reasonable $\sigma$-algebra $𝒜$.

A continuation must be defined at least on the domain of definition of the function to be continued. Therefore, the set system $𝒞$ must be contained in $𝒜$.

One possibility would be to choose by default the power set $𝒫(\mathrm{\Omega })$ as domain of definition of the continuation (i.e., the largest possible domain): It is a $\sigma$ algebra and contains $𝒞$. But this is not always a sensible choice:

• The power set is in general too ambitious a target for a continuation: the volume problem shows that with intuitive geometric volumes there can be problems defining them on the whole power set. So the power set may be too large to continue a measure to it.
• The power set may also be unnecessarily large: compared to the set system $𝒞$, $𝒫(\mathrm{\Omega })$ may contain too many sets to which continuation then makes no sense. A simple example for this case is when $\mu$ is a measure and $𝒞$ itself is already a $\sigma$ algebra, but not the power set.

A concrete example for the second point is the following: Mathe für Nicht-Freaks: Vorlage:Beispiel The $\sigma$-algebra $𝒜$ we are looking for should therefore not be larger than necessary. We have already stated above that it should, however, contain at least the set system $𝒞$. So we first consider all super-$\sigma$-algebras of $𝒞$, i.e., all $\sigma$-algebras containing $𝒞$. To find the smallest among these, we proceed as in constructing the (topological) closure of a set: The closure of a set is the smallest closed superset and is defined as a section over all closed supersets. Analogously, we choose the smallest super-$\sigma$-algebra $𝒜$ of $𝒞$ to be the intersection over all these $\sigma$-algebras.

Definition: Generated $\sigma$-algebra

The $\sigma$-algebra, which we defined in the previous section as the intersection over all super-$\sigma$-algebras of $𝒞$, is called generated $\sigma$-algebra:

We still need to verify that the generated $\sigma$-algebra is well-defined, that is, that the definition makes sense. To do this, we need to show:

• The set over which the intersection is formed is not empty. That is, there is at least one $\sigma$-algebra containing $𝒞$.
• $\sigma (𝒞)$ is indeed a $\sigma$-algebra.

The first point is clear since the power set $𝒫(\mathrm{\Omega })$ is a $\sigma$-algebra containing $𝒞$. For the proof of the second point, we have to prove that the intersection of arbitrary many $\sigma$-algebras is always a $\sigma$-algebra again. Then, we have that $\sigma (𝒞)$ as a section over certain $\sigma$-algebras is indeed a $\sigma$-algebra.

We have now shown that $\sigma (𝒞)$ is a $\sigma$-algebra. Intuitively, it should be the smallest $\sigma$-algebra containing the set system $𝒞$. We prove this in the next section "Properties of the $\sigma$-operator".

Properties of the $\sigma$-operator

We establish some useful properties of the $\sigma$-operator: Vorlage:Anker Mathe für Nicht-Freaks: Vorlage:Satz

Examples

In the section "Motivation" we have seen a first example for a generated $\sigma$-algebra: Let $\mathrm{\Omega }=\{1,2,3,4\}$ and $𝒞=\{\{1,2\},\{3,4\}\}.$ Then $\sigma (𝒞)=𝒞\cup \{\mathrm{\Omega }\}\cup \mathrm{\varnothing }\}$ is the $\sigma$-algebra generated by $𝒞$: $𝒞\cup \{\mathrm{\Omega }\}\cup \{\mathrm{\varnothing }\}$ is a $\sigma$-algebra and the smallest one containing $𝒞$. Another example for a finitely generated $\sigma$-algebra is the following:

The $\sigma$-algebra of the one-element subsets of a countable basic set often appears in discrete probability theory as a domain of definition of the distribution of discrete random variables. In this case of a discrete, i.e. countable basic set (such as $\mathrm{\Omega }=\{0,1\},\mathrm{\Omega }=\{1,\dots ,n\}$ or $\mathrm{\Omega }=\mathbb{N}$), the $\sigma$-algebra generated by these elementary events is the power set $𝒫(\mathrm{\Omega })$. So actually, introducing $\sigma$-algebras would not be necessary. However, the situation is different if the basic set is over-countable, like $ℝ$:

Some $\sigma$ algebras are so large that they cannot be written down explicitly, as in the previous examples. They can then only be characterized by the generator. An example for this is the $\sigma$-algebra generated by the intervals over $ℝ$, which is an often-used but very rich example.

Proving that two set systems generate the same $\sigma$-algebra

It is common to want to find out whether two $\sigma$-algebras $𝒜$ and $ℱ$ are equal. For this we would prefer to simply show mutual inclusion directly, i.e. to prove $𝒜\subseteq ℱ$ and $ℱ\subseteq 𝒜$. But if $𝒜,ℱ$ were defined only by generators $\sigma (𝒞),\sigma (ℰ)$, this is not an easy job. We would have to take any set $M\in 𝒜$ in the inclusion proof and show that also $M\in ℱ$ holds. The problem is that in general, $ℱ$-sets look very complicated, so we do not know what such set looks like and what properties it has. We only know that it is contained in every superset-$\sigma$-algebra of $𝒞$. However, we know what the generators look like. So it is way easier to just show that the generators are included in each other. This is what we will do now.

Subset-relations for generators

Thus we have already simplified our problem considerably. We no longer need to show for arbitrary sets $M\in 𝒜$ that $M\in ℱ$ is true (which might be a great mess to do). It suffices to prove the inclusion for sets from the generator $𝒞$ of $𝒜$.

The opposite inclusion can be simplified using the same principle. That is, instead of showing for any $M\in ℱ$ that $M\in 𝒜$ holds (again, a great mess), we take a generator $ℰ$ of $ℱ$ and show for all $M\in ℰ$ that $M\in 𝒜$ is satisfied.

Proving that a set is contained in a $\sigma$-algebra

We now know that it suffices to show only for the sets from the generator that they lie in the respective other $\sigma$-algebra. But how can we prove in general for a set $M$ that it lies in a certain $\sigma$-algebra $𝒜=\sigma (ℰ)$?

We know that $𝒜$ is closed under the operations complement and countable union (and hence also under taking differences and countable cuts). Therefore every set generated by these operations from sets of the generator $ℰ$ is again in $𝒜$. Thus, to prove that a set $M$ is in $𝒜$, it suffices to take some sets from the generator $ℰ$ and write it as an outcome of some set operations between those sets.

Since $\sigma$-algebras can be very large, however, there is no general method to find such a representation of $M$ over the sets from the generator.

Example: The $\sigma$-algebra generated by intervals

We will now demonstrate this principle with an example. Mathe für Nicht-Freaks: Vorlage:Satz

Generators of the Borel $\sigma$-algebra Vorlage:Anker

We now apply the principle of the last section to a very important example, namely the so-called Borel $\sigma$-algebra.

The Borel $\sigma$-algebra is one of the most important $\sigma$-algebras in mathematics. It plays the role of the "smallest and simplest $\sigma$-algebra, where stuff makes sense". We will encounter it later in the construction of the Lebesgue measure, again.

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