Generated sigma-algebras

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:

Example (Reasonable extension of $𝒞$)

Let $\mathrm{\Omega }=\{1,2,3,4\}$ and $𝒞=\{\{1,2\},\{3,4\}\}$. Let further $\mu$ be a function defined on the set of sets $𝒞$ with $\mu (\{1,2\})=1=\mu (\{3,4\})$. The set system

$𝒜:=𝒞\cup \{\mathrm{\Omega }\}\cup \{\mathrm{\varnothing }\}$

is a $\sigma$-algebra containing $𝒞$. But of course the power set $𝒫(\mathrm{\Omega })$ is also such a $\sigma$-algebra. Intuitively, however, $𝒫(\mathrm{\Omega })$ makes little sense as a domain of definition of a continuation $\nu$ of $\mu$. This is because the power set also contains the one-element subsets of $\mathrm{\Omega }$. However, $\mu$ does not provide any information about these at all: we could arbitrarily choose the value for $\nu (\{1\})$ from $[0,1]$. A larger value is not possible because of monotonicity, since $\nu (\{1\})\le \mu (\{1,2\})=1$ must hold. Then, because of additivity, $\nu (\{2\})=1-\nu (\{1\}).$

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:

Definition (Generated $\sigma$-algebra)

Let $\mathrm{\Omega }$ be a set and $𝒞\subseteq 𝒫(\mathrm{\Omega })$ be a set system. The $\sigma$-algebra

$\sigma (𝒞):=\bigcap \{ℱ\subseteq 𝒫(\mathrm{\Omega }):ℱ\text{ is a }\sigma \text{-algebra, }𝒞\subseteq ℱ\}$

is called the $\sigma$ algebra generated by $𝒞$. The "operator of generation $\sigma$" defined by it is called the $\sigma$ operator. The set system $𝒞$ is called generator of $\sigma (𝒞)$.

Hint

$\bigcap \{ℱ\subseteq 𝒫(\mathrm{\Omega }):ℱ\text{ is a }\sigma \text{-algebra, }𝒞\subseteq ℱ\}$

is another notation for the intersection $\bigcap _{ℱ\in ℳ}ℱ$, where $ℳ=\{ℱ\subseteq 𝒫(\mathrm{\Omega }):ℱ\text{ is a }\sigma \text{-algebra, }𝒞\subseteq ℱ\}$.

Hint

Although there is no $\mathrm{\Omega }$ in $\sigma (𝒞)$, the $\sigma$ algebra $\sigma (𝒞)$ generated of a set system $𝒞$ depends of course on the underlying basic set $\mathrm{\Omega }$. Let, for instance $𝒞=\{\mathrm{\varnothing }\}$. Then $\sigma (𝒞)=\{\mathrm{\varnothing },\mathrm{\Omega }\}$ is the $\sigma$ algebra generated by $𝒞$ over $\mathrm{\Omega }$. For another basic set $\mathrm{\Omega }$ this is a different set system. Often, the $\mathrm{\Omega }$ is clear from the context and is therefore omitted in the notation of the $\sigma$ operator.

Hint

One can also define other kinds of generated set systems according to the same principle. For example, one can define the ring or $\sigma$-ring generated by a set system $𝒞$.

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.

Theorem (The intersection of $\sigma$-algebras is again a $\sigma$-algebra.)

Let $ℳ$ be a non-empty set of $\sigma$-algebras over $\mathrm{\Omega }$. That is, every element in $ℳ$ is a $\sigma$-algebra. Then $𝒜:=\bigcap _{ℱ\in ℳ}ℱ$ is a $\sigma$-algebra.

Proof (The intersection of $\sigma$-algebras is again a $\sigma$-algebra.)

We need to prove that $𝒜$ satisfies the three properties of a $\sigma$-algebra:

1. $\mathrm{\Omega }\in 𝒜$
2. $A\in 𝒜⟹A^{\complement }\in 𝒜$
3. $A_1,A_2,\dots \in 𝒜⟹\bigcup _{n\in \mathbb{N}}{A}_n\in 𝒜$

The basic set $\mathrm{\Omega }$ is in $𝒜$: Each element of $ℳ$ is a $\sigma$-algebra over $\mathrm{\Omega }$ and thus contains the basic set. Thus $\mathrm{\Omega }$ is also contained in the section over all these elements, i.e. in $𝒜$.

Complement stability: Let $A\in 𝒜$ be arbitrary. By definition of $𝒜$, $A$ lies in the intersection of all $\sigma$ algebras from $ℳ.$ We conclude $A\in ℱ$ for all $ℱ\in ℳ$. Since every $ℱ\in ℳ$ is a $\sigma$-algebra, the complement $A^{\complement }$ also lies in $ℱ$ for all $ℱ\in ℳ$. Thus $A^{\complement }$ is also in the section over all these $\sigma$-algebras, that is, in $𝒜.$

Completeness under countable unions: Let $A_1,A_2,\dots \in 𝒜$. By definition of $𝒜$ these sets lie in the intersection of all $\sigma$-algebras from $ℳ$, so we have that $A_1,A_2,\dots \in ℱ$ for all $ℱ\in ℳ.$ Since every $ℱ\in ℳ$ is a $\sigma$-algebra and hence complete under formation of countable unions, every $ℱ$ from $ℳ$ also contains the union $\bigcup _{n\in \mathbb{N}}{A}_n.$ Thus this union also lies in the section over all these $\sigma$-algebras from $ℳ$, i.e. in $𝒜$.

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:

Theorem

Let $𝒞,ℰ\subseteq 𝒫(\mathrm{\Omega })$ be a set system. The $\sigma$-operator now satisfies the following properties:

1. Extensivity: $𝒞\subseteq \sigma (𝒞)$
2. Minimality:$\sigma (𝒞)$ is the smallest $\sigma$-algebra containing $𝒞$. If $𝒞$ is a $\sigma$-algebra, then $\sigma (𝒞)=𝒞$.
3. Idempotency: $\sigma (\sigma (𝒞))=\sigma (𝒞)$
4. Monotonicity: $𝒞\subseteq ℰ⟹\sigma (𝒞)\subseteq \sigma (ℰ)$

Proof

1. Extensivity: By definition, $𝒞$ is subset of every $\sigma$-algebra over which we take the intersection in the definition of the $\sigma$-operator. That is, for any $A\in 𝒞$, $A$ is element every $\sigma$-algebra over which we intersect. Then $A$ is also element of the intersection of all these $\sigma$-algebras, which is exactly $\sigma (𝒞)$. Since this is true $A\in 𝒞$, we have $𝒞\subseteq \sigma (𝒞)$.
2. Minimality: Let $𝒢$ be a $\sigma$-algebra with $𝒞\subseteq 𝒢$. Since $𝒢$ is one of the sets over which we intersect in the definition of $\sigma (𝒞)$, we have that $\sigma (𝒞)=\bigcap \{ℱ\subseteq 𝒫(\mathrm{\Omega }):ℱ\text{ is a }\sigma \text{-algebra, }𝒞\subseteq ℱ\}\subseteq 𝒢$. If $𝒞$ is a $\sigma$-algebra we may readily conclude $\sigma (𝒞)\subseteq 𝒞$. From extensivity we obtain the other inclusion and therefore we have $𝒞=\sigma (𝒞)$.
3. Idempotency: The idempotency follows directly from the minimality. We have that $\sigma (𝒞)$ is always a $\sigma$-algebra, and therefore we have $\sigma (𝒞))=\sigma (𝒞)$.
4. Monotonicity: Let $𝒞\subseteq ℰ$. Then, we have $𝒞\subseteq ℰ\subseteq \sigma (ℰ)$ due to extensivity. Since $\sigma (ℰ)$ is a $\sigma$-algebra, it follows from minimality that $\sigma (𝒞)\subseteq \sigma (ℰ)$ holds.

Hint

The properties 1., 3. and 4. (extensivity, idempotency and monotonicity) make the $\sigma$-operator an enveloping operator (it determines the envelope of a set, like wrapping a gift), just as the closure "$\overline{}$" of sets turning "$A$" into "$\overline{A}$".

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:

Example

If one wants to describe the probability of the occurrence of events when rolling a dice by using a measure, the domain of definition is the $\sigma$-algebra, which contains all elementary events. These are all one-element subsets $\{x\}$ of the basic set $\mathrm{\Omega }=\{1,\dots ,6\}.$ The $\sigma$-algebra generated by the set $𝒞=\{\{1\},\dots ,\{6\}\}$ generated is the power set $𝒫(\mathrm{\Omega }).$

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 $ℝ$:

Theorem ($\sigma$-algebra over $ℝ$ generated by point sets)

Let $\mathrm{\Omega }=ℝ$ be the basic set. The $\sigma$-algebra generated from the set of one-element subsets $ℰ=\{\{x\}\mid x\in ℝ\}$ is $𝒜=\{A\subseteq ℝ\mid A\text{ or }{A}^{\complement }\text{ countable}\}.$

Proof ($\sigma$-algebra over $ℝ$ generated by point sets)

We perform the proof in two steps. First, we show that $𝒜$ is a $\sigma$ algebra containing $ℰ$, i.e., $ℰ\subseteq 𝒜$ holds. Next we show $𝒜\subseteq \sigma (ℰ)$. Then we conclude $𝒜=\sigma (ℰ)$.

Proof step: $𝒜$ is a $\sigma$-algebra containing $ℰ$

The elements from $ℰ$ (which are subsets of the basic set) contain only one-element each. Thus, they are countable. It follows directly that every element from $ℰ$ is also contained in $𝒜$, so $ℰ\subseteq 𝒜$. We now show that $𝒜$ is a $\sigma$-algebra. To do this, we check the three criteria:

$\mathrm{\Omega }\in 𝒜$ is of course satisfied, since ${\mathrm{\Omega }}^{\complement }=\mathrm{\varnothing }$ is countable.

If $A\in 𝒜$, then $A$ is countable or $A^{\complement }$ is countable. In case 1, $(A^{\complement }{)}^{\complement }=A$ is countable, so it is contained in $𝒜$. In case 2, $A$ has a countable complement, so $A=(A^{\complement }{)}^{\complement }$ is contained in $𝒜$.

Let now $\bigcup _{n\in \mathbb{N}}{A}_n$ a union of sets from $𝒜$. Then we distinguish two cases. In case 1, for at least one $k\in \mathbb{N}$ the set $A_k$ has countable complement. But then ${\left(\bigcup _{n\in \mathbb{N}}{A}_n\right)}^{\complement }\subseteq A_k^{\complement }$ as a subset of a countable set is also countable and hence contained in $𝒜$. In case 2 for all $n\in \mathbb{N}$ the set $A_n$ is countable. Then, of course, their union $\bigcup _{n\in \mathbb{N}}{A}_n$ is countable and hence contained in $𝒜$.

Thus $𝒜$ is really $\sigma$-algebra and it contains $ℰ$.

Proof step: $𝒜\subseteq \sigma (ℰ)$

Let $A\in 𝒜$ be arbitrary. Then we distinguish two cases. In case 1, $A$ is countable. Consider $A=\bigcup _{n\in \mathbb{N}}\{x_n\}$ as a countable union of sets from $ℰ$. Then $A$ is in particular also a countable union of sets in $\sigma (ℰ)$ and because of the union stability of $\sigma$-algebras with respect to countable unions, it follows that $A\in \sigma (ℰ)$. In case 2 $A^{\complement }$ is countable, so according to case 1 it is contained in $\sigma (ℰ)$. From the complement stability of $\sigma (ℰ)$ it now follows that also $A^{\complement }\in \sigma (ℰ)$ is true.

We have that also $ℰ\subseteq 𝒜\subseteq \sigma (ℰ)$. Following the monotonicity of the $\sigma$-operator, we have that $\sigma (ℰ)\subseteq \sigma (𝒜)\subseteq \sigma (ℰ))$. Since $𝒜$ and $\sigma (ℰ)$ are already $\sigma$-algebras, it follows from the minimality of the $\sigma$ operator that

$\sigma (ℰ)\subseteq 𝒜\subseteq \sigma (ℰ)$ holds true, i.e. $𝒜=\sigma (ℰ)$.

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.

Example ($\sigma$-algebra generated by intervals or cuboids)

The geometric length is the function over $ℝ$ which assigns to all intervals $(a,b)$, respectively, $(a,b]$, $[a,b]$, $[a,b)$ their length $b-a$. We do not yet know whether this function can be continued to a measure on a $\sigma$-algebra. But a reasonable domain of definition of such a continuation would then be the $\sigma$-algebra generated of all such intervals, i.e. $\sigma (𝒞)\subseteq 𝒫(\mathrm{\Omega })$ with $𝒞=\{I\subseteq ℝ\mid I\text{ interval}\}$.

More generally, one can consider the geometric volume that assigns to all axis-parallel cuboids in ${ℝ}^n$ their volume, i.e., the product of the side lengths. A cuboid is a product $Q=\prod _{k=1}^n{I}_k$ of intervals $I_k\subseteq ℝ$ (open, half-open or closed). Again, we do not yet know whether this set function can be continued to a measure. But a reasonable domain of definition for a continuation would then be the $\sigma$-algebra $\sigma (𝒞)$, generated by the set system of cuboids $𝒞=\{Q\subseteq {ℝ}^n\mid Q\text{ axis-parallel cuboid}\}$.

To-Do:

Link to the article where a continuation from one to the other function above is defined.

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

Theorem

Let $𝒜,ℱ$ be $\sigma$-algebras and let $𝒞$ be a generator of $𝒜$ (that is a set $𝒞$ with $\sigma (𝒞)=𝒜$). Now if the producer $𝒞$ of $𝒜$ is a subset of $ℱ$, then also our $\sigma$-algebra $𝒜$ is already a subset of the $\sigma$-algebra $ℱ$. That is $𝒞\subseteq ℱ⟹𝒜\subseteq ℱ$.

Proof

First we see that from the minimality of the $\sigma$-operator, we get $\sigma (ℱ)=ℱ$. Now we use the monotonicity of the $\sigma$-operator: $𝒜=\sigma (𝒞)\subseteq \sigma (ℱ)=ℱ.$

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.

Theorem

Consider the set system $𝒞=\{[a,b)|a,b\in ℝ\}$, $𝒟=\{(a,b)|a,b\in ℝ\}$, $ℰ=\{[a,b]|a,b\in ℝ\}$. Then, we have $\sigma (𝒞)=\sigma (𝒟)=\sigma (ℰ)$. Now, the set system $ℱ=\{I\subseteq ℝ|I\text{ interval }\}$ generates the same $\sigma$-algebra.

Proof

We show $\sigma (𝒞)\subseteq \sigma (ℰ)\subseteq \sigma (𝒟)\subseteq \sigma (𝒞)$, since then the claim of the theorem follows.

Proof step: $\sigma (𝒞)\subseteq \sigma (ℰ)$

It is enough according to the previous theorem to show that $𝒞\subseteq \sigma (𝒟)$ holds. Let also $[a,b)\in 𝒞$. Then $[a,b]$ and also $[b,b]=\{b\}\in \sigma (ℰ)$. Because of the diference stability of $\sigma (ℰ)$ then also $[a,b)=[a,b]\setminus \{b\}\in \sigma (ℰ)$. Since $[a,b)\in 𝒞$ was arbitrary, it follows that $𝒞\subseteq \sigma (ℰ)$, and from this follows the claim of this proof step.

Proof step: $\sigma (ℰ)\subseteq \sigma (𝒟)$

We show again $ℰ\subseteq \sigma (𝒟)$. Let for this $[a,b]\in ℰ$. The sets $(-\mathrm{\infty },a)=\bigcup _{n\in \mathbb{N},n\ge 1}(a-n,a)$ and $(b,\mathrm{\infty })=\bigcup _{n\in \mathbb{N},n\ge 1}(b,b+n)$ are also in $\sigma (𝒟)$ as countable unions of sets from $\sigma (𝒟)$. The union $(-\mathrm{\infty },a)\cup (b,\mathrm{\infty })$ is contained (again because of union stability) in $\sigma (𝒟)$, and with the complement stability of $\sigma (𝒟)$ then follows $[a,b]\in 𝒟$. Since $[a,b]\in 𝒟$ was arbitrary, it follows that $ℰ\subseteq 𝒟$.

Proof step: $\sigma (𝒟)\subseteq \sigma (𝒞)$

As in the other two proof steps, we again show $𝒟\subseteq \sigma (𝒞)$. Let for this $(a,b)\in 𝒟$ be arbitrarily. We have that then for all $n\in \mathbb{N}$, $n\ge 1$ the set $[a+1/n,b)\in 𝒞$. Then, because of the union stability with respect to countable unions, $(a,b)=\bigcup _{n\in \mathbb{N},n\ge 1}[a+1/n,b)\in \sigma (𝒞)$. Since $(a,b)\in 𝒟$ was arbitrary, it follows $𝒟\subseteq \sigma (𝒞)$, and hence also $\sigma (𝒟)\subseteq \sigma (𝒞)$.

Thus $\sigma (𝒞)=\sigma (𝒟)=\sigma (ℰ)$. It makes sense in the following to define $ℬ:=\sigma (𝒞)=\sigma (𝒟)=\sigma (ℰ)$.

Proof step: $\sigma (ℱ)=ℬ$

We now show that $ℱ=\{I\subseteq ℝ|I\text{ interval }\}$ also generates this $\sigma$ algebra.

Because of the monotonicity, from $𝒞\subseteq ℱ$ directly follows $ℬ=\sigma (𝒞)\subseteq \sigma (ℱ)$. For the other set inclusion we again show, according to our principle, $ℱ\subseteq ℬ$. Let for this $I\in ℱ$ be arbitrarily. We can assume that $I$ is bounded, because if it was not, we could write $I$ as a countable union of bounded intervals and thus reduce the statement to the bounded case. That means there are $a,b\in ℝ$, so that, one of the $4$ following cases occurs

$I=[a,b)$,

$I=(a,b)$,

$I=[a,b]$,

or $I=(a,b]$.

In the first three cases $I$ is contained in a known generator from $ℬ$, and hence also in $ℬ$. In the case $I=(a,b]$ $I=\bigcup _{n\in \mathbb{N},n\ge 1}[a+1/n,b]$ as countable unions of sets in $ℬ$ lies again in $ℬ$. Since $I$ was arbitrarily chosen from $ℱ$, it follows that $ℱ\subseteq ℬ$. Thus finally obtain $\sigma (ℱ)\subseteq ℬ$.

Both inclusions are shown and we have that $\sigma (ℱ)=ℬ$.

Generators of the Borel $\sigma$-algebra

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

Theorem (Different generators of the Borel $\sigma$-algebra on real numbers)

Let $𝒞=\{\prod _{i=1}^n[a_i,b_i)\subseteq {ℝ}^n\mid a_i,b_i\in ℝ\text{ and }1\le i\le n\}\subseteq 𝒫({ℝ}^n)$. Then, we call $ℬ=\sigma (𝒞)$ the Borel $\sigma$-algebra over ${ℝ}^n$. We show that $ℬ$ is equivalently generated by the following set systems: $𝒟=\{U\subseteq {ℝ}^n|U\text{ open}\}$, $ℰ=\{A\subseteq {ℝ}^n\mid A\text{ closed}\}$. That means, $ℬ=\sigma (𝒞)=\sigma (𝒟)=\sigma (ℰ)$.

Proof (Different generators of the Borel $\sigma$-algebra on real numbers)

We prove that $\sigma (𝒞)\subseteq \sigma (ℰ)\subseteq \sigma (𝒟)\subseteq \sigma (𝒞)$. Then all these $\sigma$-algebras must be equal.

Proof step: $\sigma (𝒟)\subseteq \sigma (𝒞)$

As proved in the previous theorem, it suffices to show that $𝒟\subseteq \sigma (𝒞)$ holds. Let also $U\in 𝒟$ be chosen arbitrarily. Our idea is to represent $U$ as a countable union of sets from $\sigma (𝒞)$.

Let for $x^{\prime }=({x_1}^{\prime },{x_2}^{\prime },\dots ,{x_n}^{\prime })\in {ℝ}^n$ and $q\in \mathbb{Q}$ the set $A_{(x^{\prime },q)}:=[{x_1}^{\prime }-q,{x_1}^{\prime }+q)\times [{x_2}^{\prime }-q,{x_2}^{\prime }+q)\times \dots \times [{x_n}^{\prime }-q,{x_n}^{\prime }+q)$. Then $M=\bigcup _{q\in \mathbb{Q},x^{\prime }\in {\mathbb{Q}}^n,A_{(x^{\prime },q)}\subseteq U}{A}_{x^{\prime },q}$ is a countable union of elements of $\sigma (𝒞)$, and thus because of the stability of union with respect to countable sets of $\sigma$-algebras, also an element of $\sigma (𝒞)$.

We now show that $M=U$.

$M$, as it is as union of subsets of $U$, is of course also a subset of $U$, i.e. $M\subseteq U$.

For the opposite inclusion let $x\in U$ be arbitrary. We will now cleverly construct a half-open cube $A_{x^{\prime },q}$ with rational side length and rational center such that $x\in A_{x^{\prime },q}\subseteq U$ is fulfilled.

Since $U$ is open, $U$ is also open with respect to the maximum norm. In the following, let $U_c(x)$ always be the $c$-environment of $x$ with respect to the maximum norm. There exists then an $\epsilon >0$ with $U_{\epsilon }(x)\subseteq U$ because $U$ is open.

Let $p\in \mathbb{Q}$, $0<p<\epsilon$. Then, we have $U_p(x)\subseteq U_{\epsilon }(x)\subseteq U$. Let $q=\frac{p}{4}.$ Since ${\mathbb{Q}}^n$ is dense in ${ℝ}^n$, there is now $x^{\prime }\in U_q(x)\cap {\mathbb{Q}}^n$. It follows conversely that $x\in U_q(x^{\prime })\subseteq A_{x^{\prime },q}$. Moreover, $A_{x^{\prime },q}\subseteq U_{2q}(x^{\prime })\subseteq U_{4q}(x)=U_p(x)\subseteq U$.

Thus $A_{x^{\prime },q}$ is one of the sets over which we take the union in the definition of $M$. So $x\in A_{x^{\prime },q}\subseteq M$. Since $x$ was arbitrarily chosen from $U$, we have $U\subseteq M$ and consequently $U=M\in \sigma (𝒞)$.

Since $U\in 𝒟$ was arbitrary, $𝒟\subseteq \sigma (𝒞)$, from which $\sigma (𝒟)\subseteq \sigma (𝒞)$ follows.

Proof step: $\sigma (ℰ)\subseteq \sigma (𝒟)$

We show again $ℰ\subseteq \sigma (𝒟)$. To do this, let $A\in ℰ$ be arbitrary, i.e., $A$ is closed. We then have that by definition $A^{\complement }$ is open, so $A^{\complement }\in 𝒟\subseteq \sigma (𝒟)$. Then, because of the complement stability of $\sigma$-algebras, $A=(A^{\complement }{)}^{\complement }\in \sigma (𝒟)$.

Since $A\in ℰ$ was arbitrary, it also follows that $ℰ\subseteq \sigma (𝒟)$ and hence $\sigma (ℰ)\subseteq \sigma (𝒟)$.

Proof step: $\sigma (𝒞)\subseteq \sigma (ℰ)$

We proceed as in Step 1 and 2 and show $𝒞\subseteq \sigma (ℰ)$.

Let $A=[a_1,b_1)\times \dots \times [a_n,b_n)\in 𝒞$ be arbitrarily. Let $A^{\prime }=[a_1,b_1]\times \dots \times [a_n,b_n]$. Then $A^{\prime }$ is closed, so $A^{\prime }\in \sigma (ℰ)$. We now define $n$ sets as follows: for $i=1,\dots ,n$ let $F_i=[a_1,b_1]\times \dots \times [a_{i-1},b_{i-1}]\times \{b_i\}\times [a_{i+1},b_{i+1}]\times \dots \times [a_n,b_n]$. Then these $F_i$ are closed sets, so we have that also $F_i\in \sigma (ℰ)$. The $F_i$ are the "missing" $(n-1)$-dimensional side faces of the $n$-dimensional half-open cuboid $A$.

Further we have that $A=A^{\prime }\setminus \left(\bigcup _{n=1,\dots ,n}{F}_i\right).$

Since $\sigma (ℰ)$ is difference stable and union stable with respect to countable unions (it is a $\sigma$-algebra), it follows that $A\in \sigma (ℰ)$.

Since this is true for any $A\in 𝒞$ ,we have $𝒞\subseteq \sigma (ℰ)$ and therefore $\sigma (𝒞)\subseteq \sigma (ℰ)$.

Now we have that as previously considered, $\sigma (𝒞)\subseteq \sigma (ℰ)\subseteq \sigma (𝒟)\subseteq \sigma (𝒞)$ and from this follows $ℬ=\sigma (𝒞)=\sigma (𝒟)=\sigma (ℰ)$. That means, the Borel $\sigma$-algebra is generated from the set of half-open cuboids, or equivalently fro the set of closed sets or the set of open sets.

Hint

In the theorem we represented the Borel $\sigma$-algebra as the $\sigma$-algebra generated by the set system $𝒞=\{\prod _{i=1}^n[a_i,b_i)\subseteq {ℝ}^n\mid a_i,b_i\in ℝ\text{ and }1\le i\le n\}$ of the right-open cuboids. One can show that the following systems of cuboids also generate the Borel $\sigma$-algebra:

• the set system of open cuboids $\{\prod _{i=1}^n(a_i,b_i)\subseteq {ℝ}^n\mid a_i,b_i\in ℝ\text{ and }1\le i\le n\}$.
• the set system of closed cuboids $\{\prod _{i=1}^n[a_i,b_i]\subseteq {ℝ}^n\mid a_i,b_i\in ℝ\text{ and }1\le i\le n\}$.
• the set system of left open cuboids $\{\prod _{i=1}^n(a_i,b_i]\subseteq {ℝ}^n\mid a_i,b_i\in ℝ\text{ and }1\le i\le n\}$.
• the set system of all cuboids $\{\prod _{i=1}^n{I}_i\subseteq {ℝ}^n\mid I_i\subseteq ℝ\text{ interval and }1\le i\le n\}$.

In the section "Examples" above, we already encountered the last mentioned set system, as well as the $\sigma$-algebra generated by it.

Hint

We now know that the Borel $\sigma$-algebra on ${ℝ}^n$ is also generated by all open or by all closed subsets of ${ℝ}^n$. One can define more generally the Borel $\sigma$-algebra on a topological space as the $\sigma$-algebra generated by all open sets (note that the "topology" is just this "set of all open sets") . On ${ℝ}^n$, this agrees with our definition.

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.

To-Do:

Link to the article, where the Borel $\sigma$-algebra is treated in detail.