[Math] intersection of sigma algebras is a sigma algebra

measure-theory

I am new to $\sigma$-algebras and I am having some conceptual difficulty with the proof regarding the fact that an intersection of $\sigma$-algebras is again a sigma algebra. So formally the question is this

Let $X$ be a set.

let $\{\mathcal{H}_i\}_{i\in I}$ be a family of $\sigma$-algebras on $X$.

Prove that $\mathcal{H}=\bigcap \{\mathcal{H}_{i};i \in I \}$ is again a $\sigma$-algebra.

So the proof is to go through the Axioms 1 by 1 (though my issue with the proof revolves mainly around the first axiom).

From the first Axiom of sigma algebras as $X \in \mathcal{H}_{i}$ for all $i \in I$. Hence We see that $X \in \mathcal{H}$ by definition of the intersection $\Box $.

Here is my issue.

I don't see how we know that $X \in \mathcal{H}$ just by the definition of the intersection, i.e. how do we know that $X \notin \bigcup \{\mathcal{H}_{i};i \in I \}\setminus \bigcap \{\mathcal{H}_{i};i \in I \}$ ?

Best Answer

Recall the definition:

Let $(A_{\lambda})_{\lambda \in \Lambda}$ be an indexed family of sets. We have: $$x \in \bigcap_{\lambda \in \Lambda} A_\lambda \iff x \in A_{\lambda}, \forall \lambda \in \Lambda$$

In your case, $\mathcal{H_i}$ is a $\sigma$-algebra for every $i \in I$ so $X \in \mathcal{H}_i$ by the first axiom.

Therefore $$X \in \bigcap_{i\in I} \mathcal{H_i} = \mathcal{H}$$

Related Solutions

Descriptive Set Theory – Sigma Algebra of Subsets Generated by a Set

Let me make a general comment rather than a specific one, because the construction that you are having trouble with is one that is very common and very useful (though it does have its limitations; see below) so it is important and good to have it "down" properly.

You have the following situation: you are considering a certain type of object of interest. For simplicity, let's look at the earliest example that most students encounter, which is vector spaces. So, you are looking at vector spaces. Specifically, you are looking at a particular vector space $\mathbf{V}$.

The objects have sub-objects (subspaces). These are subsets of your original $\mathbf{V}$, which are also objects (vector spaces) in their own right. Not every subset is a subobject, but every subobject is a subset.

In this situation, it is often fruitful to consider the following problem:

Given a subset $S$ of $\mathbf{V}$, what is the smallest subspace of $\mathbf{V}$ that contains $S$?

That is, we want to find a $\mathbf{W}$ with the following properties:

$\mathbf{W}$ is a subspace of $\mathbf{V}$;
$S$ is contained in $\mathbf{W}$ ("...that contains $S$");
If $\mathbf{Z}$ is any subspace of $\mathbf{V}$ that contains $S$, then $\mathbf{W}\subseteq\mathbf{Z}$ ("... smallest ...")

This is the situation you have at hand, and it's also a very common situation that we encounter over and over again. Some examples:

Given a group $G$ and a subset $S$, to find the smallest subgroup of $G$ that contains $S$ (the "subgroup generated by $S$");
Given a group $G$ and a subset $S$, to find the smallest normal subgroup of $G$ that contains $S$;
Given a subset $S$ of the plane $\mathbb{R}^2$, to find the smallest convex set that contains $S$ (the "convex hull of $S$");
Given a set $X$ and a collection of subsets $\mathcal{S}\subseteq \mathcal{P}(X)$, find the smallest $\sigma$-algebra on $X$ that contains $\mathcal{S}$ (the case you have);
Given a set $X$ and a relation $R$ on $X$, find the smallest transitive relation on $X$ that extends $R$ (the "transitive closure");
Given a topological space $X$ and a subset $S$, find the smallest closed subset of $X$ that contains $S$ (the "closure of $S$").

and so on and so forth.

Now, in general, such a thing may not exist; or there may be minimal objects but no minimum object. For example, if in the last example above you replace "closed" with "open", there may be no such object: if $X=\mathbb{R}$ and $S=[0,1]$, there is no "smallest open set that contains $S$".

But in many situations, there is one single observation that lets you conclude that such as "smallest subobject" must exist. Namely, if you can show that the intersection of any collection of "subobjects" is again a "subobject". For the example with vector spaces: is the intersection of an arbitrary family of subspaces of $\mathbf{V}$ itself a subspace of $\mathbf{V}$? For the above examples:

Is the intersection of an arbitrary family of subgroups of $G$, itself a subgroup of $G$?
Is the intersection of an arbitrary family of normal subgroups of $G$ itself a normal subgroup of $G$?
Is the intersection of an arbitrary family of convex subsets of $\mathbb{R}^2$ itself a convex subset of $\mathbb{R}^2$?
Is the intersection of an arbitrary family of $\sigma$-algebras on $X$ itself a $\sigma$-algebra on $X$?
Is the intersection of an arbitrary family of transitive relations on $X$ itself a transitive relation on $X$?
Is the intersection of an arbitrary family of closed subsets of $X$ itself a closed subset of $X$?

When the answer is "yes", then the following construction will always show that there is such a thing as the "smallest subobject that contains $S$":

Take the family of all subobjects that contain $S$; then take the intersection of the family. That's the smallest subobject that contains $S$.

Why does this work?

Because:

(i) There is at least one subobject that contains $S$, (namely the original object itself; for $\sigma$-algebras, this would be $\mathcal{P}(X)$; for the transitive closure example, you would take the "total relation" $X\times X$).

(ii) Since the intersection of an arbitrary family of subobjects is a subobject (this is our assumption), then this intersection is a subobject.

(iii) Since each thing being intersected contains $S$, the intersection contains $S$.

This means that the intersection is indeed a subobject that contains $S$. Finally:

(iv) The intersection is always contained in each and every element of the family being intersected. So if $\mathbf{Z}$ is any subobject that contains $S$, then it is a member of the family being intersected, so the intersection is contained in $\mathbf{Z}$. This shows the intersection is indeed the "smallest subobject" with the desired properties.

So:

To find the smallest subspace of $\mathbf{V}$ that contains $S$, intersect all subspaces that contain $S$.
To find the smallest subgroup of $G$ that contains $S$, intersect all subgroups that contain $S$.
To find the smallest normal subgroup of $G$ that contains $S$, intersect all normal subgroups that contain $S$.
To find the smallest convex set that contains $S$, intersect all convex subsets of $\mathbb{R}^2$ that contain $S$.
To find the smallest $\sigma$-algebra that contains $S$, intersect all $\sigma$-algebras that contain $S$.
To find the smallest transitive relation that contains $R$, intersect all transitive relations that contain $R$.
To find the smallest closed subset that contains $S$, intersect all closed subsets that contain $S$.

And this works like magic. Voilá! You have shown that this object exists. It necessarily has the properties you want.

This is a "top-down" approach. Imagine yourself looking at the "big object", and you are "paring it down" until you get "just enough" for the object you want (intersections make things smaller; you are paring down stuff that may not be needed).

The problem? Like most magic spells, it doesn't really tell you much about the end product. The fact that the end product appeared "as if by magic" means that you are likely to be as clueless about the actual nature of the "smallest object" in question as you were when you started. You now know that there is such a thing, but you don't really know what it "looks like".

That is why in almost every situation like this, you also want a "bottom-up" description of this "smallest subobject that contains $S$". You want an explicit description of what it actually looks like. For the examples above:

The smallest subspace of $\mathbf{V}$ that contains $S$ is the set of all linear combinations of vectors in $S$.
The smallest subgroup of $G$ that contains $S$ is the set of all finite products of elements of $S$ and their inverses.
The smallest normal subgroup of $G$ that contains $S$ is the set of all finite products of conjugates of elements of $S$ and their inverses.
The smallest convex subset of $\mathbb{R}^2$ that contains $S$ is the set of all convex combinations of elements of $S$.
Asaf gives an explicit description of the smallest $\sigma$-algebra on $X$ that contains $S$ in his answer, described by starting from $S$.
The smallest transitive relation on a set $X$ that contains a given relation $R$ is the set of all pairs $(a,b)$ such that there exist a finite sequence $x_0,x_1,\ldots,x_n$ of elements of $X$ such that $x_0=a$, $x_n=b$, and $(x_i,x_{i+1})\in R$ for $i=0,\ldots,n-1$.
The smallest closed subset of a topological space $X$ that contains a given set $S$ is equal to $S\cup\partial S$ or to $S\cup S'$.

In each of these cases, one would have to show that the given description actually has the desired properties. This is a "bottom-up" approach.

The "top-down" description has the benefit of simplicity, that the "universal properties" that define the object are very clearly satisfied, and that they make proving results about how the "smallest object" relates to other objects easy. However, the "top-down" description is usually very hard to actually use to prove things about the specific smallest object. The "bottom-up" construction has the benefit of (usually) being a very concrete way of getting your hands on the object itself, making it easy to prove things about the object itself, but proving the universal properties is usually difficult. Thus, for example, the top-down definition of "subspace $\mathbf{V}$ generated by $S$" in the linear algebra setting makes it very hard to figure out things like the dimension of the subspace, or a basis, while the "bottom-up" approach makes that very easy, but then proving that the collection of all linear combinations forms a subspace is more difficult than simply taking an intersection of subspaces.

In most books or presentations, when discussing "the smallest X that contains S", you will see one of two approaches:

Define it as a big intersection, then prove a theorem that gives the "bottom-up" description; or
Give a "bottom-up" description; then prove the object described has the desired properties of being a subobject, containing S, and being the "smallest".

Whenever possible, you want both descriptions because they have complementary strengths and weaknesses.

Measure Theory – Prove Intersection of ?-Algebras is a ?-Algebra

Your proofs are OK. Well, maybe in (1) you should also state that $2^\Omega$ is nonempty. Similarly in (2) show that $\bigcap \mathcal A_i$ is nonempty.

Best Answer

Related Solutions

Descriptive Set Theory – Sigma Algebra of Subsets Generated by a Set

Measure Theory – Prove Intersection of ?-Algebras is a ?-Algebra

Related Question