Measure Theoretic Probability - Video 9 - Generated Sigma Algebras
Generated Sigma Algebras
Some \(\sigma\)-algebras are hard to describe — one of them, very importantly, is the \(\sigma\)-algebra of subsets of \(\Bbb R\) which contains the intervals “and not too much more”. That is to say, we would like to describe a \(\sigma\)-algebra which has the intervals, but also has unions and complements of intervals (so that this will be a \(\sigma\)-algebra). But then, now that the collection of subsets has grown a little, we now need to take unions and complements of those things too. And when we do, now the collection has grown even more, so we need to take unions and complements of those things too and so on.
(Note: We will require a collection of measurable subsets to be a \(\sigma\)-algebra. We will even call the elements of a \(\sigma\)-algebra “measurables” even though that connection hasn’t really been made just yet. We haven’t even talked about what a measure function is, rigorously, yet. However, a word of warning: In this section I’m going to describe an important \(\sigma\)-algebra on the real numbers, and yet it will not ultimately turn out to be the collection of measurable subsets of real numbers. It will turn out that the collection of subsets of real numbers, that I describe here, is a bit smaller than the full collection of measurable real numbers.)
Because this description goes on forever, it would be nice to have a simpler and more manageable description of what we’re talking about. What we would like to capture is the notion of having a collection which contains all the intervals, and is a \(\sigma\)-algebra.
Well, so stated, that’s actually very easy! We could just take \(\mathcal P(\Bbb R)\)! Since it has everything it is automatically a \(\sigma\)-algebra with the intervals!
Well, of course, we want to avoid this result, because we already know that we are trying to only measure certain specific subsets of real numbers. We would like a collection of subsets which contains the intervals, is a \(\sigma\)-algebra, and is smaller than \(\mathcal P(\Omega)\).
… But how do we capture the notion of “smaller than”? When dealing with sets, sometimes (often, actually) the best you can say is that the set A is smaller than the set B if and only if \( A\subset B\). So that’s what we’ll say here.
But just asking for a strict subset of \(\mathcal P(\Omega)\) which is a \(\sigma\)-algebra containing the intervals, isn’t much guidance. Besides, if we think about the first paragraph above, it really seems like we want something more extreme than that. We don’t just want something smaller than \(\mathcal P(\Omega)\). We in fact want the smallest \(\sigma\)-algebra containing the intervals. We want the intervals, and the objects that we must have by the definition of a \(\sigma\)-algebra, but nothing else.
Therefore we may want to more generally capture the notion of a “smallest” set relative to a constraint, and then think about such a thing in our particular setting.
Minimality
Let \(\mathcal B\subseteq \mathcal P(\Omega)\) where \(\Omega\) is any set. Define \(\mathcal U = \{\mathcal S:\mathcal S \text{ is a } \sigma\text{-algebra}, \mathcal B\subseteq\mathcal S\}\). This \(\mathcal U\) is like the “universe of all \(\sigma\)-algebras with \(\mathcal B\) as a subset”. From this universe, we want to “pick” or define the concept of its minimum.
The minimal collection in \(\mathcal U\) is defined as that \(\mathcal S\in\mathcal U\) such that for all \(\mathcal T\in \mathcal U\) we have \(\mathcal S\subseteq \mathcal T\). That is to say: the minimal element is the one which is a subset of every other set in the universe of \(\sigma\)-algebras containing \(\mathcal B\).
(If you know about general ordering relations, and think of inclusion \(\subseteq\) as just a kind of ordering relation \(\le\), then we are saying that a minimal element of \(\mathcal U\) is an element \(x\in \mathcal U\) such that \(x\le y\) for all \(y\in\mathcal U\).)
We define \(\sigma\langle\mathcal B\rangle\), called the \(\sigma\)-algebra generated by the basis \(\mathcal B\), to be the minimal collection of \(\mathcal U\) defined above.
This is Real Yall
In order for all of this to be meaningful, we sure hope that this minimum described above exists! Well it does, and we prove that now. To justify talking about “the” minimum, we also prove uniqueness.
Theorem: \(\sigma\langle\mathcal B\rangle = \bigcap_{\mathcal S\in\mathcal U}\mathcal S\).
Proof: Call \(\bigcap_{\mathcal S\in\mathcal U}\mathcal S = I\). I choose I here to indicate “the intersection”. I do this little notational move just so I don’t have to keep typing all the LaTeX for the intersection, since my hands are getting tired.
We need to show that I is a \(\sigma\)-algebra, that it contains \(\mathcal B\), and that it has the minimality condition. The fact that I is a \(\sigma\)-algebra follows immediately from the fact that it is defined as an intersection over a family of \(\sigma\)-algebras. I have not yet proved that such a thing is always a \(\sigma\)-algebra, and therefore I will do so in the next video, but for now we take it as settled law.
To show that \(\mathcal B\subseteq I\) we note that \(\mathcal B\subseteq \mathcal S\) for every \(\mathcal B\in\mathcal U\). Therefore, if \(b \in \mathcal B\) then \(b\in\mathcal S\) for every \(\mathcal S\in\mathcal U\) and then by definition of the intersection \(b\in I\). Therefore \(\mathcal B\subseteq I\).
To show that I has the minimality condition, just note that (from what we just showed above) \(I\in\mathcal U\) and therefore by a basic result of set theory, \(I\in \bigcap_{\mathcal T\in\mathcal U}\mathcal T\).
(To elaborate, let us agree to say that if a set is “being intersected” then it is a “parent” of the intersection. So for instance, one parent of \(A\cap B\) is A and another is B. It is a basic result of set theory that the intersection is always a subset of any parent. In the above, we are using I as both the intersection as well as the parent.)
We have now demonstrated that I satisfies the condition for being \(\sigma\langle\mathcal B\rangle\). It only remains to show uniqueness. However, any object satisfying any reasonable minimality condition will always do so uniquely. In this particular case, we can say that if \(\mathcal T\in\mathcal U\) is any other object satisfying the definition of \(\sigma\langle\mathcal B\rangle\), then both I and \(\mathcal T\) have this minimality condition. By two applications of the minimality condition we must have \(I\subseteq \mathcal T\) and \(\mathcal T\subseteq I\) whence \(I=\mathcal T\), proving uniqueness.