Uncountable increasing family of $\sigma$-algebras

Could someone give an example of what an uncountable increasing family of $\sigma$-algebras $\{\mathcal{F}_t\}_{t\geq 0}$, $(\mathcal{F}_s \subset \mathcal{F}_t$ for $s<t)$ might look like?

For the discrete parameter case, I always have in mind the filtration induced on $([0,1),\mathbf{B}_{[0,1)})$ by the sequence of independent random variables $(X_n)_{n \geq 1}$ where $X_k(\omega) = \omega_k$ for $\omega = 0.\omega_1\omega_2\omega_3...$ represented in binary system. For a given $n$, $\mathcal{F}_n = \sigma(X_1,X_2,...,X_n)$ is just the $\sigma$-algebra whose atoms are the dyadic intervals $[k/2^n, (k+1)/2^n)$ for $0 \leq k \leq 2^n-1$. As $n$ increases, $\mathcal{F}_n$ gets finer and finer and ultimately "converges" to $\mathcal{F}=\mathbf{B}_{[0,1)}$.

Are there any explicit examples in the continuous parameter case ?

• Do these $\sigma$-algebras have to concentrate on $\Bbb R$ or something? Also, do they have to be order isomorphic (under inclusion) to $[0,1]$ or some other interval on $\Bbb R$? Jul 30 '15 at 0:57
• I was hoping they might be easy to "visualize", but it seems there are hardly any examples. Jul 30 '15 at 1:10
• You can't really visualize anything which is not the finite union of intervals and rays; or some "simple enough" union of connected sets on the plane/in the 3D space. Let alone closed sets, or even open sets, which are more complex than that. Imagine the rational numbers as an ordered set; now imagine the irrational numbers as an ordered sets. Both sets have the same properties as far as the order is concerned, so you shouldn't be able and see any difference in your visualization. And yet... Jul 30 '15 at 1:22

Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. $\mathcal{F}_t=\sigma(W_s:s \in [0,t])$. This sort of thing is usually how filtrations are created.

• Yes, but could you imagine what $\mathcal{F}_t$ "looks like" for a given $t$? Jul 30 '15 at 1:02
• Explicitly, no. Can you imagine what the Borel $\sigma$-algebra on $\mathbb R^n$ looks like? Jul 30 '15 at 1:02
• @Nocturne While you can't really visualize, the intuition is that it encapsulates all the information that you could measure by knowing the trajectory up to time t.
– Ian
Jul 30 '15 at 1:37
• @Ian Yes, thanks. $\mathcal{F}_t$ contains precisely the events $A$ for which you can tell with certainty whether they occurred or not just by knowing the values of $X_s(\omega), 0 \leq s \leq t$. One gets a partial knowledge of which (unknown) $\omega$ might have been realized. Jul 30 '15 at 9:55

An example of a filtered probability space is $$([0,1), \mathbf{B}_{[0,1)}, \{\mathcal{F}_t\}_{t\geq 0} , \mathbb{P})$$, where, for each $$t\in \mathbb{R}, t\geq 0$$, define $$\mathcal{F}_t = \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right]}=\textrm{ the Borel \sigma-algebra defined in } \left [0\,,\,1-\frac{1}{t+1} \right]$$

Note that, by definition of filtered probability space, it is not required that $$\bigcup_{t\geq 0} \mathcal{F}_t$$ be a $$\sigma$$-algebra. It is just required that for each $$t\in \mathbb{R}, t\geq 0$$, $$\mathcal{F}_t \subseteq \mathbf{B}_{[0,1)}$$.

Note also that in the discrete case you mentioned, the union of the $$\sigma$$-algebras $$\mathcal{F}_n$$ is NOT a $$\sigma$$-algebra and so such union is not $$\mathbf{B}_{[0,1)}$$. However $$\mathbf{B}_{[0,1)}=\sigma \left( \bigcup_{n\in \mathbb{N}} \mathcal{F}_n \right)$$.

The same happens in this example: the union of the $$\sigma$$-algebras $$\mathcal{F}_t$$ is NOT a $$\sigma$$-algebra and so such union is not $$\mathbf{B}_{[0,1)}$$. However $$\mathbf{B}_{[0,1)}=\sigma \left( \bigcup_{t\geq 0} \mathcal{F}_t \right)$$.

• What did you post a new answer instead of editing this into the previous one? Jul 30 '15 at 15:56
• @Asaf For clarity. The question is not clear what is the example Nocturne is looking for. One of my answer addresses the "question" of an example of an uncountable family of $\sigma$-algebra. My other answer addresses the "question" of an example of a filtered probability space with continuous parameter. Althought, somehow related they are completely different answer (and both are too long to be just a comment). Jul 30 '15 at 16:22
• @Asaf There are still other ways to consider the question. For instance, maybe Nocturne is looking for an uncountable family of independent random variables which would produce a continuos parameter filtration on the probability space. But, I don't plan to provide other answers to this question. Jul 30 '15 at 16:28
• Thank you guys! @RamiroGuerreiro Is there a process which induces your constructed filtration, i.e. a family of random variables $(X_t)_{t \geq 0}$ such that $\mathcal{F}_t = \mathbf{B}_{[0,1-\frac{1}{1+t}]} = \sigma(X_s, 0 \leq s \leq t)$ ? Jul 30 '15 at 18:49
• @Nocturne If you want the random variables defined in the $[0,1)$, then we need to adapt the example as follows: For each $t\in \mathbb{R}, t\geq 0$, let $\mathcal{F}_t =\left\{A+B \,|\, A\in \mathbf{B}_{\left[0\,,\,1-\frac{1}{t+1}\right)} \textrm{ and } B\in \{\emptyset, [1-\frac{1}{t+1},1)\} \right\}$. Now define: for each $t≥0$, $X_t$ to be the function from [0,1) to [0,1], such that, if $t>0$, if $\omega\in [0,1-\frac{1}{t+1})$, $X_t(ω)=\frac{t+1}{t}ω$ and, if $\omega\in [1-\frac{1}{t+1},1)$, $X_t(ω)=1$. For $t=0$, $X_0=1$. Jul 31 '15 at 16:01

You can look at $$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$

For $r\in[0,1]$. Where Borel means the standard Borel sets of $[0,1]\times[0,1]$. Then for $r\leq s$ we have $\cal B_r\subseteq B_s$, since whenever $A\in\cal B_r$, we have that $A\cap[r,s]\times[0,1]\in\cal B_s$, and both of these sets are Borel sets.

You should note, however, that if you can find a countable chain without an upper bound, then it is most likely that the union of the these $\sigma$-algebras is not a $\sigma$-algebra itself. Simply $\cal B_n$ to be such witnessing chain, and pick $A_n\in\cal B_{n+1}\setminus\cal B_n$. Then $\bigcup A_n$ is not in any $\cal B_n$, so it is not necessarily in any other member of the family. (See the comment by hot_queen below, a countable union of increasing chain of $\sigma$-algebras is never a $\sigma$-algebra.)

Therefore indexing your $\sigma$-algebras using $[0,1]$ is not always a great idea.

• You mean $A\cap(r,1]\times[0,1]$ is either $(r,1]\times[0,1]$ or empty. Also, I don't follow what you're saying in the penultimate paragraph. Why can't $\bigcup A_n$ be in any $\mathcal{B}_n$? In any case, the use of nested $\sigma$-algebras indexed by reals is standard when talking about continuous-time stochastic processes. Jul 30 '15 at 2:02
• Eric, yes, thanks. And in the penultimate paragraph, it could be a $\sigma$-algebra, or it could not be a $\sigma$-algebra. That depends on what $\bigcup A_n$ is. But if you try to prove that in general an increasing union of $\sigma$-algebras is a $\sigma$-algebra, you will run into troubles exactly at that point. But I will be more accurate there. Thank you. Jul 30 '15 at 7:09
• @EricWofsey An increasing union of $\omega$-chain of sigma algebras is never a sigma algebra - See A comment on unions of sigma fields, A. Broughton, B. Huff, American mathematical monthly, 1977 Vol. 84 No. 7, pp 553-54 Jul 30 '15 at 13:49
• @AsafKaragila The comment above from hot_queen is about $\omega$-chain of $\sigma$-algebras, NOT about countable unions of $\sigma$-algebras in general. In fact, it is easy to build examples of a countable union of increasing chain of $\sigma$-algebras which is a σ-algebra. For instance, from a any increasing $\omega$-chain of $\sigma$-algebras, build a $\omega + 1$-chain (which is still countable) of $\sigma$-algebras by adding as the $\omega + 1$ element in the chain the $\sigma$-algebra generated by all the previous $\sigma$-algebras in the chain. Aug 23 '15 at 21:00
• @Ramiro: You're right. I had in mind any limit-type chain. With successor type it's not much of a contest, the result is just the maximum element of the chain. Aug 23 '15 at 22:31

You can do the same thing as you do in the discrete-time case by just thinking of the $X_n$ as being indexed by the rationals. Explicitly, choose a bijection $f:\mathbb{N}\to\mathbb{Q}$, and for each $r\in\mathbb{R}$, let $\mathcal{F}_r$ be the $\sigma$-algebra generated by the $X_n$ such that $f(n)<r$.

• @PedroTamaroff: This is uncountable; $\mathcal{F}_r$ is defined and different for each real $r$. Jul 31 '15 at 0:30
• Oh, for a second I thought $r$ was rational. Deleted. Jul 31 '15 at 0:41