Uncountable increasing family of $\sigma$-algebras

Question

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 ?

Answer 1

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.

Answer 2

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 $$ \mathbf{B}_{[0,1)}=\sigma \left( \bigcup_{t\geq 0} \mathcal{F}_t \right)$$.

Answer 3

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.

Answer 4

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 $f(n)<r$.