Sigma Field up to Stopping Time

Last updated Nov 1, 2022

Let $\mathcal{F} = {\mathcal{B}{n} : n \in \mathbb{N}}$ be a Discrete-Time Filtration over $\Omega$ and let $\nu : \Omega \to \bar{\mathbb{N}}$ be a Stopping Time on $\mathcal{F}$. Then the Sigma Field up to Stopping Time is the Sigma Algebra $$\mathcal{B}{\nu} := {A \in \mathcal{B}{\infty}: A \cap [\nu = n] \in \mathcal{B}{n} \text{ }\forall n \in \mathbb{N}}$$

# The Sigma Field up to Stopping Time is a Sigma Algebra

We verify the axioms of a Sigma Algebra.

1. Observe that $\Omega \cap [\nu = n] = [\nu = n] \in \mathcal{B}{n}$ for all $n \in \mathbb{N}$ by definition of Stopping Time. Thus $\Omega \in \mathcal{B}\nu$. $\checkmark$
2. Suppose $A \in \mathcal{B}_\nu$. Then for all $n \in \mathbb{N}$ $$A^{C} \cap [\nu = n] = (A \cap [\nu = n])^{C} \cap [\nu =n] \in \mathcal{B}n$$ since $A \cap [\nu = n], [\nu = n] \in \mathcal{B}n$ and $\mathcal{B}{n}$ is a Sigma Algebra. Thus, $A^{C} \in \mathcal{B}\nu$. $\checkmark$
3. Suppose ${A_{i}}{i \in \mathbb{N}} \subset \mathcal{B}\nu$. Then $\forall n \in \mathbb{N}$ $$[\nu = n] \cap\bigcup\limits_{i \in \mathbb{N}} A_{i} = \bigcup\limits_{i \in \mathbb{N}} (A_{i} \cap [\nu = n]) \in \mathcal{B}{n}$$ since $A{i} \cap [\nu = n] \in \mathcal{B}{n}$ $\forall i \in \mathbb{N}$ and $\mathcal{B}{n}$ is a Sigma Algebra. $\checkmark$

Thus, $\mathcal{B}_\nu$ is a Sigma Algebra. $\blacksquare$

1. Sigma Field up to Constant Stopping Time is just the Sigma Field at that time
2. A Stopping Time is Measurable with respect to the Sigma Field up to it
3. Corollary of (2): A Stopping Time is Measureable with respect to the Maximal Sigma Field of its Filtration
4. Equivalent Conditions for being a Sigma Field up to a Stopping Time
5. Comparing two Stopping Times is in Sigma Field up to either Stopping Time

1. If $A \in \mathcal{B}{\nu}\subset \mathcal{B}\infty$ then since $[\nu = \infty] \in \mathcal{B}\infty$, we get that $A \cap [\nu = \infty] \in \mathcal{B}\infty$. Thus, the defining property can be extended to $\bar{\mathbb{N}}$.