Stopped Sigma-Algebra is Sigma-Algebra

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Theorem

Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.

Let $T$ be a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$.

Let $\FF_T$ be the stopped $\sigma$-algebra associated with $T$.


Then $\FF_T$ is a $\sigma$-algebra.


Proof

We show that $\Omega \in \FF_T$, that $\FF_T$ is closed under countable intersection, and relative complement.

For each $t \in \Z_{\ge 0}$ we have:

$\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} = \set {\omega \in \Omega : \map T \omega \le t}$

Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$ we have:

$\Omega \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

So $\Omega \in \FF_T$.

Now let $\sequence {A_n}_{n \mathop \in \N}$ be a sequence of sets in $\FF_T$.

Then for each $t \in \Z_{\ge 0}$, we have:

$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} = \bigcap_{n \in \N} \paren {A_n \cap \set {\omega \in \Omega : \map T \omega \le t} }$

from Intersection is Associative.

Since $A_n \in \FF_T$ for each $n \in \N$, we have:

$A_n \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

for each $n \in \N$.

Since $\FF_t$ is closed under countable intersection, we have:

$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

for each $t \in \Z_{\ge 0}$.

So:

$\ds \paren {\bigcap_{n \in \N} A_n} \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_T$

Finally we look to prove that if $A \in \FF_T$ then $A^c \in \FF_T$.

Let $t \in \Z_{\ge 0}$.

Note that:

\(\ds \set {\omega \in \Omega : \map T \omega \le t} \cap \paren {A^c \cup \set {\omega \in \Omega : \map T \omega > t} }\) \(=\) \(\ds \paren {A^c \cap \set {\omega \in \Omega : \map T \omega \le t} } \cup \paren {\set {\omega \in \Omega : \map T \omega \le t} \cap \set {\omega \in \Omega : \map T \omega > t} }\) Intersection Distributes over Union
\(\ds \) \(=\) \(\ds \paren {A^c \cap \set {\omega \in \Omega : \map T \omega \le t} } \cup \O\)
\(\ds \) \(=\) \(\ds A^c \cap \set {\omega \in \Omega : \map T \omega \le t}\) Union with Empty Set

From De Morgan's Laws (Set Theory): Set Difference: Family of Sets: Difference with Intersection:

$A^c \cup \set {\omega \in \Omega : \map T \omega > t} = \paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c$

Since $A \in \FF_T$ we have:

$A \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

Since $\FF_t$ is closed under relative complement we have:

$\paren {A \cap \set {\omega \in \Omega : \map T \omega \le t} }^c \in \FF_t$

Since $T$ is a stopping time with respect to $\sequence {\FF_n}_{n \ge 0}$, we have:

$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

Since $\FF_t$ is closed under finite intersection, we have:

$\set {\omega \in \Omega : \map T \omega \le t} \cap \paren {A^c \cup \set {\omega \in \Omega : \map T \omega > t} } \in \FF_t$

so that:

$A^c \cap \set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$

Since this holds for all $t \in \Z_{\ge 0}$, we have $A^c \in \FF_T$.

$\blacksquare$