= Solution
{c}
The whole space belongs to $\mathcal F_T$, since $\Omega\cap\{T\leq n\}=\{T\leq n\}\in\mathcal F_n$. If $A\in\mathcal F_T$, then
$$
A^c\cap\{T\leq n\}=\{T\leq n\}\setminus(A\cap\{T\leq n\})\in\mathcal F_n.
$$
If $A_j\in\mathcal F_T$, then $(\bigcup_jA_j)\cap\{T\leq n\}=\bigcup_j(A_j\cap\{T\leq n\})$ is also in $\mathcal F_n$. These are precisely the sigma-algebra closure properties, so $\mathcal F_T$ is a sigma-algebra.
For every integer $m\geq0$,
$$
\{T\leq m\}\cap\{T\leq n\}=\{T\leq\min(m,n)\}\in\mathcal F_n.
$$
Hence $\{T\leq m\}\in\mathcal F_T$. Such threshold sets generate the Borel sigma-algebra on $\mathbb N_0\cup\{\infty\}$, proving \b[$T$ is $\mathcal F_T$-measurable]. This is the <stopping-time sigma-algebra>.
Back to article page
