defining the stopping time sigma algebra For a stopping time T, define $\mathcal{F}_T$ by 
$\mathcal{F}_T={A \in \mathcal{F}:A \cap \{T \le t\} \in \mathcal{F}_t, \text{for each t.}}$ 
Verify that $\mathcal{F}_T$ is a $\sigma$-algebra.
To prove this I know we must show that $\mathcal{F}_T$ satisfies the three properties of  $\sigma$-algebras, ie that  
A $\sigma$-algebra $\mathcal{F}$ of subsets of a set X is a collection $\mathcal{F}$ of subsets satisfying the following properties;
(1) $\emptyset$ $\in$ $\mathcal{F}$
(2) if A $\in$ $\mathcal{F}$ then it's compliment $A^{c}$ is also in $\mathcal{F}$
(3) if $A_1,A_2,...$ is a countable collection of sets in $\mathcal{F}$  then their union $\cup^{\infty}_{n=1}A_n \in \mathcal{F}$  
I have attempted this myself but am concerned I am not being precise enough. Any hints or comments are gladly welcome.  
For (3):
Suppose $A_1,A_2,...$ is a countable collection of sets in $\mathcal{F}_T$
Then $A_n \in \mathcal{F}$ : $A_n \cap \{T \le t\} \in \mathcal{F}_t$ for all n by definition
Then $\cup^{\infty}_{n=1}(A_n \cap \{T \le t\})\in \mathcal{F}_t$
Then $(\cup^{\infty}_{n=1}A_n) \cap (\{T \le t\})\in \mathcal{F}_t$ 
Then $(\cup^{\infty}_{n=1}A_n) \in \mathcal{F}_T$
For(2) I am less convinced
My reasoning would be that if $A \in \mathcal{F}_T$ then $A \in \mathcal{F}: A \cap \{T \le t\} \in \mathcal{F}_t$
We know that $A \cap \{T \le t\} \in \mathcal{F}_t$
Now $A^c \in \mathcal{F}$ since $A^c = \mathcal{F}/A$
Then $A^{c} \cap \{T \le t\} = \mathcal{F} \cap \{T \le t\} - A \cap \{T \le t\}$
Since $\mathcal{F} \cap \{T \le t\} = \mathcal{F}_t$
Then $A^{c} \cap \{T \le t\} = \mathcal{F}_t - A \cap \{T \le t\}$
ie $A^{c} \in \mathcal{F}_T$
However for this to hold we must consider the situation where $A \cap \{T \le t\} = \mathcal{F}_t$
In other words here $A^c$ would be $\emptyset$ and this is where I'm stuck. 
If I could prove $\emptyset \in \mathcal{F}_T$ (2) would follow so any hints to get me started would be really appreciated as I really don't know where to start, or if I am completely wrong that would also be good to know! Thanks in advance!
 A: \begin{multline}
1) \emptyset : \emptyset \cap \{T\leq n\} =\emptyset \in \mathcal{F}_n \Rightarrow \emptyset \in \mathcal{F}_n
\end{multline}
\begin{multline}
2) \Omega : \Omega \cap \{T\leq n\} =\{T\leq n\} \in \mathcal{F}_n \Rightarrow \Omega \in \mathcal{F}_n
\end{multline}
\begin{multline}
3) A \in \mathcal{F}_n : A \cap \{T\leq n\} \in \mathcal{F}_n \forall n
\end{multline}
We have to prove that $A^c \cap \{T\leq n\}\in \mathcal{F}_n$
\begin{multline}
4) A^c \cap \{T\leq n\} = \underbrace{(A^c \cap \{T\leq n\})^c}_{\in \mathcal{F}_n} \cap \underbrace{\{T\leq n\}}_{\in \mathcal{F}_n}\in \mathcal{F}_n
\end{multline}
\begin{multline}
5) A_k \in \mathcal{F}:(\bigcup^{\infty}_{k=1}A_k) \cap \{T\leq n\} = \bigcup^{\infty}_{k=1} \underbrace{(A_k \cap \{T\leq n\})}_{\in \mathcal{F}_n}\in \mathcal{F}_n
\end{multline}
A: Your argument for (2) doesn't make sense - in particular, the statement $A^c=\mathcal{F}-A$ is not defined: On the lefthand side is a set; on the righthand side is a collection of sets.
Start with $A^c=\Omega-A$ (if $\Omega$ is the set on which the filtration is taken), and note $A=(A\cap\{T\le t\})\cup(A^c\cap \{T\le t\})$. 
