Unclear step ($T \wedge n = T$ a.s) in Doob's optional sampling theorem for discrete martingales So leading up in my class we defined and proved the following things inorder to proof Doob's optional sampling theorem

A family $(X_i)_{i\in I}$ of real random variables is called uniformly integrable, if
  
  
*
  
*$\sup_{i\in I} E(|X_i|)<\infty$
  
*$\sup_{i\in I}E\left(|X_i|\cdot 1_{\{|X_i|\geq k\}}\right)\rightarrow0$ for $k\rightarrow\infty$.

Then we proved this theorem which is supposed to be an extension of the dominated convergence theorem

Let $\left(X_{n}\right)_{n \in \mathbb{N}}$ be a uniformly integrable sequence. If $X_{n}$ converges to $X$ in distribution, then $E\left(X_{n}\right) \rightarrow E(X)\hspace{15cm}(1)$

Using this theorem we got to the proof that I'm interested in

Definition Let $T$ be a stopping time. $X^{T}$ is the process defined by $X_{n}^{T}=X_{T \wedge n}$ or $X_{n}^{T}(\omega)=X_{T(\omega) \wedge n}(\omega)$
(Doob's optional sampling theorem) Let $X$ be a submartingale and let $T$ be a stopping time such that $T<\infty$ almost surely and $\left(X_{T \wedge n}\right)_{n=1,2, \ldots,}$ is uniformly integrable. Then $E\left(X_{T}\right) \geq E\left(X_{0}\right) .$ The analogue holds for supermartingales and martingales.
Proof. $T \wedge n=T$ almost surely for eventually all $n$ and thus $X_{T \wedge n} \stackrel{\text { a.s. }}{\longrightarrow} X_{T}$ and (1) implies $E\left(X_{T}\right)=\lim E\left(X_{T \wedge n}\right) \geq E\left(X_{0}\right)$

Now my questions are


*

*Why is $T \wedge n = T$ a.s ? I mean I get that when $n$ goes to infinity $T \wedge n = T$ eventually is just equal to $T$ but doesn't the almost sure part on the probability measure we use ?

*How is (1) a "extension of the dominated convergence theorem" ?
 A: *

*I think that the proof was trying to say that the probability that $T\wedge n \ne T$ tends to zero as $n\rightarrow\infty$ and that therefore $X_{T\wedge n}$ converges in distribution to $X_T$.

*(1) is called the Vitali convergence theorem. 
The statement of the dominated convergence theorem is as follows:

Let $f_n$ be a sequence of complex-valued measurable functions on a
  measure space (S, Σ, μ). 
Suppose that the sequence converges pointwise to a function f and is
  dominated by some integrable function g in the sense that $|f_n(x)|\le
g(x)$ for all numbers n in the index set of the sequence and all
  points x ∈ S. 
Then f is integrable and $\lim_{n\rightarrow\infty}\int_S
 f_n\mathrm{d}\mu=\int_S f\mathrm{d}\mu$.

Now imagine that $f_n$ is the probability distribution of $X_n$ and $f$ is the probability distribution of $X$. Then (1) is saying that if $f_n\rightarrow f$ and $(X_n)_n$ is uniformly integrable then $\lim_{n\rightarrow \infty}\int_Sf_n\mathrm{d}\mu=\int_Sf\mathrm{d}\mu$. So the condition $|f_n(x)\le g(x)|$ is replaced with the weaker condition that $(X_n)_n$ is uniformly integrable.
