I encountered an argument that I could not follow. Here are the setting (that I simplified) and my question. Thank you in advance!
Time is discrete: $t \in \mathbb{N}$. Consider an adapted stochastic process $(X_{t})_{t \in \mathbb{N}}$ on a (standard) filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_{t})_{t \in \mathbb{N}},\mathbb{P})$. For some $N \in \mathbb{N}$, we also have a finite probability space: \begin{align*} \left( \tilde{\Omega}=\left\{1,2,\ldots,2^{N}\right\},~ \tilde{\mathcal{F}}=2^{\tilde{\Omega}},~ \tilde{\mathbb{P}}=\left\{p_{1},p_{2},\ldots,p_{2^{N}}\right\} \right). \end{align*} Assume that a random variable $Y$ defined on $(\tilde{\Omega},\tilde{\mathcal{F}},\tilde{\mathbb{P}})$ takes one of the $2^{N}$ different values, denoted $y^{1} < y^{2} < \cdots < y^{2^{N}}$.
Suppose that there is an almost surely finite stopping time $\sigma \ge 0$ such that: \begin{align*} \mathbb{P}(X_{\sigma} \ge y^{i}) > \tilde{\mathbb{P}}(Y \ge y^{i}) \qquad \forall\, i = 1,2,\ldots,2^{N}. \end{align*} Note that $\mathbb{E}[X_{\sigma}] < \infty$ (as pointed out by Math1000).
I encountered the following argument:
The Dominated Convergence Theorem implies $T \in \mathbb{N}$ such that \begin{align*} \mathbb{P}(X_{\sigma \wedge T} \ge y^{i}) \ge \tilde{\mathbb{P}}(Y \ge y^{i}) \qquad \forall\, i = 1,2,\ldots,2^{N}. \end{align*} Hence, we have the bounded stopping time $\tau \equiv \sigma \wedge T$.
I had no idea how the dominated convergence theorem is used in this context? Thank you.