Let $Z=\left(Z_{n}\right)_{n \in \mathbb{N}}$ be an $\left(\mathcal{F}_{n}\right)$ -adapted sequence of random variables with $\mathbb{E} \sup _{n \in \mathbb{N}}\left|Z_{n}\right|<\infty$. Let $\mathcal{T}$ be the set of all stopping times with respect to the filtration $\left(\mathcal{F}_{n}\right)_{n \in \mathbb{N}}$. We will use the following notation: $$ \begin{aligned} \mathcal{T}_{n, N} &=\{\nu \in \mathcal{T} \mid \mathbb{P}(\nu \in[n, N])=1\}, \quad 0 \leq n \leq N \\ \mathcal{T}_{n, \infty} &=\{\nu \in \mathcal{T} \mid \mathbb{P}(\nu \in[n,+\infty))=1\}, \quad n \in \mathbb{N} \end{aligned} $$
The Snell envelope of $\left(Z_{n}\right)_{n \in \mathbb{N}}$ is the sequence $\left(U_{n}\right)_{n \in \mathbb{N}}$ defined by $$ U_{n}=\operatorname{ess\,sup}_{\nu \in \mathcal{T}_{n, \infty}} \mathbb{E}\left(Z_{\nu} | \mathcal{F}_{n}\right) $$
Then we have the following results:
The author then presents Optimal stopping times:
My problem is how to show $$\mathbb{E} U_{\nu^{*}}=\mathbb{E} U_{0} \implies \forall n \in \mathbb{N}: \mathbb{E} U_{\nu^{*} \wedge n}=\mathbb{E} U_{0}$$
The reverse direction is easy, but this direction is unclear to me. Could you please elaborate on this point?
Because $U$ is a u.i. supermartingale and $0\le\nu^*\wedge n\le\nu^*<\infty$ (a.s.), you have $\Bbb E[U_0]\ge\Bbb E[U_{\nu^*\wedge n}]\ge\Bbb E[U_{\nu^*}]$ for all $n$.