Sigma Field up to Constant Stopping Time is just the Sigma Field at that time
# Statement
Let $\mathcal{F} = {\mathcal{B}{n} : n \in \mathbb{N}}$ be a Discrete-Time Filtration over $\Omega$, let $k \in \bar{\mathbb{N}}$, and let $\nu : \Omega \to \bar{\mathbb{N}}$ be the Stopping Time on $\mathcal{F}$ defined as $\nu(\omega) = k$. Then $\mathcal{B}{\nu} = \mathcal{B}_{k}$.
# Proof
Suppose $A \in \mathcal{B}{\nu}$. Then $\forall n \in \bar{\mathbb{N}}$, $A \cap [\nu = n] \in \mathcal{B}{n}$. Note that $$[\nu = n] = \begin{cases} \emptyset &\text{if }n \neq k\\ \Omega &\text{if }n=k \end{cases}$$ so $\mathcal{B}{k} \ni A \cap [\nu = k] = A$ and $\mathcal{B}{\nu}\subset \mathcal{B}_{k}$. $\checkmark$
On the other hand, if $B \in \mathcal{B}{k}$, then $B \in \mathcal{B}{\infty} \supset \mathcal{B}{k}$. Observe that $B \cap [\nu = k] = B \in \mathcal{B}{k}$ by construction and $B \cap [\nu = n] = \emptyset \in \mathcal{B}{n}$ by definition of Sigma Algebra, for $n \neq k$. Thus, $B \in \mathcal{B}\nu$ and $B_{\nu} \supset \mathcal{B}_{k}$. $\checkmark$
Therefore $\mathcal{B}{\nu}= \mathcal{B}{k}$. $\blacksquare$