I want to prove that given two ($\mathcal{F}_t$)-stopping times $\tau$ and $\nu$ and a $\mathcal{F}_\tau$-measurable random variable $Y$
\begin{equation} Y \mathbb{1}_{\{\tau \leq \nu \}} \text{ is } \mathcal{F}_\nu\text{-measurable}. \end{equation} Recall that $\mathcal{F}_\tau = \{ A \in \mathcal{F} : A \cap \{\tau \leq t\} \in \mathcal{F}_t$ for all $t\geq 0 \}$.
The proof I have come up with goes as follows:
Let $B \in \mathcal{B} (\mathbb{R})$ and $t\geq 0$.
\begin{equation} \{Y \in B\} \cap \{ \tau \leq \nu\} \cap \{ \nu \leq t\} = \Big(\{Y \in B\} \cap \{ \tau \leq t\}\Big) \cap \Big(\{ \tau \leq \nu\} \cap \{ \nu \leq t\} \Big) \end{equation}
Now $\{Y \in B\} \cap \{ \tau \leq t\}$ is in $\mathcal{F}_t$ because $Y$ is $\mathcal{F}_\tau$-measurable. $\{ \tau \leq \nu\}$ is in $\mathcal{F}_{\tau \wedge \nu} \subset \mathcal{F}_{\nu}$ and the right part is therefore in $\mathcal{F}_t$, too. This gives us that $\{Y \in B\} \cap \{ \tau \leq \nu\} \cap \{ \nu \leq t\} \in \mathcal{F}_t$ which completes the proof.
Is the above proof correct? I feel a bit weird about it.
Thanks!