Let
- $(\Omega,\mathcal A)$ be a measurable space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$ and $$\mathcal F_\infty:=\sigma(\mathcal F_t,t\ge0);$$
- $\tau$ be an $(\mathcal F_t)_{t\ge0}$-stopping time on $(\Omega,\mathcal A)$
- $(E,\mathcal E)$ be a measurable space;
- $(X_t)_{t\ge0}$ be an $(\mathcal F_t)_{t\ge0}$-progressive $(E,\mathcal E)$-valued process on $(\Omega,\mathcal A)$;
- $X_\infty:\Omega\to E$.
Am I missing something or is $X_\infty$ always $\mathcal F_\tau$-measurable; without any measurability assumption on $X_\infty$?
In fact, this should be true, since $(\tau\wedge t)_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-adapted, we easily see that $(X_{\tau\wedge t})_{t\ge0}$ is $(\mathcal F_t)_{t\ge0}$-adapted as well.
By introduction of $X_\infty$, $X_\infty$ is a well-defined function; even on $\{\tau=\infty\}$. And the former adaptedness result shows that $$\{X_\tau\in B\}\cap\{\tau\le t\}=\{X_{\tau\wedge t}\in B\}\cap\{\tau\le t\}\in\mathcal F_t\;\;\;\text{for all }t\ge0\tag1$$ and hence $$\{X_\tau\in B\}\in\mathcal F_\tau\tag2$$ for all $B\in\mathcal E$.
(the natural assumption would clearly be that $X_\infty$ is $\mathcal F_\infty$-measurable, since then $(X_t)_{t\in[0,\:\infty]}$ is $(\mathcal F_t)_{t\in[0,\:\infty]}$-adapted)
You have argued that the restriction of $X_\tau$ to $\{\tau<\infty\}$ is $\mathcal F_\tau\cap\{\tau<\infty\}$ measurable. But you have shown nothing about $X_\tau$ ($=X_\infty$) on $\{\tau=\infty\}$.