I am struggling to understand a Lemma from Chapter 3 Section 5 of Protter's Stochastic Integration and Differential Equations.
Here, we take $\tau$ to be a random time on a complete probability space $(\Omega, \mathscr{F},P)$. Then set $\mathscr{F}_t^{00} = \sigma\{\tau \wedge t\}$, i.e. the smallest sigma algebra making $\tau \wedge t$ measurable. Then adjoin the null sets of $\mathscr{F}$ to get $\mathscr{F}_t^0$. Finally, we take $\mathscr{F}_t$ to be the right continuous version of $\mathscr{F}_u^0$. The Lemma states that the $\mathscr{F}_t$ so constructed forms the smallest filtration making $\tau$ a stopping time.
I am trying to understand why this is as they claim. So by definition, $\sigma \{ \tau \wedge t \}$ is generated by sets of the form $\{\tau \wedge t \ge r\}= \{\tau \ge r\} \cap [t,r] .$ From this, we can see that $\mathscr{F}_t^0$ is isomorphic to the Borel $\sigma$-algebra on $[0,t]$ together with the indivisible atom $(t,\infty)$. Hence indeed, we would have $\mathscr{F}_t^0 = \sigma\{ \tau \wedge u; u \le t\}$.
But how do we get the final identity for $E[Y| \mathscr{F}_t]$ from this? I cannot understand how we are able to replace the condition on $\mathscr{F}_t$ by $\tau$ here. It seems like we split $Y$ into $Y1_{\{t \ge \tau\}}+Y1_{\{\tau > t\}}$ but I cannot see why the conditional expectation of each given $\mathscr{F}_t$ are the expressions on the right. I would greatly appreciate some explanation on this.
You know from your considerations that $\{\tau>t\}$ is an atom of $\mathcal{F}_t$. Since $\mathbb{E}(Y \mid \mathcal{F}_t)$ is $\mathcal{F}_t$-measurable, the random variable needs to be almost surely on the atom $\{\tau>t\}$, i.e. there is a constant $c \in \mathbb{R}$ such that
$$\mathbb{E}(Y \mid \mathcal{F}_t) =c \quad \text{a.s. on $\{\tau>t\}$}.$$
In order to determine $c$, note that, by the definition of the conditional expectation,
$$c \int_{\{\tau>t\}} \, d\mathbb{P} = \int_{\{\tau>t\}} \mathbb{E}(Y \mid \mathcal{F}_t) \, d\mathbb{P} = \int_{\{\tau>t\}} Y \, d\mathbb{P},$$
i.e.
$$c \, \mathbb{P}(\tau>t) = \mathbb{E}(Y 1_{\{\tau>t\}}).$$ Dividing both sides by $\mathbb{P}(\tau>t)$ yields $c$, and we get
$$\mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau>t\}}= \frac{\mathbb{E}(Y 1_{\{\tau>t\}})}{\mathbb{P}(\tau>t)} 1_{\{\tau>t\}}. \tag{1}$$
It remains to determine the conditional expectation on $\{\tau \leq t\}$. To this end, note that
$$Z = \mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau \leq t\}} = \mathbb{E}(Y 1_{\{\tau \leq t\}} \mid \mathcal{F}_t),$$
is the $\mathcal{F}_t$-measurable random variable such that
$$\forall F \in \mathcal{F}_t\::\: \int_F Z \, d\mathbb{P} = \int_F (Y 1_{\{\tau \leq t\}}) \, d\mathbb{P}. \tag{2} $$
Consequently, we are done if we can show that $Z=\mathbb{E}(Y \mid \tau) 1_{\{\tau \leq t\}}$ has the above properties. To this end, recall that by the factorization lemma $\mathbb{E}(Y \mid \tau) = h(\tau)$ for some measurable mapping $h:\mathbb{R}\to \mathbb{R}$. From
$$Z = h(\tau) 1_{\{\tau \leq t\}} = h(\tau \wedge t) 1_{\{\tau \leq t\}},$$
we see that $Z$ is $\mathcal{F}_t$-measurable (because $\{\tau \leq t\} \in \mathcal{F}_t$ and $\tau \wedge t$ is $\mathcal{F}_t$-measurable). It remains to check $(2)$. Note that $\mathcal{F}_t \subseteq \sigma(\tau)$. Hence, if $F \in \mathcal{F}_t$, then $F \cap \{\tau \leq t\} \in \mathcal{F}_t \subseteq \sigma(\tau)$, and so
$$\int_{F \cap \{\tau \leq t\}} \mathbb{E}(Y \mid \tau) \, d\mathbb{P} = \int_{F \cap \{\tau \leq t\}} Y \, d\mathbb{P}.$$
Equivalently,
$$\int_F Z \, d\mathbb{P} = \int_F 1_{\{\tau \leq t\}} Y \, d\mathbb{P},$$
which is exactly $(2)$. Combining both considerations, we conclude that
\begin{align*} \mathbb{E}(Y \mid \mathcal{F}_t) &=\mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau \leq t\}} + \mathbb{E}(Y \mid \mathcal{F}_t) 1_{\{\tau>t\}} \\ &= \mathbb{E}(Y \mid \tau) 1_{\{\tau \leq t\}} + \frac{\mathbb{E}(Y 1_{\{\tau>t\}})}{\mathbb{P}(\tau>t)} 1_{\{\tau>t\}}, \end{align*}
as claimed.