Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $(X_t)_{t\in[0,\:\infty]}$ be a real-valued process on $(\Omega,\mathcal A,\operatorname P)$;
- $\tau$ be an $[0,\infty]$-valued random variable on $(\Omega,\mathcal A,\operatorname P)$.
Now let $$\varphi(t):=\operatorname E\left[X_\tau\mid\tau=t\right]\;\;\;\text{for }t\in[0,\infty].$$ How can we prove the intuitively trivial claim $$\varphi(t)=\operatorname E\left[X_t\mid\tau=t\right]\tag1$$ for all $t\in[0,\infty]$?
By definition, $\varphi$ is the function from the factorization lemma; i.e. it is the Borel measurable function $[0,\infty]\to\mathbb R$ with $\varphi\circ\tau=\operatorname E[X_\tau\mid\tau]$. Now, let $$\tilde\varphi(t):=\operatorname E\left[X_t\mid\tau=t\right]\;\;\;\text{for }t\in[0,\infty].$$
Now, by the local property of the conditional expectation and from noting that $$X_\tau=X_t\;\;\;\text{on }\{\tau=t\}\tag2,$$ we obtain $$\varphi(t)=\varphi(\tau)=\operatorname E[X_\tau\mid\tau]=\operatorname E[X_t\mid\tau]=\tilde\varphi(\tau)=\tilde\varphi(t)\tag3$$ almost surely on $\{\tau=t\}$ for all $t\in[0,\infty]$. But is this really enough to conclude?
You cannot conclude anything in general if $P[\tau=t] = 0$ since the probability $P[\cdot|\tau=t]$ are not well-defined. Moreover, using regular version of conditional probability does not solve the problem.
A counterexample. Let $\tau$ be a random time with diffuse law and set $X_t=1_{[\tau=t]}$ for all $t \ge 0$. Since $X_t$ is null with probability $1$, $E[X_t|\sigma(\tau)] = 0$ almost surely, i.e. $E[X_t|\tau=s] = 0$ for almost (for the law of $\tau$) every $s$. Yet, $X_\tau=1$ so $E[X_\tau|\sigma(\tau)] = 1$ almost surely, i.e. $E[X_\tau|\tau=s] = 1$ for almost (for the law of $\tau$) every $s$. Hence, the most natural choices when $s=t$ are $E[X_t|\tau=t] = 0$ and $E[X_\tau|\tau=t] = 1$.