Recall that the product sigma-algebra $\mathcal{B}[0,T] \otimes \mathcal{F}_T$ is generated by
$$\{[0,t] \times B; t \leq T, B \in \mathcal{F}_T\}.$$
Consequently, it suffices to prove that $M^{-1}([0,t] \times B) \in \mathcal{B}[0,T] \otimes \mathcal{F}_T$ for $t \leq T$, $B \in \mathcal{F}_T$. This follows from
$$\begin{align*} M^{-1}([0,t] \times B) &= \{(s,\omega) \in [0,T] \times \Omega; s \wedge \tau(\omega) \leq t, \omega \in B\} \\ &= \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau \leq t]\} \\ &\quad \cup \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau>t]\} \\ &= ([0,T] \times (B \cap[\tau \leq t])) \cup ([0,t] \times (B \cap [\tau>t])).\end{align*}$$
I know the perils of this exercise, having been on the end of it at the start of my journey. I'll give my best possible shot at putting everything in it completely beyond doubt.
So we start with $\Omega = [0,2)$, $\mathscr F = \mathscr B[0,2)$ and $P(F) = \mu(F \cap [0,1])$, which is a weird measure since it's concentrated on $[0,1]$. We'll understand why this is the case when we investigate the answer further.
The definition of $X_t(\omega)$ is very arbitrary-looking : we will be able to explain it once we're done.
$$
X_t(\omega) = \begin{cases}
1 & \omega \in (1,2), t = \omega \\
0 & \text{otherwise}
\end{cases}
$$
We take $t_0 = 2$, and will prove that $A = \{X_t \text{ is continuous on } [0,2)\}$ does not belong to $\mathscr F^x_{2}$. Call this the continuity-characterisation of $A$.
The first is the result on the representation of $\mathscr F^X_{t_0}$ as the set of preimages $$\{(X_{t_1},X_{t_2},\ldots,) \in B : B \in \mathscr B(\mathbb R^d) \otimes \mathscr B(\mathbb R^d) \otimes \ldots, 0 \leq t_1<t_2<\ldots , t_i \leq t_0\}$$
The proof of this representation uses the definition that $\mathscr F^x_{t_0}$ is generated by the preimages of $X_i$ at finitely many time points $i \leq t_0$. If this is known, then the set described above is a sigma-algebra and is contained in $\mathscr F^X_{t_0}$ as each set is the countable intersection of its finite truncations, so it must be equal to $\mathscr F^X_{t_0}$.
Once that is done , $A \in \mathscr F^X_{t_0}$ implies that $A = \{(X_{t_1},\ldots) \in B\}$ for appropriate $t_k \in [0,2]$ and image set $B$. This is the second characterisation of $A$.
Now we pick a $\bar{t} \neq t_k$ for all $k$ with $\bar{t} \in (1,2)$. This can be done since $\{t_k\}$ is countable and $(1,2)$ is uncountable.
We now claim that $\bar{t} \notin A$. Indeed, to prove this, we use the continuity-based characterisation of $A$ and therefore must prove that the path corresponding to $\omega = \bar{t}$ given by $t \to X_t(\bar{t})$ is not continuous on $[0,2)$.
This is clear since it's not continuous at the point $\bar{t} \in (0,2)$ : $X_{\bar{t}}(\bar{t}) = 1$ but $X_t(\bar{t}) = 0$ for $t \neq \bar{t}$.
Once we know that $\bar{t} \notin A$, we use the second characterisation of $A$ which tells us that $(X_{t_1}(\bar{t}),X_{t_2}(\bar{t}), \ldots) \notin B$. However, that tuple is equal to $(0,0,\ldots)$ since $t_i \neq \bar{t}$ for all $i$. Therefore , $(0,0,\ldots ) \notin B$.
However, let $\omega \in [0,1]$ be any sample space element. We can show that $\omega \in A$ and $\omega \notin A$ using both characterisations.
Indeed, using the continuity characterisation, $X_t(\omega) = 0$ for all $t$, so obviously $X_t$ is continuous on $[0,2)$.
However, using the second characterisation, $(X_{t_1}(\omega),X_{t_2}(\omega),\ldots) = (0,0,\ldots)$, which we know isn't in $B$. So $\omega \notin A$! Which is true for every $\omega \in [0,1]$.
That's obviously a contradiction : the way Karatzas-Shreve state it is that according the the first characterisation, $[0,1] \subset A$ so $P(A) = 1$, but according to the second characterisation, $[0,1] \cap A = \emptyset$ so $P(A) = 0$. This means that $A$ can't be in $\mathscr F$.
The only way out of this contradiction is that $A \notin \mathscr F^X_{2}$, completing the proof.
So, how exactly was this example come up with?
Well, $\mathscr F^X_{t_0}$ essentially consists of all countably $X$-determinable subsets of $\mathscr F$, till time $t_0$. That is, for every $A$ in $\mathscr F^X_{t_0}$, there should be countably many indices $t_1,t_2,... \leq t_0$ such that membership of a sample space element in $A$ depends entirely upon the values of $X_{t_1},X_{t_2},\ldots$ at that point.
In this example, what occurs is that the set of sample elements with continuous paths $A$ is obviously $[0,1]$, but the reason why $A$ doesn't belong to $\mathscr F^X_{2}$ is because if you take an element outside $A$, say $\omega$, then showing that $\omega \notin A$ requires us to look exactly at the time $X_{\omega}$. That is, to determine that the elements outside $A$ don't belong in $A$, you have to end up looking at all the time points in $(1,2)$, which means that $A$ can't be countably determined!
Furthermore, the set of points playing truant are all coming from a set of probability $0$ : because $(1,2)$ has probability zero. Therefore, it's a typical example of a null set playing havoc : the point is, the continuity behaviour of sample paths on this null set cannot be captured by countably many points.
That's why controlling null sets is so important : and the condition of putting all $P$-null sets in $\mathscr F^X_{t_0}$ essentially controls those misbehaving null sets.
Best Answer
Fubini's theorem is enough, since $f(s,\omega)$ belongs to $L^1([0,t] \times \Omega, \mathcal{B}[0,t] \otimes \mathcal{F}_t)$, plus $\int_{0}^{t}f(s,\omega)ds$ exists for all $\omega$ thus it's a well-defined random variable. Fubini's theorem says it should be in $L^1(\Omega, \mathcal{F}_t,P)$.