The measurable projection theorem (see the George Lowther blog) asserts the following.
Theorem. If $(\Omega,\mathcal{F},\mathbb{P})$ is a complete probability space and $A\in\mathcal{B}(\mathbb{R})\otimes\mathcal{F}$ then $\pi_{\Omega}(A)\in\mathcal{F}$, where $\pi_{\Omega}:\mathbb{R}\times\Omega\rightarrow\Omega$ is defined as $\pi_{\Omega}(x,\omega)=\omega$.
In the same post, the author uses the theorem to prove this version of the Debut theorem:
Theorem. Let $A\subseteq [0,\infty)\times\Omega$ be a progressively measurable set and let the filtration be right-continuous. Then its debut time $$ T_A(\omega) = \inf\{t\in[0,\infty) | (t,\omega)\in A \} $$ is a stopping time.
The proof proceeds in this way. First note that $$ \{ T_A<t \} = \pi_{\Omega}\left([0,t)\times\Omega \bigcap A\right), $$ which is easy to be verified, for example with a graphical representation. By the measurable projection theorem, being, by progressive measurability, $A\in \mathcal{B}([0,t])\otimes\mathcal{F}_t$, we have that the projection $\{T_A<t\}\in\mathcal{F}_t$, so $T_A$ is optional. Having assumed the right-continuity of the filtration this is enough to guarantee that $\{T_A\leq t\}\in\mathcal{F}_t$, so $T_A$ is a stopping time.
My problem is, couldn't we just say that
$$ \{ T_A\leq t \} = \pi_{\Omega}\left([0,t]\times\Omega \bigcap A\right)\in\mathcal{F}_t $$
and remove the hypothesis of right-continuity?
ps = I guess that in the Debut theorem we also have to assume that the filtration $\mathcal{F}_t$ is complete for all $t$, otherwise the measurable projection theorem cannot be used.
I think I figured out the problem. I'll explain with a simple diagram. Consider the example in the figure below
The set A is represented in the simplified case $\Omega=\mathbb{R}$. The dotted line indicates that the corresponding boundary is not included. The graph of $T_A$ is represented by the shaded light blue line. It is clearly false that $\{ T_A\leq t \} = \pi_{\Omega}\left([0,t]\times\Omega \bigcap A\right)$, this is why the set $\{ T_A\leq t \}$ includes the $\omega$ for which $T_A(\omega)=t$, which is clearly not included in $\pi_{\Omega}\left([0,t]\times\Omega \bigcap A\right)$, being the lower boundary (again, the dotted line) of $A$ not included in $A$ and, accordingly, the intersection between the horizontal line $y=t$ with $A$ is empty (a part the unique intersection on the left side), and so $\pi_{\Omega}\left([0,t]\times\Omega \bigcap A\right)=\pi_{\Omega}\left([0,t)\times\Omega \bigcap A\right)$ (the latter is represented as a shaded green area).