I found this proof of the Debut theorem and it's an amazing resource that is both short and straightforward (for almost every part of course, since I have a problem with it), but I am struggling tremendously to understand the inequality I've circled below. It appears to falsely use $\pi \left( \bigcap_i K_i \right) = \bigcap_i \pi (K_i)$. This is true for the way they define the $L_n$, i.e. $\pi \left( \bigcap_i L_i \right) = \bigcap_i \pi (L_i)$, but that inequality just makes absolutely no sense to me since this does not apply to the $K_n$.
Well the following is true $\pi(A\cap B)\subset \pi(A)\cap \pi(B)$ so as $\pi(\cap_{i=1}^n K_i)\subset \cap_{i=1}^n \pi(K_i)$ so
$ \cap_{i=1}^n \pi(K_i)^c\subset \pi(\cap_{i=1}^n K_i)^c$ and
$\pi(A_n) \cap \cap_{i=1}^n \pi(K_i)^c\subset \pi(A_n) \cap \pi(\cap_{i=1}^n K_i)^c$
But this is
$\cap_{i=1}^n\pi(A_n)\backslash \pi(K_i)\subset \pi(A_n) \backslash\pi(\cap_{i=1}^n K_i)$
So this cannot work because there is no specification in this proof as to how choose the $K_n$ so that
$\pi(A_n) \backslash\pi(\cap_{i=1}^n K_i)\subset \cap_{i=1}^n\pi(A_i)\backslash \pi(K_i)$
To do that is very hard and I let you look at the very readable but still hard to understand proof of measurable projection here where the point is to define a proper strategy to get the rights $K_n$ from a: https://almostsuremath.com/2019/01/10/proof-of-the-measurable-projection-and-section-theorems/