Problem: $(X_i, M_i)$ be measurable spaces. Then $M_1 \otimes M_2 \otimes M_3 = M_1 \otimes (M_2 \otimes M_3)$ under the set identification.
I was wondering why none gave the following proof. I believe this would be a neater proof (if correct). I think Timo (linked above) has implicitly assumed the projection map is measurable.
Proof: It suffices to show any measurable set in $M_1 \otimes (M_2 \otimes M_3)$ is measurable in $M_1 \otimes M_2 \otimes M_3$, i.e. the map $$ id: (X:=)X_1 \times X_2 \times X_3 \rightarrow X_1 \times (X_2 \times X_3)$$ is measurable.
This is measurable, iff it is measurable under composition by $\pi_1:X \rightarrow X_1, \pi_{2,3}:X \rightarrow X_2 \times X_3$ which are measurable iff $\pi_1:X \rightarrow X_1$, $\pi_2: X \rightarrow X_2$ and $\pi_3:X \rightarrow X_3$ are measurable.
This is the exact definition of product measure.
Is this proof right?