Let $(\Omega, \mathcal{F}, {\bf P})$ be a probability space. Let $A, B \in \mathcal{F}$ be two events and $Z$ a random variable with values in a nice space $(S, \mathcal{S})$. Suppose that $A$ and $B$ are conditionally independent given $Z$ (i.e., independent of the $\sigma$-algebra generated by $Z$). It is known that $A$ and $B$ are not necessarily independent.
However, it seems possible to show that if $A$ and $B$ are themself independent of $Z$, then $A$ and $B$ are also independent. Is the proof below correct ? If yes could you give some intuition of why ? Thanks a lot !
${\bf Proof:}$ Denote by ${\bf P}( . | Z = z)$ the regular conditional probability of ${\bf P}$ given $Z$ and by ${\bf P}(Z \in dz)$ the law of $Z$. We have $${\bf P}(A \cap B) = {\bf P}((A \cap B) \cap \{Z \in S\}) = \int_S{\bf P}(A \cap B | Z = z){\bf P}(Z \in dz)$$ $$= \int_S{\bf P}(A | Z = z){\bf P}(B | Z = z){\bf P}(Z \in dz)$$ $$= \int_S{\bf P}(A){\bf P}(B){\bf P}(Z \in dz) = {\bf P}(A){\bf P}(B).$$