Given a probability space $(\Omega, \mathscr {B}, P)$, then $\sigma : \mathscr{B}\times \mathscr{B} \to [0,1]^2$ is defined as, for any $A, B \in \mathscr{B}$ $$(A,B) \mapsto (P(A),P(B))$$ Now take $[0,1]^2$ as a measure space endowed with Lebesgue measure $\lambda$ , which induces a Borel $\sigma$-algebra on $2^{\mathscr{B} \times \mathscr{B}}$ and a measure $\mu$. For any $\mathscr{K} \subseteq \mathscr{B} \times \mathscr{B}$, $\mathscr{K}$ is Borel measurable, iff $\sigma(\mathscr{K}) $ is Borel measurable in $[0,1]^2$ with $\mu(\mathscr{K}) = \lambda(\sigma(\mathscr{K}))$.
Consider, an event $\mathscr{I}$ which consists of all pairs of events that are independent, which is, for any $C, D \in \mathscr{B}$, $(C,D) \in \mathscr{I}$, iff $P(C\cap D) = P(C)P(D)$.
My question is what is the probability of $\mathscr{I}$, $\mu(\mathscr{I})$?It seems to me $\mu(\mathscr{I}) = 1$, but I don't know how to show it.
Added: As pointed out by Tim, it's not obvious that $\mathscr{I}$ is measurable. Is it true?