# Probability of two events are indepedent

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?

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I'd say it's not obvious that $\mathscr J$ is measurable. I can't think of a way of proving it for a general space.

I can think of two examples where it's definitely measurable.

First if $\mathscr B$ is a finite $\sigma$-algebra. But in this case the image of $\mathscr B^2$ would be a finite set of points, hence $\mu\left(\mathscr J\right)\leq \mu\left(\mathscr B^2\right) = 0$.

Secondly if $\Omega = \mathbb N \setminus\{0\}$ and $\mathbb P$ the distribution of a shifted geometric random variable mean $2$ (so $\mathbb P(N=n) = 2^{-n}$. Then an event may be interpreted as a binary sequence and $\sigma$ is the uniform map if the sequence is interpreted as a binary sequence. The $n$th digit of $E$ an event $E$ is $1$ if $n\in E$.

So $\mathscr J$ is the set pairs of binary strings such that $xy = x\cap y$, where $\cap$ is the intersection of the events considered as strings. I'm pretty sure this would be probability $0$ as well.

There's a chance that it's always $0$, I can't think of how you'd prove that off hand. Try finding more cases where you can show the map is measurable, see if you can gain some intuition about this map.

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Thank you for your answer and apologize for untenable guess made in the problem. – Metta World Peace May 1 '13 at 4:31
Never apologise for a wrong guess. If you always guess right it's because you're not thinking about anything interesting. – Tim May 1 '13 at 5:41