Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.