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

Consider a sample space $\Omega$ and a sigma-algebra $F$ over it. Let $A$ be a class of $F$-sets. $A$ then generates a partition of $\Omega$ based on the following equivalence relation: Points $w_{1}$ and $w_{2}$ of $\Omega$ are related if, for every set $B$ in $A$, they both belong or they both don't belong to $B$ [ref: Billingsley, 3rd edition, Chapter 1, Section 4, Subfields]. Further, this partition is the same as that generated by the sigma-algebra of $A$ [$\sigma(A)$] defined using the same relation.

Q1. Does it mean that $\sigma(A)$ gives exactly the same information that $A$ gives? In other words, is knowing either $A$ or its sigma-algebra enough to extract any information that we could possibly seek?

Q2. Is there any condition for the above to hold? For example, if $A$ is not a pi-system (not closed under finite intersections) does it hold?

Q3. Can independence of 2 classes $C$ and $D$ (or the lack of it) be inferred by studying their sigma-algebras? For example, if a set in $D$ lies in $\sigma(C)$, is it enough to ensure that $C$ and $D$ are not independent? If so, why?

share|cite|improve this question

Q3. Independence is related to what probability measure you are using. So I'd say it cannot be inferred from the sigma-algebras. Regarding your example $\Omega$ is always in $\sigma(C)$, but it can also be in an independent sigma algebra.

share|cite|improve this answer

Q1: The answer is no. Let $\Omega=[0,1]$. Let $A_1$ be the Borel $\sigma$-algebra on $\Omega$ and let $A_2$ be the $\sigma$-algebra consisting of subsets of $\Omega$ that are countable or have a countable complement. Both lead to the partition into singletons, yet are quite different. In particular, a $A_2$-measurable real valued function is constant on a set with Lebesgue measure $1$. To see this, let $f:\Omega\to\mathbb{R}$ be $A_2$-measurable. Note that for each $n$, there is a set of the form $[z/2^n,(z+1)/2^n]$ for some $z\in\mathbb{Z}$ such that $\mu f^{-1}\big([z/2^n,(z+1)/2^n]\big)=1$. Picking such a set for each $n$ and taking their countable intersection, you get a singleton set $\{x\}$ such that $\mu f^{-1}(\{x\})=1$.

Q2: There is a class of measurable spaces for which the answer is almost yes. A countably generated measurable space in which all points are measurable is strongly Blackwell if any two countably generated sub-$\sigma$-algebras that generate the same partition are equal. David Blackwell has shown (corollary 1) that any analytic set is strongly Blackwell.

share|cite|improve this answer

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.