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

Let $i\in\{1,2\}$. The Measure Product Theorem states that, given the measure spaces $(X_i,\Sigma_i,\mu_i)$, there is at least one product measure $\pi$ such that $\pi(A_1\times A_2)=\mu_1(A_1)\;\mu_2(A_2)$, for $A_i\in\Sigma_i$.

It also states that if the $\mu_i$'s are $\sigma$-finite, then $\pi$ is unique.

I'd like an example of a non-$\sigma$-finite pair of measures from which, nevertheless, we only obtain one product, thus showing "if" cannot be "iff" on the paragraph above.

share|cite|improve this question
@DavideGiraudo Hmmm, you're right. I had actually found that question before, but not read it. I thought it was a "counter-counter-example", but it's about a counter. I will edit my question and ask about a double counter instead. – Luke Nov 20 '11 at 19:34
@All: This is no longer a duplicate. – t.b. Nov 20 '11 at 19:50
@t.b. You're right, I voted to close, but now I can't remove my vote. – Davide Giraudo Nov 20 '11 at 19:55
@Davide: You can't (I voted to close, too, before the question was changed). You could remove your "possible duplicate" comment, though (in the hope that people no longer vote for closure). – t.b. Nov 20 '11 at 20:00
up vote 5 down vote accepted

Trivial example: $X = \{x\}$ has one point, $\mu(\{x\}) = \infty$, $X_1 = X_2 = X$, and $\mu_1 = \mu_2 = \mu$. Not $\sigma$-finite, but the unique product measure is $\lambda(\{(x,x)\}) = \infty$.

share|cite|improve this answer
+1 And I was thinking my example was a cheat :) – t.b. Nov 20 '11 at 20:12

Take an uncountable set $X$ with counting measure defined on the power set and consider $X \times X$. Then the usual product measure and the complete locally determined product measure coincide with counting measure on the product and we conclude by the result that any measure $\lambda$ on the product $\sigma$-algebra satisfying $\lambda(A \times B) = \mu(A) \nu(B)$ for all sets $A$ and $B$ of finite measure must lie between the complete locally determined product and the usual product, as was explained in this answer.

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.