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  1. On Planetmath, product measure is roughly defined as follows:

    Let $(E_i, \mathbb{B}_i, u_i)$ be measure spaces, where $i\in I$ an index set, possibly infinite.

    When each $u_i$ is totally finite, there is a unique measure on the product measurable space $(E, \mathbb{B})$ of $(E_i, \mathbb{B}_i, u_i)$. such that "taking measure" and "taking product" can be "exchanged" for any $B=\prod_{i \in I} B_i$ with $B_i \in \mathbb{B_i}$ and $B_i=E_i$ for all $i \in I$ except on a finite subset $J$ of $I$.

    I was wondering if there is also a unique measure on the product measurable space, such that "taking measure" and "taking product" can be "exchanged" for any $B=\prod_{i \in I} B_i$ with $B_i \in \mathbb{B_i}$, without requiring "$B_i=E_i$ for all $i \in I$ except on a finite subset $J$ of $I$"? This is not used in the definition of product measure, and is it only because the product might be for infinite number of terms?

  2. If $I$ is infinite, one sees that the total finiteness of $u_i$ can not be dropped. For example, if $I$ is the set of positive integers, assume $u_1(E_1) < \infty$ and $u_2(E_2)=\infty$ . Then $u(B)$ for $$B:=B_1 \times \prod_{i>1} E_i=B_1\times E_2 \times \prod_{i>2} E_i,$$ where $B_1 \in \mathbb{B}_1$ would not be well-defined (on the one hand, it is $u_1(B_1)<\infty$ , but on the other it is $u_1(B_1)u_2(E_2)=\infty$ ).

    I don't understand the example. Specifically how does the last sentence in parenthesis show that the measure $u$ is not well-defined on $B$?

Thanks and regards!

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    Don't you have to assume anyway that your measures are probability measures in order for this all but finitely many $E_i$ stuff to prevent infinite products?2011-02-14
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    @Stefan: (1) Is product measure only defined for probability measures? Can it be defined for totally finite measures? (2) By "infinite products", you mean product of infinite terms? Does the definition of product measure require to prevent "infinite products?2011-02-14
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    It's not clear how you define an infinite (maybe uncountable!) product of real numbers. However, it's no problem if all but finitely many of those numbers are equal to 1. If you look at the Planetmath article again, they do assume that each $u_i$ is a probability measure. They just pretend that this is some kind of "without loss of generality" assumption. As far as I can tell, this is nonsense. But on the other hand, in practice, once your measure is finite, you simply need to multiply it by a constant to get a probability measure which is in many respects very similar to the original measure.2011-02-14
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    @Stefan: I agree. @Tim: Note also that on planetmath (and in the article Pete has shown you) quite a few things are swept under the rug: ..it can be shown that there is a *unique* measure.. Yes, I think it's even true, but here some non-trivial facts enter (like the Carathéodory-Hahn extension and Vitali-Hahn-Saks uniqueness theorem) and it's not so clear a priori that all possible constructions will give the same resulting measure.2011-02-14

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