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? – Stefan Feb 14 '11 at 15:50
@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? – Tim Feb 14 '11 at 17:32
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. – Stefan Feb 14 '11 at 18:27
@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. – t.b. Feb 14 '11 at 18:56

Let $(\beta_i)_{i \in I} \in [0, +\infty]^{I}$ be an arbitrary family of non-negative real numbers. We set $ln(0)=-\infty$.

Definition 1. A standard product of the family of numbers $(\beta_i)_{i \in I}$ denoted by ${\bf (S)}\prod_{i \in I}\beta_i$ is defined as follows:

~${\bf (S)}\prod_{i \in I}\beta_i=0$ if ~$\sum_{i \in I^{-}}\ln(\beta_i)=-\infty$, where $I^{-}=\{i:ln(\beta_i)<0\}$ , and ${\bf (S)}\prod_{i \in I}\beta_i=e^{\sum_{i \in I}\ln(\beta_i)}$ if $\sum_{i \in I^{-}}\ln(\beta_i) \neq -\infty$.

Now we will try to give answers to Stefan's questions when a set of induces $I$ is countable.

Question 1(Stefan). Is product measure only defined for probability measures?

Let $(E_i,\mathbb{B}_i,u_i)_{i \in I}$ be a family of totally finite, continuous measures.

Theoretically there are possible the following three cases:

Case 1. ${\bf (S)}\prod_{i \in I}u_i(E_i)=0.$

In that case we define $\prod_{i \in I}u_i$ as zero measure, i.e. $$(\forall X)(X \in \prod_{i \in I}\mathbb{B}(E_i) \rightarrow (\prod_{i \in I}u_i)(X)=0).$$

Case 2. $0< {\bf (S)}\prod_{i \in I}u_i(E_i)< +\infty.$

In that case we define $\prod_{i \in I}u_i$ as follows:

$$(\forall X)(X \in \prod_{i \in I}\mathbb{B}(E_i) \rightarrow (\prod_{i \in I}u_i)(X)= ({\bf (S)}\prod_{i \in I}u_i(E_i))\times (\prod_{i \in I}\frac{u_i}{u_i(E_i)})(X)).$$

Case 3. ${\bf (S)}\prod_{i \in I}u_i(E_i)= +\infty.$

In that case we define $\prod_{i \in I}u_i$ as a standard product of measures $(u_i)_{i \in I}$ construction of which is given below:

Without loss of generality, we can assume that $u_i(E_i)\ge 1$ when $i \in I$.

Let $L$ be a set of rectangles $R:=\prod_{i \in I}R_i$ where $R_i \in \mathbb{B}(E_i)(i \in I)$ and $0\le{\bf (S)}\prod_{i \in I}u_i(R_i)<+\infty$

Note that a rectangle $R$ with $0<{\bf (S)}\prod_{i \in I}u_i(R_i)<+\infty$ exists because $u_i$ is continuous and $u_i(E_i)\ge 1$.

Let $\mu_R$ be a measure defined on $\prod_{i \in I}\mathbb{B}(R_i)$ as follows

$$(\forall X)(X \in \prod_{i \in I}\mathbb{B}(R_i) \rightarrow \mu_R(X)= ({\bf (S)}\prod_{i \in I}u_i(R_i))\times (\prod_{i \in I}\frac{u_i}{u_i(R_i)})(X)).$$ For each $R \in L$ we have a measure space $(R,S_R(:=\prod_{i \in I}\mathbb{B}(R_i)),\mu_R)$. That family is consistent in the following sense: if $R=R_1 \cap R_2$ then $$(\forall X)(X \in S_R \rightarrow \mu_R(X)=\mu_{R_1}(X)=\mu_{R_2}(X)).$$

If a measurable subset $X$ of $\prod_{i \in I}E_i$ is covered by a family $\{R_k : R_k \in L ~\&~k=1,2, \cdots\}$ then we set $$\Lambda(X)=\mu_{R_1}(R_1 \cap X)+\mu_{R_2}((R_2\setminus R_1)\cap X)+\cdots+\mu_{R_n}([R_n\setminus \cup_{1 \le i \le n-1}R_i]\cap X)+ \cdots.$$ If a measurable subset $X$ of $\prod_{i \in I}E_i$ is not covered by a countable family of elements of $L$, then we set $\Lambda(X)=+\infty$.

Note that $\Lambda$ is measure on $\prod_{i \in I}\mathbb{B}(E_i)$ and $\Lambda(R)={\bf (S)}\prod_{i \in I}u_i(R_i)$ for each $R \in L$.

This measure is called standard product of measures $(u_i)_{i \in I}$ and is denoted by ${\bf (S)}\prod_{i \in I}u_i$.

As we see product can be defined for totally finite continuous measures. Here we need no a requirement of totally finiteness(they may be infinite (i.e. $u_i(E_i)=+\infty$) as well we do not require their sigma-finiteness.

I think that it gives a partially solution of that problem when $card(I)=\aleph_0$ and the measure ${\bf (S)}\prod_{i \in I}u_i$ is well defined on $\prod_{i \in I}E_i.$

P.S. I agree with Mister Stefan Walter remark that there may be a situation when product measures are not defined uniquelly.

Indeed, let $(n_k)_{k \in N}$ be a family of strictly increasing natural numbers such that $n_0=0$ and $n_{k+1}-n_k \ge 2$. We set $\mu_k=\prod_{i \in [n_k,n_{k+1}]}u_i$. Let us consider ${\bf (S)}\prod_{k \in N}\mu_k$. Then that measure will be defined on $\prod_{i \in I}\mathbb{B}(E_i)$ and $({\bf (S)}\prod_{k \in N}\mu_k)(R)={\bf (S)}\prod_{i \in I}u_i(R_i)$ for all $R \in L^{+}$, where
$$L^{+}=\{ R:R \in L~\&~0<{\bf (S)}\prod_{i \in I}u_i(R_i)<+\infty\}$$

Note that the measure ${\bf (S)}\prod_{k \in N}\mu_k$ is called $(n_{k+1}-n_k)_{k \in N}$-standard product of measures $(u_i)_{i \in I}$.

It is natural that both measures $({\bf (S)}\prod_{i \in I}u_i)$ and $({\bf (S)}\prod_{k \in N}\mu_k)$ can be considered as products of measures $(u_i)_{i \in I}$ but they(in general) are different.

Indeed, let $u_i=l_1$ for $i \in I$, where $l_1$ denotes a linear Lebesgue measure on real axis. Let $n_{k+1}-n_k=2$ for $k \in N$. Consider a set $D$ defined by $$D=[0,2]\times [0,\frac{1}{2}]\times [0,3]\times [0,\frac{1}{3}]\times \cdots.$$ Then $({\bf (S)}\prod_{i \in I}u_i)(D)=0$ and $({\bf (S)}\prod_{k \in N}\mu_k)(D)=1.$

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Thanks, @Gogi! +1. I will get back to this thread. – Tim Dec 31 '12 at 12:35

You can found answers on yours questions in the following articles:

[1] G.Pantsulaia , On ordinary and standard products of infinite family of $\sigma$ -finite measures and some of their applications. Acta Math. Sin. (Engl. Ser.) 27 (2011), no. 3, 477--496

[2] G.Pantsulaia , On Strict Standard and Strict Ordinary Products of Measures and Some of Their Applications, Georg. Inter. J. Sci. Tech., Nova Science Publishers, Volume 2, Issue 3 (2010).

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While the references may be helpful, it's better form to give at least some of the answer in a self-contained way. For instance, sometimes posters state results and leave the proof to a reference. Journal articles are expensive, and many MSE members are not associated with universities. – Kevin Carlson Oct 23 '12 at 20:20
I downvoted. This answer does little more than advertising. – Giuseppe Negro Dec 8 '12 at 4:27