Questions about product measure 
*

*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?


*

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!
 A: 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 $({\bf (S)}\prod_{k \in N}\mu_k)(D)=1.$ 
A: 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).  
