6
$\begingroup$
  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!

  • 1
    @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

2 Answers 2

1

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

  • 0
    Thanks, @Gogi! +1. I will get back to this thread.2012-12-31
-3

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).

  • 3
    I downvoted. This answer does little more than advertising.2012-12-08