infinite product measure
Let $({E}_{i},{\mathcal{B}}_{i},{\mu}_{i})$ be measure spaces^{}, where $i\in I$ an index set^{}, possibly infinite^{}. We define the product^{} of $({E}_{i},{\mathcal{B}}_{i},{\mu}_{i})$ as follows:

1.
let $E=\prod {E}_{i}$, the Cartesian product of ${E}_{i}$,

2.
let $\mathcal{B}=\sigma ({({\mathcal{B}}_{i})}_{i\in I})$, the smallest sigma algebra containing subsets of $E$ of the form $\prod {B}_{i}$ where ${B}_{i}={E}_{i}$ for all but a finite number of $i\in I$.
Then $(E,\mathcal{B})$ is a measurable space^{}. The next task is to define a measure $\mu $ on $(E,\mathcal{B})$ so that $(E,\mathcal{B},\mu )$ becomes in addition a measure space. Before proceeding to define $\mu $, we make the assumption^{} that
each ${\mu}_{i}$ is a totally finite measure, that is, $$.
In fact, we can now turn each $({E}_{i},{\mathcal{B}}_{i},{\mu}_{i})$ into a probability space by introducing for each $i\in I$ a new measure:
$${\overline{\mu}}_{i}=\frac{{\mu}_{i}}{{\mu}_{i}({E}_{i})}.$$ 
With the assumption that each $({E}_{i},{\mathcal{B}}_{i},{\mu}_{i})$ is a probability space, it can be shown that there is a unique measure $\mu $ defined on $\mathcal{B}$ such that, for any $B\in \mathcal{B}$ expressible as a product of ${B}_{i}\in {\mathcal{B}}_{i}$ with ${B}_{i}={E}_{i}$ for all $i\in I$ except on a finite subset $J$ of $I$:
$$\mu (B)=\prod _{j\in J}{\mu}_{j}({B}_{j}).$$ 
Then $(E,\mathcal{B},\mu )$ becomes a measure space, and in particular, a probability space. $\mu $ is sometimes written $\prod {\mu}_{i}$.
Remarks.

•
If $I$ is infinite, one sees that the total finiteness of ${\mu}_{i}$ can not be dropped. For example, if $I$ is the set of positive integers, assume $$ and ${\mu}_{2}({E}_{2})=\mathrm{\infty}$. Then $\mu (B)$ for
$$B:={B}_{1}\times \prod _{i>1}{E}_{i}={B}_{1}\times {E}_{2}\times \prod _{i>2}{E}_{i}\text{, where}{B}_{1}\in {\mathcal{B}}_{1}$$ would not be welldefined (on the one hand, it is $$, but on the other it is ${\mu}_{1}({B}_{1}){\mu}_{2}({E}_{2})=\mathrm{\infty}$).

•
The above construction agrees with the result when $I$ is finite (see finite product measure^{} (http://planetmath.org/ProductMeasure)).
Title  infinite product measure 

Canonical name  InfiniteProductMeasure 
Date of creation  20130322 16:23:14 
Last modified on  20130322 16:23:14 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  13 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 28A35 
Classification  msc 60A10 
Related topic  ProductSigmaAlgebra 
Defines  totally finite measure 