infinite product measure

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

1. let $E=\prod E_i$, the Cartesian product of $E_i$,

2. let $ℬ=\sigma ({({ℬ}_i)}_{i\in I})$, 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 $(E,ℬ)$ is a measurable space. The next task is to define a measure $\mu$ on $(E,ℬ)$ so that $(E,ℬ,\mu )$ 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 $(E_i,{ℬ}_i,{\mu }_i)$ 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 $(E_i,{ℬ}_i,{\mu }_i)$ is a probability space, it can be shown that there is a unique measure $\mu$ defined on $ℬ$ such that, for any $B\in ℬ$ expressible as a product of $B_i\in {ℬ}_i$ 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 $(E,ℬ,\mu )$ 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

  $$B:=B_1\times \prod _{i>1}{E}_i=B_1\times E_2\times \prod _{i>2}{E}_i\text{, where }{B}_1\in {ℬ}_1$$

  would not be well-defined (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)).