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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, $\mu_{i}(E_{i})<\infty$.
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 $\mu_{1}(E_{1})<\infty$ and $\mu_{2}(E_{2})=\infty$. Then $\mu(B)$ for
$B:=B_{1}\times\prod_{{i>1}}E_{i}=B_{1}\times E_{2}\times\prod_{{i>2}}E_{i}% \mbox{, where }B_{1}\in\mathcal{B}_{1}$ would not be welldefined (on the one hand, it is $\mu_{1}(B_{1})<\infty$, but on the other it is $\mu_{1}(B_{1})\mu_{2}(E_{2})=\infty$).

The above construction agrees with the result when $I$ is finite (see finite product measure).
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