# Product measure and independent measure

We have a product space $X= \prod_{s \in S} X_i$, where the spaces $X_i$ are finite and all the same, $S$ is finite or infinite countable.

I am reading a book where the following definition is provided: "We term a measure on a product space $X= \prod_{i \in S} X_i$ Bernoulli or independent if it equals the product of its projections onto single $X_s$ sets." (in the contest of the book we are speaking about Probability Measures.)

I have the following questions:

1) Does this definition of independent measure is equivalent to the definition of product measure?

2) Is the projection of a measure $\mu$ on $X_s$ defined as $\mu_s =\mu \circ \pi_s^{-1}$, where $\pi_s$ is the projection $\pi_s : X \longmapsto X_s$?

3) Define the function $f : X \mapsto \mathcal{R}$ in the following way: $f= 1$ if $x_h =x_s=1$, $f=0$ otherwise. If $\mu$ is an independent measure, are the following equalities all true? $$\mu (f) = \mu(x_s=1 , x_h=1) = \mu (x_s=1)\mu (x_h=1) = \mu_s(x_s=1)\mu_h(x_h=1).$$ With $\mu(f)$ I mean $\mu(f) = \int f d\mu$, with $\mu(x_s=1 , x_h=1)$ I mean the probability of the cylinder set specified by $x_s=x_h=1$, with $\mu(x_i=1)$ I mean the probability of the cylinder set specified by the condition $x_i=1$.

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Of course the first subscript of X is s.... a mistake... –  Lorenzo Sep 28 '12 at 19:56
try asking on stackexchange... –  Anthony Quas Sep 28 '12 at 20:17
answers: yes, yes, yes. –  Pietro Majer Sep 28 '12 at 21:38