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I am reading this answer: https://math.stackexchange.com/a/246469/1078713 and I don't understand it.

What is the domain of $X_0$,...,$X_n$?

Is $P(X_0 \in A_0,...,X_n \in A_n)=P(A_0,...,A_n)$? I think so because $P$ is defined on sets like $A_0$x...x$A_n$ and sets like {$X_0 \in A_0$} could be weird

What is set {$X_0 \in A_0$} (This should be clear when I understand two previous questions)

I think these $X_1$,...,$X_n$ are not constructed, only its distributions (via $P$) but I have to be wrong, but I don't see it.

Where can I read about this construction, where is it described more deeply?

romperextremeabuser
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The measurable space is $(E^n, \mathcal{E}^{\otimes n})$, and we define $$X_j(x) = x_j,$$ so $X_j$ is just the $j$th coordinate function. Then the measure that makes $X_1, \dots, X_n$ i.i.d. with distribution $\mu$ is $$\mu^{n}(A) = \int_{E}\dots\int_{E}1_A(x)\,\mu(dx_1) \dots \,\mu(dx_n).$$

Mason
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  • does this measure has something in common with product measure? – romperextremeabuser Oct 22 '23 at 18:28
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    @romperextremeabuser It is the product of $n$ copies of $\mu$. By the $\pi$-$\lambda$ theorem, any measure which assigns measure $\mu(A_1)\dots\mu(A_n)$ to $A_1 \times \dots \times A_n$ is the product of $n$ copies of $\mu$. – Mason Oct 22 '23 at 18:31
  • wikipedia says that product measure is unique – romperextremeabuser Oct 22 '23 at 18:33
  • So in the answer constructed measure $P$ happens to be product measure which is unique so we can do it backwards, take product measure and prove that $X_1$,...,$X_n$ are iid right?

    I imagine it would be harder but it is possible right? I don't know if I am getting it right.

     – romperextremeabuser Oct 22 '23 at 18:38
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    It's not backward. Existence of i.i.d. $X_1, \dots, X_n$ is equivalent to existence of the product measure of $n$ copies of $\mu$ since $P_{X_1, \dots, X_n} = \mu^n$ if and only if $X_1, \dots, X_n$ are i.i.d. with distribution $\mu$. – Mason Oct 22 '23 at 19:34
  • thank you, do you know how to prove that $P$ is pre-measure? I can't prove that if countable union of sets from algebra of finite union of measurable rectangles is in that algebra then $P$ is countably additive – romperextremeabuser Oct 22 '23 at 23:38
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    @romperextremeabuser You don't have to use Caratheodory's theorem to construct $P$. You can just define $P((X_1, \dots, X_n) \in A) = \int_{E}\dots\int_{E}1_A(x_1, \dots, x_n),\mu(dx_1) \dots \mu(dx_n)$ for $A \in \mathcal{E}^{\otimes n}$. – Mason Oct 24 '23 at 18:34
  • yeah I realized that after some time I posted my question, thank you – romperextremeabuser Oct 24 '23 at 19:58