On non-existence of two independent random experiments on a same probability space I have a question on Probability as follows:
"Let $(\Omega,\mathcal{F},P)$ be a probability space. Then, on $\Omega$ we cannot construct two independent random experiments in which each experiment is that we select randomly an element $\omega$ from $\Omega$ with replacement".
In below, Michael Hardy gave an answer for the problem. It's very difficult to understand this answer. Can someone help me to explain the answer in detail.
 A: In one trivial case, the thing can be done.  That is the case where one member of $\Omega$ has probability $1$ and all others have probability $0$.  Otherwise it is impossible when $\Omega$ is finite.
One can construct such "experiments" on a product space whose underlying set is the product $\Omega\times\Omega$ and the probability measure is a product measure.  One randomly chooses a point $(\omega_1,\omega_2)\in\Omega\times\Omega$, distributed according to the product measure, and then the projection functions
$$
(\omega_1,\omega_2)\mapsto\omega_1
$$
and
$$
(\omega_1,\omega_2)\mapsto\omega_2
$$
are independent.
But the problem is to choose a single $\omega\in\Omega$, distributed according to $P$, and then find two functions
$$
\omega\mapsto\omega_1\in\Omega\text{ and } \omega\mapsto\omega_2\in\Omega
$$
that are independent.  Now if the space is $\{1,2,3,4,5,6\}$ and $P$ is the uniform distribution, then one can indeed construct independent random variables, for example
\begin{align}
& \omega\mapsto\text{even or odd, as the case may be}, \\
& \omega\mapsto\begin{cases}
\text{small} & \text{if }\omega\in\{1,2\}, \\
\text{medium} & \text{if }\omega\in\{3,4\}, \\
\text{large} & \text{if }\omega\in\{5,6\}.
\end{cases}
\end{align}
But in this case the value of each random variable is not a member of $\Omega$.  You would need to find a way to roll the die once and map the outcome, just one number, to one of the $36$ pairs of numbers in the set $\{1,2,3,4,5,6\}$ in such a way that the two components of the pair are independent.  Each pair would have probability $1/36$.  This is clearly impossible when $\Omega$ is finite.
The question is whether there is a mapping $\Omega\to\Omega\times\Omega$ for which the two projections are independent according to the measure $P$ on the domain.
In the case of the uniform distribution on the interval $[0,1]$ I'm not sure it can't be done by means of a Peano curve.
To show that the same is true of every finite probability space that assigns probability strictly between $0$ and $1$ to at least one point, consider the one $\omega_0\in\Omega$ with the smallest positive probability, and call that smallest positive probability $p$.  The map $\omega\mapsto(\omega_1,\omega_2)$ would have to take some element to the pair $(\omega_0,\omega_0)$, which should have probabilty $p^2$.  But there is no one element whose probability is $p^2$.
Mercio's comment below gives an example of one case in which such an experiment exists when $\Omega$ is infinite.
