Prove that two measurable maps X and Y are not independent? Let $(\Omega,\mathcal{F},P)$ be a probability space, where these elements of $\Omega$ are $\omega_1,\ldots, \omega_n$ and there exists an $\omega\in\Omega$ such that $0<P(\omega)<1$. We pick randomly an element from $\Omega$ with replacement. After that, we pick randomly an element from $\Omega$. So, we have two random experiments (or trials), are denoted $T_1$, $T_2$, respectively. We see each experiment as a measurable map from $\Omega$ to $\Omega$. Thus we get two measurable maps $X$, $Y$. Two experiments $T_1$, $T_2$ are independent if two maps $X:\Omega\to\Omega$, $Y:\Omega\to\Omega$ are independent (definition). 
Prove that in the above situation, $X$ and $Y$ are not independent. I stuck this problem. Please help me. Thank so much for helping.
 A: An incomplete answer, perhaps a hint
$(\Omega,\mathcal{F},P)$ is not the model of the experiment you've described above. The two random variables you described both are functions on the same $\Omega$ and cannot be "independently" defined on pairs of $\omega$'s.
The remark above is not a proof that one cannot create two independent random variables the way the OP suggested.
Here is a failed attempt:
In general:  $$P(X=\omega_j,Y=\omega_k)=\begin{cases}
0 \text{ if there is no } \omega \in \Omega \text{ for which } X(\omega)=\omega_j \text{ and } Y(\omega)=\omega_k\\
\sum_{ \{\omega :\ X(\omega)=\omega_j,\ Y(\omega)=\omega_k\}}P(\omega),  \text{ otherwise,}
\end{cases}$$
$$P(X=\omega_j)=\begin{cases}
0 \text{ if there is no } \omega \in \Omega \text{ for which } X(\omega)=\omega_j \\
\sum_{ \{\omega \ :\ X(\omega)=\omega_j \} }P(\omega),  \text{ otherwise.}
\end{cases},$$
$$P(Y=\omega_k)=\begin{cases}
0 \text{ if there is no } \omega \in \Omega \text{ for which } X(\omega)=\omega_j \\
\sum_{ \{\omega \ :\ Y(\omega)=\omega_k \} }P(\omega),  \text{ otherwise.}
\end{cases},$$
Thus
$$P(X=\omega_j)P(Y=\omega_k)$$
is either $0$ or 
$$\left(\sum_{ \{u \ :\ X(u))=\omega_j \} }P(u)\right)\left(\sum_{ \{v :\ Y(v)=\omega_k\}}P(v)  \right)=$$
$$=\sum_{ \{u \ :\ X(u))=\omega_j \} }\sum_{ \{v :\ Y(v)=\omega_k\}}P(u)P(v)$$
as opposed to 
$$\sum_{ \{\omega :\ X(\omega)=\omega_j,\ Y(\omega)=\omega_k\}}P(\omega).$$
Unfortunately the argumentation above is not enough. One still has to show that  $P$ cannot be defined so that 
$$\sum_{ \{\omega :\ X(\omega)=\omega_j,\ Y(\omega)=\omega_k\}}P(\omega)=\sum_{ \{u \ :\ X(u))=\omega_j \} }\sum_{ \{v :\ Y(v)=\omega_k\}}P(u)P(v)$$
for all $\omega_j, \omega_k$.
Here is an example: Let $\Omega=\{1,2,3,4\}$ and let $P(i)=\frac{1}{4}$ for all $i$. Let $$X,Y=\begin{cases}1,2\text{ if }\omega=1\\
1,1\text{ if }\omega=2\\
2,2\text{ if }\omega=3\\
3,3\text{ if }\omega=4.\\
\end{cases} $$
Now,
$$P(X=1,Y=2)=P(1)=\frac{1}{4}=P(X=1)P(Y=2)$$
$$=(P(1)+P(2))(P(1)+P(3))=\frac{1}{4}.$$
(At the same time
$$P(X=1,Y=1)=P(2)=\frac{1}{4}\not=P(X=1)P(Y=1)$$
$$=(P(1)+P(2))P(2)=\frac{1}{8}.$$
$X,Y$ are not independent. Also, my argumentation is not complete.)
