Confusion on Random Variables, Ω vs X(Ω) I am having a little confusion on random variables. Namely, the difference between X(Ω) and Ω itself. I will use the dice roll as an example.
So let X be the values the dice can take. Ω={1,2,3,4,5,6}, the sample space of values. According to this Wikipedia page: https://en.wikipedia.org/wiki/Expected_value, the possible values of X are 
X(Ω)={1,2,3,4,5,6}. So is X just a kind of identity random variable and X(Ω)=Ω? 
This question arose because in the definition of expectation in the discrete case. That is, summing over the values of x*P(X=x) for each x in X(Ω). However I thought you were meant to be summing over the values of the sample space, not X(Ω).
Is it because X can be anything we want, including "The values the dice can take multiplied by 2", yet Ω would stay the same, but X(Ω)={2,4,6,8,10,12}?
 A: You've decided to model the "dice rolls" by naming the number that comes up on the top of the die. You could, instead, consider all possible positions of a die on the table (the $xy$ location, the rotation of the die on the table, the rotation of the die from its original position (i.e., the thing that determine which face is "up"). That'd give you, for $\Omega$, something like $\Bbb R^2 \times [0, 2\pi) \times Q$, where the set $Q \subset  SO(3)$ consists of all rotations formed by products of rotations of 90 degrees about the three principal axes. 
Now you can see that in this case, $X$ is much more complicated than the identity. 
Perhaps as a simpler example, you could decide to identify each die-roll with the number of pips on the face that's hidden. Then it turns out that $X(\omega) = 7-\omega$. 
As for your question about expectation: you can form two sums. The first
is 
$$
A = \sum_{\omega \in \Omega} p_\omega X(\omega)
$$
the second is 
$$
B = \sum_{x \in \Bbb R} x * P(X = x)
$$
These end up being the same. For $P(X = x)$ is the sum, over all $\omega \in \Omega$ with $X(\omega) = x$, of $p_\omega$. When you expand this out for all possible $x$ that are in the image of $X$, you end up summing over all $\omega \in \Omega.$ 
Let me illustrate with an example. Suppose you play a game where you roll a (fair) die, and if it comes up odd (1, 3, or 5) you win 7 dollars; if it comes up even, you win 13 dollars. (You should definitely play this game!). 
Then we can model the die-rolls as $\Omega = \{1, 2, 3, 4, 5, 6 \}$, and the probabilities with 
$p(1) = p(2) = \ldots = p(6) = \frac{1}{6}$. The random variable, $X$, representing your winnings, is defined by 
$$
X(1) = X(3) = X(5) = 7\\
X(2) = X(4) = X(6) = 13.
$$
Let's compute sum $A$: 
\begin{align}
A &= \sum_{\omega \in \Omega} p_\omega X(\omega)\\
 &= p(1)X(1) + p(2)X(2) + p(3) X(3) + p(4) X(4) + p(5)X(5) + p(6) X(6)\\
 &= p(1)\cdot 7 + p(2)\cdot 13 + p(3) \cdot 7 + p(4) \cdot 13 + p(5)\cdot 7 + p(6) \cdot 13
\end{align}
We could write in $\frac{1}{6}$ for each of the probabilities, but I'm going to skip that step. 
Now let's compute $B$:
\begin{align}
B 
&= \sum_{x \in \Bbb R} x * P(X = x)\\
&= \sum_{x = 7, 13} x * P(X = x)\\
&= 7 \cdot P(X = 7) + 13 \cdot P(X = 13)\\
&= 7 \cdot P(\{1, 3, 5\}) + 13 \cdot P(\{2, 4, 6\})\\
&= 7 \cdot (p(1) + p(3) + p(5)) + 13 \cdot (p(2)+ p(4) + p(6))\\
&= 7 \cdot p(1)\cdot 7 + 13 \cdot p(2) + 7 \cdot p(3) + 13 \cdot p(4)  + 7 \cdot p(5) + 13 \cdot p(6)\end{align}
which is evidently the same as $A$. 
