$X$ and $Y$ are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$? Can they have different CDFs? Suppose $X$ and $Y$ are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Is possible that $X$ and $Y$ have different CDF?
I know the answer is yes, I don't understand why.
I don't understand the link between the measure $\mathbb{P}$ and the CDFs $P_X(X \le x), P_Y(Y \le y) $
Intuitively what doest it mean for two RVs to be defined on the same probability space?
Could you provide an example?

If $X:\Omega\to\mathbb R$ is measurable in the sense that $X^{-1}(B)\in\mathcal F$ for every element $B$ of Borel $\sigma$-algebra $\mathcal B(\mathbb R)$ then its CDF $F_X$ is defined by:
$$F_X(x):=\mathbb P(\{\omega\in\Omega\mid X(\omega)\leq x\})$$
The RHS is abbreviated by $\mathbb P(\{X\leq x\})$ or (a bit shorter) by $\mathbb P(X\leq x)$.
The expression $P_X(X\leq x)$ that appears in your answer is wrong. 
The random variable $X$ induces a probability measure $P_X$ on measurable space $(\mathbb R,\mathcal B(\mathbb R))$ by:$$P_X(B)=\mathbb P(\{\omega\in\Omega\mid X(\omega)\in B\})=\mathbb P(X\in B)$$ where the utmost RHS is again an abbreviation.
Applying that on $B=(-\infty,x]$ we arrive at: 
$$F_X(x):=\mathbb P(\{\omega\in\Omega\mid X(\omega)\leq x\})=\mathbb P(X\leq x)=P_X((-\infty,x])$$
So the argument of $P_X$ is a subset of $\mathbb R$ and not necessarily a subset of $\Omega$ as the (wrong) notation $P_X(X\leq x)$ suggests.
If $Y:\Omega\to\mathbb R$ is another measurable function then it will have likewise its CDF and there is no reason at all to think that this function will coincide with the CDF of $X$.
E.g. define $X$ by $\omega\mapsto 0$ and define $Y$ by $\omega\mapsto1$. 
Then it can be deduced that $F_X(x)=1_{[0,\infty)}(x)\neq1_{[1,\infty)}(x)=F_Y(x)$.
If $X$ and $Y$ are defined on the same probability space then expressions like $X+Y$ make sense and are new random variables on that probability space.
