Consider the probability space with $3$ possible outcomes, $a$, $b$, $c$, each of which occurs with probability $1/3$. Suppose that $X$ and $Y$ are random variables such that $X(a) = −1$, $X(b) = 0$, $X(c) = 1$, and $Y(a) = 0$, $Y(b) = 1$, and $Y(c) = 0$.
Having hard time figuring out how to get $P(X(a))$. Isn't that what I need to prove that $P(X(a),Y(a)) \neq P(X(a))P(Y(a))$?
To show that $X$ and $Y$ are not independent, you need to show that there are some sets $A$ and $B$, such that $$P(X\in A, Y\in B)\neq P(X\in A)P(Y\in B).$$ We can try to take $A=B=\{0\}$. Looking at the right-hand-side first, we have $$P(X\in A) = P(X=0) = P(\text{outcome } b) = \frac{1}{3},$$ since $b$ is the only outcome with $X(b) = 0$. Similarly
$$P(Y\in B) = P(Y=0) = P(\text{outcome } a\text{ or }c)=\frac{2}{3}.$$ The right-hand-side is thus the product of these two numbers, i.e. $\frac{2}{9}$. Let's look at the left-hand-side. We are looking at $$P(X\in A, Y\in B) = P(X=0, Y=0).$$ We see that none of the outcomes $a$, $b$, or $c$ lets $X$ and $Y$ be equal to $0$ at the same time. Thus $$P(X=0, Y=0) = 0,$$ which is different from $\frac{2}{9}$.
