Explanation of Joint Probability and Independence 
Can anyone explain further to me the solution for the second instance where $f(x,y) = 24xy$. Specifically, I don't really understand the part "because the region in which the joint density is nonzero cannot be expressed in the form $x ∈ A, y ∈ B$".
I don't really understand how illustration with $I(x,y)$ are doing in terms of proving dependency between the variables?
 A: Informally, independence mean that knowing the value of one random variable gives you no extra information about the other
But if $0 \lt X+Y \lt 1$, then knowing $X=\frac34$ tells you $Y < \frac14$.  Meanwhile  knowing $X=\frac13$ tells you $Y$ can take values up to $\frac23$.  So the value of $X$ is affecting the distribution of possible values of $Y$ and thus they are not independent.  The indicator function has this effect, because it cannot be seperated into an $X$ part and a $Y$ part 
A: The necessary (not sufficient) condition for independence is that $f(x,y)$ should be factored into something like $g(x)h(y)$. For that to happen, $I(x,y)$ should be factored like $I_A(x)I_B(y)$, but the author says that there is no way to do it, basically because of the line $0<x+y<1$.
Assume $I(x,y)=I_A(x)I_B(y)$, where $A=(0,1)$, $B=(0,1)$ (it's $(0,1)$ because there is density for $x,y$ everywhere in $(0,1)$). So, $I_A(x)$ and $I_B(y)$ should be non-zero, if for example $x=0.3,y=0.8$, but $I(x,y)$ is zero.
A: A sufficient test for detecting non-independence of random variables is the eyeball test (described briefly in this answer of mine on stats.SE and in more detail in an answer on math.SE) which says that if the support of the joint density is not a rectangle with sides parallel to the coordinate axes, then the random variables are dependent. Here, the support of $f_{X,Y}(x,y)$ is a triangle and so we can assert that the random variables are dependent without the need for laboriously calculating $f_X(x)$ and $f_Y(y)$ and then checking whether $f_{X,Y}(x,y)$ equals the product $f_X(x)f_Y(y)$ for all real numbers $x$ and $y$ as the definition of independence says (or should say). That $f_{X,Y}(x,y)$ can be expressed as $g(x)h(y)$ for some real numbers $x$ and $y$ (as it does in this instance) is not enough to claim independence.
A: I don't know if this makes it any easier to understand, but consider a function that's equal to $x^2y^3$ when $x<y$ and $0$ otherwise. Now consider the function $x^2y^3 (|y-x|+y-x)/(2y-2x)$. If you plot out the second function, you'll find that it's the same as the first function. It's in more complicated form, but it's in closed form rather than having a conditional definition.
While the first function appears to factor, the second one doesn't. If you want to factor out the $y^3$, you still have the $(|y-x|+y-x)/(2y-2x)$ to deal with. 
