Finding $f_Y(y)$ where $Y$ has a non-simple space? I have the solution to this problem but I don't really understand it. 

Let $f_X(x) = \dfrac {1}{10}, \ x = 0, 1, 2, ... 9$ and $h(Y|x) = \dfrac {1}{10-x}, y = x, x+1, x+2, ..., 9$. 
a) Find $f(x, y)$.
b) Find $f_Y(y)$



This is the solution my professor posted:

I am having trouble understanding part b). As I recall, $\displaystyle f_Y(y) = \sum_{x \in S_x}f(x, y)$. I see that $S_Y$ is not fixed, but $S_X$ is. So why did my professor write $f_Y(y) = \displaystyle \sum_{x=0}^y \dfrac {1}{10(10-x)}$ instead of  $f_Y(y) = \displaystyle \sum_{x=0}^{9} \dfrac {1}{10(10-x)}?$ The first one seems to imply that the domain of $x$ is not fixed.

EDIT: As I was writing this question, I think I may have figured it out, but I would still like confirmation please that I am correct:
Take $y = 3$ as an example. 
$\displaystyle f_Y(y) = \sum_{x \in S_x}f(x, y) = f(0, 3) + f(1, 3) + f(2, 3) + f(3, 3) + f(4, 3) + f(5, 3) + f(6, 3) + f(7, 3) + f(8, 3) + f(9, 3) = \sum_{x=1}^3 \dfrac {1}{10(10-x)} + 0 + 0 + 0 + 0 + 0+0$
 A: One of the best ways to reason with joint distributions is to draw a picture.

Because this situation involves a discrete distribution, the picture plots the individual probabilities, $f_{X,Y}(x,y)$.  It represents the probability of each possible $(x,y)$ with a square symbol whose area is proportional to the probability.  All nonzero probabilities are posted beneath each symbol, as computed in part $(a)$ of the solution. 
This picture was constructed by taking each possible value $x=0, 1, \ldots, 9$ in turn and plotting the values of $y$ where the probabilities are nonzero.  For instance, when $x=4$ the possible values of $y$ are $y=4,5,\ldots, 9$: these appear as a vertical line of dots ascending from $(4,4)$ to $(4,9)$.
The marginal probabilities for $Y$, $f_Y(y)$, are found by summing across all $x$ for each possible value $y$, as computed in part $(b)$ of the solution.  Those sets of possibilities are outlined in darker strips.  The sums of the probabilities within those strips are posted at the right. 
Finding the marginal distribution $f_Y$ is thereby exhibited as a geometrical process of cutting the picture into horizontal strips and collecting the probabilities within each strip.

This is a common theme in many bivariate probability problems: probabilities are specified or constructed by fixing one variable and changing the other (here, fixing $X$ and changing $Y$) but the question requires fixing a different variable (here, fixing $Y$ and changing $X$).  A two-dimensional plot like this helps coordinate those operations correctly.
