Why does the bounds of this integral not consider both equalities? I'm trying to show that this function is a joint probability density function.
$$f(x,y)=\begin{cases}1/x&:& 0<y<x<1\\0&:&\text{otherwise}\end{cases}$$
To do this, I need to integrate the function over the bounds of x and the bounds of y and check that the area equals one. According to the problem, 0 < y < x, and y < x < 1.
However, in the solution, only the inequality for y is used when integrating over the bounds for y (from 0 to x). However, the bounds for the integral with respect to x is from 0 to 1.
According to the problem specifications though,y < x < 1 and thus x > y, so I thought that the bounds of the outside integral with respect to x would be from y to 1. Why is this not the case?
 A: Intuitively:
$$\begin{align}\iint_{\text{all supported values}}\dfrac 1x \,\mathrm d\langle x,y\rangle &= \int_{\text{all values $x$ may take}}\left[\int_{\text{all values $y$ may take for a particular $x$}}\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\\[2ex]\iint_{0\leq y\leq x\leq 1}\dfrac 1x \,\mathrm d\langle x,y\rangle &= \int_{0\leq x\leq 1}\left[\int_{0\leq y\leq x}\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\\[1ex]&= \int_0^1\left[\int_0^x\dfrac 1x\,\mathrm d y\right]\,\mathrm d x\end{align}$$

Because the domain is $\{\langle x,y\rangle: 0\leq y\leq x\leq 1\}=\{\langle x,y\rangle: 0\leq x\leq 1\text{ and }0\leq y\leq x\}$

Consider a count of the integer sequence $\{(1,1),(2,1),(2,2),(3,1),(3,2),(3,3)\}$ which is $\{\langle x,y\rangle\in\Bbb N^2:1\leq y\leq x\leq 3\}$.  This is clearly 6, and taking the series agrees:
$$\begin{align}\sum_{\small\langle x,y\rangle\in\Bbb N^2:1\leq y\leq x\leq 3}1&=\sum_{x=1}^3\sum_{y=1}^x1\\&=\sum_{x=1}^3x\\&=6\end{align}$$
But if we do not "sum out" the inner variable and bound the outer series as $\{x\in\Bbb N:y\leq x\leq 3\}$:
$$\begin{align}\sum_{x=y}^3\sum_{y=1}^x 1&=\sum_{x=y}^3x\\&=y+\ldots+3\end{align}$$
Which is clearly incorrect, as we have left $y$ as a free variable.
The same principle applies.
A: When you write a definite integral such as
$$ \int_a^b f(x,y)\, \mathrm dy, $$
The variable $y$ wherever it appears in that integral comes from the $y$ in
$\mathrm dy.$
All of the instances of the variable named $y$ inside the integral are invisible and inaccessible to any expressions outside the integral.
In fact, technically the choice of the name $y$ is irrelevant and any variable name that is not already in use in the integral could have been chosen:
$$ \int_a^b f(x,y)\, \mathrm dy = \int_a^b f(x,z)\, \mathrm dz = \int_a^b f(x,v)\, \mathrm dv. $$
The fact that $y$ was chosen in the written solution is merely a helpful hint to you that this $y$ is somehow related to the $y$ in the definition of the distribution.
So the idea that we could somehow use $y$ outside the inner integral,
for example to set bounds of the variable $x$ in the outer integral,
simply will not work.
Fortunately, when we set the boundary values of the inner integral as shown in the solution,
$$ \int_0^x f(x,y)\, \mathrm dy, $$
this enforces the condition that $y < x$ by requiring $y$ to be integrated only over values between $0$ and $x.$
This is enough. The condition, once enforced, stays enforced.
