I'm in a formal statistics class and it is driving me crazy. Here's the problem, I'm given that $X_1$ and $X_2$ are independent random variables uniformly distributed on $[0, \theta]$. I'm being asked to show that $P(X_1 = X_2) = 0$. Obviously, I can see that this is true, so I'm trying to figure out what I'm even being expected to prove.
I'm provided with the hint $P(X_1 = X_2) = E[1_{X_1=X_2}] = 1/\theta^2\int_{[0,\theta]^2}1_{x_1=x_2}dx_2dx_2$.
I assume the 2nd part of the hint is because $E[1_{X_1=X_2}] = \int_{\mathbb{R}^2}1_{X_1=X_2}dP(X_1,X_2)$ where the notation $dP(X_1, X_2)$ basically refers to the joint PDF of $X_1$ and $X_2$ (which for some reason we now instead refer to as the probability measure of $X_1$ and $X_2$), which gives us the $1/\theta^2$ term.
But from here I'm confused. What exactly do I do to prove that $\int_{[0,\theta]^2}1_{x_1=x_2}dx_2dx_2 = 0$? Is there some way to break it into a standard double integral? I think the problem is that I basically have no idea how to evaluate Lebesgue integrals except by turning them into some combination of Reimann integrals and then evaluating those, and in this case there seems to be no obvious way of doing so.
The double integral is $$ \int_0^\theta dx_1 \int_0^\theta dx_2 \; 1_{x_1=x_2} $$
But $$\int_0^\theta dx_2 \; 1_{x_1=x_2} = \int_0^\theta dx_2\; 0 = 0$$ because changing an integrand at a single point never changes the integral.