Let the joint distribution $f(x,y) = \frac{2}{3}(2x + y)$, $0\leq x \leq 1$, and $0\leq y \leq 1$.
What is the probability $P(x + y = \alpha)$? Let's say $\alpha = \frac{1}{2}$.
My gut feeling says that $P(x + y = \alpha) = 0$.
For a single random variable, I know that $P(x = c) = 0$ because $x = c$ only happens at a single point and so it's an area of a zero width.
I am confused however for multiple random variables because there are multiple cases where $x + y = \frac{1}{2}$, not just a single point.
I think I can write, for example, for $\alpha = \frac{1}{2}$:
$$ P(x+y=\alpha) = \int_{0}^{\frac{1}{2}}\int_{\frac{1}{2}-x}^{{\frac{1}{2}-x}}f(x,y)dydx = \int_{0}^{\frac{1}{2}}0dydx = 0. $$
Is this correct?
What about if there are, say, infinitely many random variables? If there are infinitely random variables and if I need to find $P(x_1 + x_2 + \dots + x_n = \alpha)$ as $n \rightarrow \infty$ then is it still zero?
This is strange to me because wouldn't there be infinitely many cases where the sum could equal to $\alpha$?