I assume two continuous independent random variables $x$, $y$ with the same PDF $p(x)=\frac{A}{|x|^a} (|x|<c)$, $\ p(x)=0(|x|>c)$ and $\frac{1}{2}<a<1$.
What is the probability that $x=y$? I got the following result:
$\iint p(x)p(y) \delta (x-y)dx dy=\int p(x)^2dx=\int \frac{A^2}{|x|^{2a}}dx=\infty$
What is wrong?
By simple inspection, we can see that because both variables ar I.I.D, then the set {(x,y),x=y} is non-empty.
The measure of the non-empty set $x=y$ (a line with length $\sqrt c$ and zero area) is neglectable compared with the measure of the non-empty set $|x|<c \cup |y|<c$, (a surface with area $c^2$) hence the probability is almost surely zero, which very roughly means, the probability is zero, but the set being measured is non-empty.
Actually, the individual distribution is: $$ \int p(x)dx=\int_0^c{A\over x^a}=A\left.{x^{-a+1}\over -a+1}\right|_{0^+}^c=A{c^{1-a}\over 1-a}=1 $$ $$ A={1-a \over c^{1-a}} $$
And the requested joint distribution is: $$ \int_{x=0}^c \int_{y=0, x=y}^c p(x)p(y)dxdy=\int_{x=0}^c p(x)\int_{y=0, x=y}^c p(y)dxdy $$ The last term on the integral must be applied on the set $y=x$, hence the inner integral is zero. No deltas shall be introduced in there, because the set is getting constrained, just adding the values in $x=y$. This is different than the case when the set is getting conditioned to stay on the $x=y$ line. $$ \int_{y=0, x=y}^c p(y)dy =0 $$
