If $X_1$ and $X_2$ are two iid copies of $X$, how to show that $P_{X_1,X_2}[P_X[X_1The question is in the title. How do I go about proving it? The distribution function of $X$ is continuous, but may not be strictly increasing.
I have some heuristics. For example, if $X$ follows a mixture of unif$(0,1)$ distribution and $3+\chi^2$ distribution, then its cdf is constant between $1$ and $3$. Now $X$ realizing two values between which it has $0$ probability of occurrence is same as probability of $X$ taking values $1$ and $3$, which is 0. Now I think we cannot take mixture of more than a countable number of distributions so this probability should always be 0, but this is not a rigorous proof.

If we’re taking about continuous values, if $x_1 < x_2$, there would always be some $x$ such that $x_1 < x< x_2$. Just take something slightly higher than $x_1$ and lower than $x_2$. In such a case, it boils down to asking if $\Pr(X_1 < X_2) > 0$ for i.i.d. variables.
