Suppose $X_n \to X$ in probability and $P[X_n \leq X_{n+1}] = 1$. Then show that $P[ \cup_{n\geq1}\{X_n > X\}] = 0$.
I can prove this by assuming that $\{X_n>X\}$ is monotone increasing sets. Then using continuity theorem I can show the given probability converges to $0$. But this is not clear from $P[X_n \leq X_{n+1}] = 1$. How should I use $P[X_n \leq X_{n+1}] = 1$ in this problem?
Let $E =\cup_n \{X_n > X_{n+1}\}$. Then $P(E) \leq\sum_n P(X_n >X_{n+1})=0$. On $E^{c}$ the sequence $X_n$ is increasing so it converges almost surely to $Y=\sup_n X_n$. This, together with the fact that $X_n \to X$ in probability implies that $X=\sup_n X_n$ almost surely from which the conclusion follows.