Let $(X_n)$ a sequence of real random variables on the probability space $(\Omega,\mathcal F,P)$ increasing almost surely to the real random variable $X$. Show that
$$P(X<t)=\sup_{s<t} \inf_{n\in\mathbb N} P(X_n<s)$$
for all $t \in \mathbb R$.
My attempt:
Fix $t\in \mathbb R$. For each $s\in \mathbb R$ and $n\in \mathbb N$ we have $P(X<s)\leq P(X_n<s)$ by monotonicity. Therefore $P(X<s)\leq \inf_{n\in\mathbb N} P(X_n<s) $ for each $s\in \mathbb R$ which implies $\sup_{s<t} P(X<s) \leq \sup_{s<t} \inf_{n\in\mathbb N} P(X_n<s)$. Since the map $s\mapsto P(X<s)$ is nondecreasing and left continuous we have $\sup_{s<t} P(X<s)=P(X<t)$, and so we get the inequality
$$P(X<t)\leq\sup_{s<t} \inf_{n\in\mathbb N} P(X_n<s) \quad \quad(1)$$
Now suppose by contradiction that $P(X<t)<\sup_{s<t} \inf_{n\in\mathbb N} P(X_n<s)$. Then there exists $s<t$ such that $P(X<t)<\inf_{n\in\mathbb N} P(X_n<s)$. Choose any $s'$ such that $s<s'<t$. Then, since the sequence of events $\{X_n<s\}$ is decreasing almost surely, we have $\inf_{n\in\mathbb N} P(X_n<s)\leq P(X\leq s)\leq P(X<s')$. But this means that $P(X<t)<P(X<s')$ with $s'<t$, a contradiction. Therefore the inequality in $(1)$ is in fact an equality.
Is this correct? Thanks a lot for your help.