Let $\{X_n\}$ be a collection of random variable with $X_{n+1} \geq X_n$ for all $n$ and $X_n \rightarrow X$ in probabilty. How to prove that $X_n \rightarrow X$ almost surely.
My partial answer:
Consider a probability space $(\Omega, B, P)$ Let $\varepsilon$ be a abritary positive constant. Define $E_n=\{\omega\in \Omega: |X_n(\omega) -X(\omega)|\geq \varepsilon\}$. Let $S=\cap_{n=1}^{\infty} \cup_{k=n}^{\infty} E_k$. Since $X_n \rightarrow X$ in probabilty, we have \begin{align} P(S)=\lim_{n\rightarrow \infty} P(\cup_{k=n}^{\infty} E_k)= \lim_{n\rightarrow \infty}P(E_n)=0. \end{align} For $y \in \Omega \backslash S$, we have $|X_n(y)-X(y)|<\varepsilon$. Hence, $X_n \rightarrow X$ almost surely.
My poblem: I'm not sure about $\lim_{n\rightarrow \infty} P(\cup_{k=n}^{\infty} E_k)= \lim_{n\rightarrow \infty}P(E_n)$ and how to argue this by monotonicity of random variable.