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.
Best Answer
If you proved $X \geq X_n$ ($n \in \mathbb{N}$), this would imply $$\mathbb{P} \left( \bigcup_{k=n}^{\infty} E_k \right) = \mathbb{P}(E_n)$$ since $E_k \subseteq E_n$ for $k \geq n$ (this follows directly from the monotonicity).
Here is an alternative proof: Since $X_{n+1} \geq X_n$ we know that $$Y := \sup_{n \in \mathbb{N}} X_n = \lim_{n \to \infty} X_n \in (-\infty,\infty]$$ exists. This implies in particular $X_n \to Y$ in probability and from the uniqueness of the limit, we conclude $X=Y$ a.s.. Thus $X_n \to X$ almost surely.