Prove that if the sequence Xn converges to X almost surely then Xn converges to X in probability

Question

I am still getting the hang of almost surely convergence, and I just learned convergence in probability. Any help would be appreciated.

Answer 1

Let $\epsilon >0$. Let $A_m=\cap_{n \geq m} \{|X_n-X| \leq \epsilon\}$. Then $A_m$ increases to $A=\cup_n \cap_{n \geq m} \{|X_n-X| \leq \epsilon\}$. Hence $P(A_m) \to P(A)$ as $m \to \infty$. Form definition of almost sure convergence check that $P(A)=1$. Hence $P(A_m) \to 1$ which implies $P(\{|X_m-X| \leq \epsilon\}) \to 1$. [Because $A_m$ is a subset of $\{|X_m-X| \leq \epsilon\}$]. Taking complement we get $P(\{|X_n-X| >\epsilon\}) \to 0$.