Let $(X_n)$ be a sequence of real random variables on $(\Omega,\mathcal A,\mathbb P)$. Then 1. and 2. are equivalent:
- There exists a random variable $X$, s.t. $X_n\to X$ $P$-almost sure for $n\to \infty$.
- $\sup_{m>n} |X_m-X_n|\to 0$ in probability for $n\to\infty$.
I tried showing $1.\Rightarrow 2.$:
I know that $X_n\to X$ $P$-almost sure means that $$P(\lim_{n\to\infty} X_n=X)=1$$ or equivalently $$(*)\quad \lim_{n\to\infty} P(\sup_{m\geq n} |X_m-X|\geq\varepsilon)=0\quad\forall\varepsilon>0$$
$\sup_{m>n} |X_m-X_n|\to 0$ in probability means \begin{equation}\lim_{n\to\infty} P(\sup_{m>n}|X_m-X_n|\geq\varepsilon)=0\quad\forall\varepsilon>0\end{equation}
This looks like a Cauchy-sequence in probability, but I don't know if I can deduce the convergence of this from $(*)$.
How can I go on proving this? Thanks for any input!