A characterization of almost sure convergence Suppose we have a sequence of positive random variables $X_1,X_2,...,X$. I am trying to prove a characterization of almost sure convergence.
It states that $X_n \rightarrow X$ almost surely iff for every $\epsilon >0$, $\lim_{n \rightarrow \infty} P[\sup_{k \ge n} \frac{X_k}{X} > 1+ \epsilon]=0$ and $\lim_{n \rightarrow \infty} P[\sup_{k \ge n} \frac{X}{X_k} > 1+ \epsilon]=0$.
If I assume almost sure convergence, then the implication is easy but I am not being able to prove the other way round.
 A: Since all the random variables are positive,
\begin{align}
&\{\omega:X_n(\omega)\not\to X(\omega)\}=\{\omega:X_n(\omega)/X(\omega)\not\to 1\} \\
&\qquad=\{\limsup X_n/X > 1\} \cup \{\liminf X_n/X < 1\}.
\end{align}
Thus,
$$
\mathsf{P}(X_n\not\to X)\le\mathsf{P}(\limsup X_n/X>1)+\mathsf{P}(\liminf X_n/X<1).
$$
But
\begin{align}
\mathsf{P}(\limsup X_n/X>1)&=\lim_{m\to\infty}\mathsf{P}(\limsup X_n/X\ge 1+ m^{-1}) \\
&=\lim_{m\to\infty}\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X_k/X\ge 1+ m^{-1}\right).
\end{align}
That is, $\mathsf{P}(\limsup X_n/X>1)=0$ is equivalent to
$$
\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X_k/X\ge 1+\epsilon\right)=0 \quad\forall \epsilon>0,
$$
and, similarly, $\mathsf{P}(\liminf X_n/X<1)=0$ is equivalent to
$$
\lim_{n\to\infty}\mathsf{P}\!\left(\inf_{k\ge n} X_k/X\le 1-\epsilon\right)=0 \quad\forall \epsilon>0.
$$

The last condition is equivalent to $\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n} X/X_k\ge 1+\epsilon\right)=0 \quad\forall \epsilon>0$.
A: Let
$$A_n^{\epsilon} = \bigl\{\sup_{k\geq n}\frac{X_k}{X}>1+\epsilon\bigr\} \,\text{ and }\, B_n^{\epsilon} = \bigl\{\sup_{k\geq n}\frac{X}{X_k}>1+\epsilon\bigr\}.$$ Then $A_{n+1}^\epsilon\subseteq A_n^\epsilon$ and $B_{n+1}^\epsilon\subseteq B_{n}^\epsilon$, thus
$$\lim_{n\to \infty}P(A_n^\epsilon)=P\biggl(\bigcap_{n=1}^{\infty}A_n^\epsilon\biggr)=P\biggl(\limsup_{n\to\infty}\frac{X_n}{X}>1+\epsilon\biggr)$$
and
$$\lim_{n\to \infty}P(B_n^\epsilon)=P\biggl(\bigcap_{n=1}^{\infty}B_n^\epsilon\biggr)=P\biggl(\limsup_{n\to\infty}\frac{X}{X_n}>1+\epsilon\biggr).$$
Since the above hold for every $\epsilon>0$ we conclude that
$$P\biggl(\limsup_{n\to \infty}\frac{X_n}{X}>1\biggr) = 0 \, \text{ and }\, P\biggl(\limsup_{n\to \infty}\frac{X}{X_n}>1\biggr) = 0.$$
These two are equivalent to the following two
$$P\biggl(\limsup_{n\to \infty}\frac{X_n}{X}\leq 1\biggr)=1\, \text{ and }\, P\biggl(\limsup_{n\to \infty}\frac{X}{X_n}\leq 1\biggr)=1.$$
Now observe that $\limsup\limits_{n\to \infty}\frac{X_n}{X}=\frac{1}{X}\cdot\limsup\limits_{n\to \infty}X_n$ and $\limsup\limits_{n\to \infty}\frac{X}{X_n}=X\cdot \limsup\limits_{n\to \infty}\frac{1}{X_n}=X\cdot \frac{1}{\liminf\limits_{n\to \infty}X_n}.$
Therefore,
$$P\biggl(\limsup_{n\to \infty}X_n\leq X\biggr)=1\, \text{ and }\, P\biggl(\liminf_{n\to \infty}X_n\geq X\biggr)=1,$$
taking intersections we conclude that $X_n\overset{\text{a.s}}{\longrightarrow}X$.
