Is stopping time directed convergence in probability equivalent to a.s. convergence? Let $X,X_1,X_2,...$ be random variables on the probability space $(\Omega, \mathcal{F}, P)$. Let $S$ denoted the set of all bounded stopping times on $(\Omega, \mathcal{F}, (\mathcal{F}_n))$, where $\mathcal{F}_n = \sigma(X_1,...,X_n)$.
$S$ is a directed set under the pointwise partial order $\leq$, so we can speak of stopping time directed nets of random variables.
We write $X_\tau \xrightarrow{p} X$ to mean: for all $\epsilon, \delta > 0$, there exists $\sigma \in S$ such that $\tau \geq \sigma$ implies
$$P\Big( \Big\{\omega: |X_{\tau(\omega)}(\omega) - X(\omega)| > \delta  \Big\} \Big) < \epsilon. $$
It's clear that if $X_n \xrightarrow{a.s.} X$, then $X_\tau \xrightarrow{p} X$.
A paper I was reading seemed to claim that the converse is also true and obvious, but I cannot see it. The argument given is that if $X_\tau \xrightarrow{p} X$, then there exists $\tau_1 < \tau_2 < ... \in S$ such that 
$$X_{\tau_n} \xrightarrow{a.s.} \limsup_n X_{\tau_n} = \limsup_n X_n.$$
I don't understand how this proves the claim, and I guess I'm just confused about something simple. Any pointers or hints would be appreciated. 
 A: For fixed $\delta,\epsilon>0$ let $\sigma$ be a bounded stopping time such that
$$\forall \tau \in S, \tau \geq \sigma: \quad \mathbb{P}(|X_{\tau}-X|>\delta) < \epsilon. \tag{1}$$
Without loss of generality, we may assume $\sigma \geq 1$; otherwise we replace $\sigma$ by $\sigma+1$.  Note that $(1)$ implies, in particular,
$$\mathbb{P}(|X_{\sigma}-X|>\delta) < \epsilon. \tag{2}$$
For fixed $k \in \mathbb{N}$ we define a bounded stopping time $\tau_k$ by
$$\tau_k(\omega) := \inf\{n \geq \sigma(\omega); |X_n(\omega)-X_{\sigma}(\omega)| \geq 2 \delta\} \wedge (k \sigma(\omega)).$$
Clearly,
$$\begin{align*} \mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) &\leq \mathbb{P}(|X_{\sigma}-X|>\delta) + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_k-X|>3 \delta \, \, \text{i.o.}) \\ &\stackrel{\text{(2)}}{\leq} \epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_k-X_{\sigma}|>2 \delta \, \, \text{i.o.}). \end{align*}$$
By the definition of $\tau_k$ this implies
$$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq \epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_{\tau_k}-X_{\sigma}|> 2\delta \, \, \text{i.o.}).$$
As $k \sigma \to \infty$ as $k \to \infty$ it follows (from the definition of $\tau_k$) that we can choose $K \gg 1$ sufficiently large such that
$$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq 2\epsilon + \mathbb{P}(|X_{\sigma}-X| \leq \delta, |X_{\tau_K}-X_{\sigma}|>2\delta).$$
Hence,
$$\mathbb{P}(|X_k-X|>3 \delta \, \, \text{i.o.}) \leq 2\epsilon+ \mathbb{P}(|X_{\tau_K}-X|>\delta) \stackrel{\text{(1)}}{\leq} 3 \epsilon.$$
Since $\epsilon>0$ and $\delta>0$ are arbitrary, this proves $X_k \to X$ almost surely.
