Suppose we are given a filtered probability $(\mathcal{F},\{\mathcal{F}\}_t,\mathbb{P})$ space fulfilling the "usual conditions" (.i.e. the filtration is right-continuous and complete). Let $\tau$ be a stopping time with respect to this space, let $a$ be a positive real number and assume $a<\tau$. Is it true that for any $\epsilon>0$ and $\delta>0$ there exists a stopping time $\sigma_{\epsilon,\delta}$ such that $\mathbb{P}\left(\sigma_{\epsilon,\delta}\in\left[\tau-a,\tau-a+\delta\right]\right)>1-\epsilon$.
Since the set $\left[\tau-a,\tau-a+\delta\right]$ is not optional I guess the optional section theorem will not yield anything useful.
Any ideas?
No, no such stopping time exists in general. Let $B$ be a Brownian motion, $\tau := \inf\{t > 2: B_t - B_2 \ge 1\} \wedge 3$, and $a = 2$. Then if $\sigma$ is any stopping time, we will show there is an upper bound on $\mathbb{P}(\sigma \in [\tau - a, \tau - a + \delta])$.
First, we may assume WLOG that $\sigma \in [0,1 + \delta]$, so that $\mathcal{F}_\sigma \subset \mathcal{F}_2$ which is independent of $\tau$. Then, because $\{ \sigma \in [\tau - a, \tau - a + \delta]\} = \{\tau \in [\sigma + a - \delta, \sigma + a]\},$ by the independence lemma we have \begin{align*} \mathbb{P}(\sigma \in [\tau - a, \tau - a + \delta]) &= \mathbb{P}(\tau \in [\sigma + a - \delta, \sigma + a]) \\ &\le \sup_{b \in \mathbb{R}} \mathbb{P}(\tau \in [b-\delta,b]). \end{align*}
Since $\sup_{b \in \mathbb{R}} \mathbb{P}(\tau \in [b-\delta,b]) < 1$ for all $\delta < 1$, one can therefore choose $\varepsilon > 0$ such that $\mathbb{P}(\sigma \in [\tau - a, \tau - a + \delta]) < 1-\varepsilon$ for all stopping times $\sigma$.