 Solution

For $m\ge 1$, the increments have exponential momentThus $Z_n=2^{S_n}$ is an exponential martingale of a random walk relative to the natural filtration. We justify its stopping limit. From any surviving state in $(-m,m)$, a block of $2m$ successive increments all equal to $-1$ forces an exit. The block has probability $q=(6/7{)}^{2m}>0$, independently of the preceding increments. The geometric tail bound from a uniform escape probability givesso $T$ is finite almost surely and has finite expectation.

At the lower exit $S_T=-m$ exactly; at the upper exit $S_T$ is either $m$ or $m+1$. Before exit the same overall bound $-m\le S_{n\wedge T}\le m+1$ holds. Hence $Z_{n\wedge T}$ is bounded by $2^{m+1}$, uniformly in $n$. The bounded optional stopping theorem gives $\mathbb{E}{Z}_{n\wedge T}=Z_0=1$, and the dominated convergence theorem now yieldsAllowing the upper overshoot by one is essential; the terminal state need not equal $m$ on upper exit.