 Solution

The hitting time $T$ is a stopping time for the natural filtration, because $\{T\le n\}={\bigcup }_{k\le n}\{S_k=b\}$. The integer-valued increments are at most one. A path started at zero therefore cannot reach or exceed $b$ without first visiting $b$, soDefine $Y_n=b-S_{n\wedge T}$. It is nonnegative and integrable: $|S_{n\wedge T}|\le {\sum }_{j\le n}|X_j|$. Moreover,The indicator function is ${\mathcal{F}}_n$-measurable, and $X_{n+1}$ is independent of ${\mathcal{F}}_n$ with zero expected value. Taking conditional expectation therefore proves directly that $(Y_n)$ is a nonnegative martingale.

The general result used here is that a nonnegative discrete-time martingale has a finite almost-sure limit; this is the martingale convergence theorem, since $\mathbb{E}{Y}_n=\mathbb{E}{Y}_0=b$ bounds its L1 norm. On $\{T=\infty \}$ it follows that $S_n=b-Y_n$ converges finitely. A convergent sequence of integers is eventually constant, so $X_n=S_n-S_{n-1}$ would eventually be zero on this event.

However, the independent events $\{X_n=1\}$ all have the same strictly positive probability. The second Borel-Cantelli lemma says that they occur infinitely often almost surely. This contradicts eventual constancy on any positive-probability subset of $\{T=\infty \}$. ThusNotice that no lower bound on the negative increments was needed.