Let $S\_t = M\_t + D\_t$ be the sum of a martingale $\left(M\_t\right)\_{t=1,2,\ldots}$ and a predictable process $(D\_t)\_{t=1,2,\ldots}$ such that the variance of the increments of $M$ is uniformly bounded $$\mathbb{E}\left[\left(M\_{t+1}-M\_t\right)^2 | \mathcal{F}\_t\right] \leq v$$
and the increments of $D$ are uniformly bounded from below. i.e. there exists $\Delta>0$ such that
$$
D\_{t+1} - D\_t \geq \Delta > 0 \,.
$$
I want to show that for all $\alpha<\Delta$
$$
\lim\_{t \to \infty} \mathbb{P}[S\_t \leq \alpha \, t ] = 0\,.
$$



Large deviation/concentration inequality for submartingale
For convenience, suppose that $D\_0 = 0$ and $M\_0=0$. The lower bound $D\_{t+1} - D\_t \ge \Delta$ implies that $D\_{t} \ge \Delta t$ a.s., i.e., $D\_t$ grows at least linearly with $t$. Thus, for any $t \in \mathbb{N}$ we have that
$$
\{ S\_t = M\_t + D\_t \le \alpha t \} \subset \{ M\_t + \Delta t \le \alpha t \} \tag{$\star$}
$$  

Also, note that
$$
\mathbb{E} \left\{ (M\_{t+1}-M\_t)^2 \mid \mathcal{F}\_t \right\} \le \nu \implies \mathbb{E} M\_{t}^2 \le \nu t
\tag{$\diamond$}
$$
uniformly in $t$. We apply these two simple facts below.


Let $\tilde M\_t = -M\_t$ and $\tilde M\_t^\* = \max\_{1 \le s \le t} \tilde M\_s$. Then, ($\star$) implies that
\begin{align\*}
\mathbb{P}( S\_t \le \alpha t) &\le \mathbb{P}( M\_t \le - (\Delta - \alpha) t ) \\
&\le \mathbb{P}( \tilde M\_t \ge (\Delta - \alpha) t ) \\
&\le \mathbb{P}( \tilde M\_t^\* \ge (\Delta - \alpha) t) \\
&\le \frac{1}{(\Delta - \alpha) t} \mathbb{E} \tilde M\_t^+ \\
&\le \frac{1}{(\Delta - \alpha) t} \frac{\sqrt{\mathbb{E}M\_t^2 }}{2} \\
&\le \frac{\sqrt{\nu}}{2 (\Delta - \alpha) \sqrt{t}} 
\end{align\*}
where we used a discrete-time martingale inequality in terms of $(a)^+ = \max(a,0) = (a+|a|)/2$, Jensen's inequality to bound $\mathbb{E} |M\_t|$, and the bound ($\diamond$) on the second moment of $M\_t$.


Pass to the limit as $t \to \infty$ in the final inequality to obtain the desired asymptotic result.


