Show that for martingale and predictable process, it is not possible to gain almost surely in some step

Question

Let $X_t, t = 0, 1,\ldots, T$ be a martingale and $V_t, t = 1,2,\ldots, T$ a predictable process, I want to show that for $t = 1,2,\ldots, T$ we have $$ V_t\cdot (X_t - X_{t-1}) \ge 0 \textrm{ P.-a.s.} \Rightarrow V_t \cdot (X_t - X_{t-1}) = 0 \textrm{ P.-a.s.} $$ So any hints how to do that?

Answer 1

Since $V_t(X_t-X_{t-1})\ge 0$ a.s. it suffices to show that $\mathbb{E}[V_t(X_t-X_{t-1})]=0$. But (assuming that $X_t$ is $(\mathcal{F}_t)$-adapted)

$$\mathbb{E}[V_t(X_t-X_{t-1})]=\mathbb{E}[V_t\cdot \mathbb{E}[X_t-X_{t-1}|\mathcal{F}_t]]=0$$