Let
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space
- $T>0$
- $I:=(0,T]$
- $(\mathcal F_t)_{t\in\overline I}$ be a filtration on $(\Omega,\mathcal A)$
- $\mathcal P$ denote the predictable $\sigma$-algebra on $\Omega\times\overline I$
- $A:\Omega\times\overline I\to\mathbb R$ be $\mathcal A\otimes\mathcal B(\overline I)$-measurable such that $A_0(\omega)=0$ and $A(\omega)$ is continuous and of bounded variation for all $\omega\in\Omega$ and $$\operatorname E\left[|A|_T\right]<\infty\tag1,$$ where $|A(\omega)|$ denotes the total variation function of $A(\omega)$ for $\omega\in\Omega$
- $B:\Omega\times\overline I\to\mathbb R$ satisfy all assumptions imposed on $A$ and assume that $B(\omega)$ is nondecreasing for all $\omega\in\Omega$
Assume $$\int_{(0,\:t]}\xi\:{\rm d}A=0\;\;\;\text{almost surely for all }t\in I\tag1$$ for all bounded nonnegative $\mathcal P$-measurable $\xi:\Omega\times\overline I\to\mathbb R$ with $$\int_{(0,\:t]}\xi\:{\rm d}B=0\;\;\;\text{almost surely for all }t\in I.\tag2$$
Now, let $$\mu(P):=\operatorname E\left[\int1_P\:{\rm d}A\right]$$ and $$\nu(P):=\operatorname E\left[\int1_P\:{\rm d}B\right]$$ for $P\in\mathcal P$. By assumption, $\mu$ is absolutely continuous with respect to $\nu$ and hence the Radon-Nikodým derivative $$\alpha:=\frac{{\rm d}\mu}{{\rm d}\nu}$$ is well-defined and $\mathcal P$-measurable with $$\operatorname E\left[\int|\alpha|\:{\rm d}B\right]<\infty\tag3$$ and $$\operatorname E\left[\int\xi\:{\rm d}B\right]=\int\xi\:{\rm d}\mu=\int\alpha\xi\:{\rm d}\nu\tag4$$ for all bounded nonnegative $\mathcal P$-measurable $\xi:\Omega\times\overline I\to\mathbb R$.
Let $$M:=A-\int\alpha\:{\rm d}B$$ and note that $$\operatorname E\left[\int\xi\:{\rm d}M\right]=0\tag5$$ for all bounded nonnegative $\mathcal P$-measurable $\xi:\Omega\times\overline I\to\mathbb R$. Why does $(5)$ yield that $M$ is an $\mathcal F$-martingale?
Fix $s \leq t$. For fixed $F \in \mathcal{F}_s$ the process
$$\xi(u,\omega) := 1_F(\omega) 1_{(s,t]}(u)$$
is bounded, non-negative and progressively measurable. By assumption, this implies that
$$\mathbb{E} \left( \int \xi \, dM \right)=0$$
which means that
$$\mathbb{E}(1_F (M_t-M_s)) =0.$$
Since $F \in \mathcal{F}_s$ is arbitrary, this is equivalent to saying that
$$\mathbb{E}(M_t-M_s \mid \mathcal{F}_s)=0,$$
and so $(M_t,\mathcal{F}_t)_{t \geq 0}$ is a martingale.