The esistence of a martingale with given diffusion functional

Let $\mathbf R_+=[0,\infty)$ and $C$ be the space of continuous functions on $\mathbf R_+$ “starting” from zero with the topology of uniform convergence on compacts.
Let $A\colon \mathbf R_+\times C\mapsto \mathbf R_+$ be a Borel functional such that
(i) for each $\mathbf x\in C$, $A(\,\cdot\,,\mathbf x)\in C$ and is non-decreasing,
(ii) the set
$$\{\{A(t,\mathbf x)\}_{t\in \mathbf R_+}\mid\mathbf x\in C\}$$
is relatively compact in $C$,
(iii) for each $t\in \mathbf R_+$, $A(t,\,\cdot\,)$ is continuous, and
(iv) for each $t\in \mathbf R_+$, $x_s=y_s$ $(0\le s\le t)$ implies
$$A(t,\mathbf x)=A(t,y)\quad(\mathbf x=\{x_s\}_{s\in \mathbf R_+},y=\{y_s\}_{s\in \mathbf R_+}).$$
Then we prove that (on some probability space) there exists a continuous martingale $\mathbf X$ such that its Meyer squared variation process
$$\langle\mathbf X\rangle=A(\,\cdot\,,\mathbf X)\quad\text{a.s.}$$

In particular, in case
$$A(t,\mathbf x)=\int_0^ta^2(t,\mathbf x)\,ds$$
where $a^2$ is a bounded non-anticipative function, it follows that in the conditions of D. W. Stroock and S. R. S. Varadhan [12] continuity in $(s,\mathbf x)$ may he replaced by that in $\mathbf x$ only.