No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int\_0^ta_s\\,d\rho\_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup\_{s\le t}\rho_s$ is integrable. In fact, for any cadlag martingale $\rho$, the following are equivalent.

1. $\sup\_{s\le t}\vert\rho\_s\vert$ is integrable for all $t\in\mathbb{R}^+$.
2. $\int\xi\_s\\,d\rho\_s$ is a proper martingale for all bounded predictable processes $\xi$.

The implication (1) ⇒ (2) is a consequence of the Burkholder-Davis-Gundy inequality. This states that there exists positive constants $c < C$ such that $$ c\mathbb{E}\left[[M]_t^{1/2}\right]\le\mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]\le C\mathbb{E}\left[[M]_t^{1/2}\right] $$ for all cadlag martingales M with $M_0=0$, where $[\cdot]$ is the quadratic variation (I have written a blog post with a proof of the BDG inequality here). If $M=\int\xi\\,d\rho$ for bounded predictable $\vert\xi\vert\le1$ and (1) is satisfied then $$ \begin{align} \mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]&\le C\mathbb{E}\left[\left(\int\_0^t\xi^2\\,d[\rho]_s\right)^{1/2}\right]\\\\&\le C\mathbb{E}\left[[\rho]_t^{1/2}\right]\\\\&\le Cc^{-1}\mathbb{E}\left[\sup\_{s\le t}\vert\rho\_s-\rho\_0\vert\right]\\\\& < \infty. \end{align} $$ This is enough to guarantee that the local martingale M is a proper martingale. Conversely, to prove (2) ⇒ (1) then, assuming that $\sup\_{s\le t}\rho\_s$ is not integrable, you just need to find a bounded predictable process $\xi$ such that $\int\_0^t\xi\\,d\rho$ is not a martingale. Supposing that $\rho$ is continuous and writing $\rho^\ast\_t=\sup\_{s\le t}\rho\_s$, you can use the identity $$ \int\_0^t f^\prime(\rho^\ast\_s)\\,d\rho\_s=f(\rho^\ast\_t)-f(\rho\_0)+f^\prime(\rho^\ast\_t)(\rho\_t-\rho^\ast\_t). $$ This holds for all absolutely continuous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$. For example, choosing $f^\prime(x)=\sum\_{n\in\mathbb{Z}}(-1)^n1\_{\{n\le x < n+1\}}$ then $f(x)=\int\_0^x f^\prime(y)\\,dy$ will satisfy $0\le f\le1$. So, up to a bounded value, the right hand side of the above identity has absolute value $\vert\rho\_t-\rho^\ast\_t\vert$, which is not integrable, so is not a martingale. The case where $\rho$ is not continuous can be handled similarly, but is a bit messier.

No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int_0^ta_sd\rho_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup_{s\le t}\rho_s$ is integrable. In fact, for any cadlag martingale $\rho$, the following are equivalent.

1. $\sup_{s\le t}\vert\rho_s\vert$ is integrable for all $t\in\mathbb{R}^+$.
2. $\int\xi_sd\rho_s$ is a proper martingale for all bounded predictable processes $\xi$.

The implication (1) ⇒ (2) is a consequence of the Burkholder-Davis-Gundy inequality. This states that there exists positive constants $c < C$ such that $$ c\mathbb{E}\left[[M]_t^{1/2}\right]\le\mathbb{E}\left[\sup_{s\le t}\vert M_s\vert\right]\le C\mathbb{E}\left[[M]_t^{1/2}\right] $$ for all cadlag martingales M with $M_0=0$, where $[\cdot]$ is the quadratic variation (I have written a blog post with a proof of the BDG inequality here). If $M=\int\xi d\rho$ for bounded predictable $\vert\xi\vert\le1$ and (1) is satisfied then $$ \begin{align} \mathbb{E}\left[\sup_{s\le t}\vert M_s\vert\right]&\le C\mathbb{E}\left[\left(\int_0^t\xi^2d[\rho]_s\right)^{1/2}\right]\\\\&\le C\mathbb{E}\left[[\rho]_t^{1/2}\right]\\\\&\le Cc^{-1}\mathbb{E}\left[\sup_{s\le t}\vert\rho_s-\rho_0\vert\right]\\\\& < \infty. \end{align} $$ This is enough to guarantee that the local martingale M is a proper martingale. Conversely, to prove (2) ⇒ (1) then, assuming that $\sup_{s\le t}\rho_s$ is not integrable, you just need to find a bounded predictable process $\xi$ such that $\int_0^t\xi d\rho$ is not a martingale. Supposing that $\rho$ is continuous and writing $\rho^\ast_t=\sup_{s\le t}\rho_s$, you can use the identity $$ \int_0^t f^\prime(\rho^\ast_s)d\rho_s=f(\rho^\ast_t)-f(\rho_0)+f^\prime(\rho^\ast_t)(\rho_t-\rho^\ast_t). $$ This holds for all absolutely continuous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$. For example, choosing $f^\prime(x)=\sum_{n\in\mathbb{Z}}(-1)^n1_{\{n\le x < n+1\}}$ then $f(x)=\int_0^x f^\prime(y)dy$ will satisfy $0\le f\le1$. So, up to a bounded value, the right hand side of the above identity has absolute value $\vert\rho_t-\rho^\ast_t\vert$, which is not integrable, so is not a martingale. The case where $\rho$ is not continuous can be handled similarly, but is a bit messier.

No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int\_0^ta_s\\,d\rho\_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup\_{s\le t}\rho_s$ is integrable. In fact, for any cadlag martingale $\rho$, the following are equivalent.

1. $\sup\_{s\le t}\vert\rho\_s\vert$ is integrable for all $t\in\mathbb{R}^+$.
2. $\int\xi\_s\\,d\rho\_s$ is a proper martingale for all bounded predictable processes $\xi$.

The implication (1) ⇒ (2) is a consequence of the Burkholder-Davis-Gundy inequality. This states that there exists positive constants $c < C$ such that $$ c\mathbb{E}\left[[M]_t^{1/2}\right]\le\mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]\le C\mathbb{E}\left[[M]_t^{1/2}\right] $$ for all cadlag martingales M with $M_0=0$, where $[\cdot]$ is the quadratic variation (I have written a blog post with a proof of the BDG inequality here). If $M=\int\xi\\,d\rho$ for bounded predictable $\vert\xi\vert\le1$ and (1) is satisfied then $$ \begin{align} \mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]&\le C\mathbb{E}\left[\left(\int\_0^t\xi^2\\,d[\rho]_s\right)^{1/2}\right]\\\\&\le C\mathbb{E}\left[[\rho]_t^{1/2}\right]\\\\&\le Cc^{-1}\mathbb{E}\left[\sup\_{s\le t}\vert\rho\_s-\rho\_0\vert\right]\\\\& < \infty. \end{align} $$ This is enough to guarantee that the local martingale M is a proper martingale. Conversely, to prove (2) ⇒ (1) then, assuming that $\sup\_{s\le t}\rho\_s$ is not integrable, you just need to find a bounded predictable process $\xi$ such that $\int\_0^t\xi\\,d\rho$ is not a martingale. Supposing that $\rho$ is continuous and writing $\rho^\ast\_t=\sup\_{s\le t}\rho\_s$, you can use the identity $$ \int\_0^t f^\prime(\rho^\ast\_s)\\,d\rho\_s=f(\rho^\ast\_t)-f(\rho\_0)+f^\prime(\rho^\ast\_t)(\rho\_t-\rho^\ast\_t). $$ This holds for all absolutely continuous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$. For example, choosing $f^\prime(x)=\sum\_{n\in\mathbb{Z}}(-1)^n1\_{\{n\le x < n+1\}}$ then $f(x)=\int\_0^x f^\prime(y)\\,dy$ will satisfy $0\le f\le1$. So, up to a bounded value, the right hand side of the above identity has absolute value $\vert\rho\_t-\rho^\ast\_t\vert$, which is not integrable, so is not a martingale. The case where $\rho$ is not continuous can be handled similarly, but is a bit messier.

No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int\_0^ta_s\\,d\rho\_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup\_{s\le t}\rho_s$ is integrable. In fact, for any cadlag martingale $\rho$, the following are equivalent.

1. $\sup\_{s\le t}\vert\rho\_s\vert$ is integrable for all $t\in\mathbb{R}^+$.
2. $\int\xi\_s\\,d\rho\_s$ is a proper martingale for all bounded predictable processes $\xi$.

The implication (1) ⇒ (2) is a consequence of the Burkholder-Davis-Gundy inequality. This states that there exists positive constants $c < C$ such that $$ c\mathbb{E}\left[[M]_t^{1/2}\right]\le\mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]\le C\mathbb{E}\left[[M]_t^{1/2}\right] $$ for all cadlag martingales M with $M_0=0$, where $[\cdot]$ is the quadratic variation (I have written a blog post with a proof of the BDG inequality here). If $M=\int\xi\\,d\rho$ for bounded predictable $\vert\xi\vert\le1$ and (1) is satisfied then $$ \begin{align} \mathbb{E}\left[\sup\_{s\le t}\vert M_s\vert\right]&\le C\mathbb{E}\left[\left(\int\_0^t\xi^2\\,d[\rho]_s\right)^{1/2}\right]\\\\&\le C\mathbb{E}\left[[\rho]_t^{1/2}\right]\\\\&\le Cc^{-1}\mathbb{E}\left[\sup\_{s\le t}\vert\rho\_s-\rho\_0\vert\right]\\\\& < \infty. \end{align} $$ This is enough to guarantee that the local martingale M is a proper martingale. Conversely, to prove (2) ⇒ (1) then, assuming that $\sup\_{s\le t}\rho\_s$ is not integrable, you just need to find a bounded predictable process $\xi$ such that $\int\_0^t\xi\\,d\rho$ is not a martingale. Supposing that $\rho$ is continuous and writing $\rho^\ast\_t=\sup\_{s\le t}\rho\_s$, you can use the identity $$ \int\_0^t f^\prime(\rho^\ast\_s)\\,d\rho\_s=f(\rho^\ast\_t)-f(\rho\_0)+f^\prime(\rho^\ast\_t)(\rho\_t-\rho^\ast\_t). $$ This holds for all absolutely continuous functions $f\colon\mathbb{R}\rightarrow\mathbb{R}$. For example, choosing $f^\prime(x)=\sum\_{n\in\mathbb{Z}}(-1)^n1\_{\{n\le x < n+1\}}$ then $f(x)=\int\_0^x f^\prime(y)\\,dy$ will satisfy $0\le f\le1$. So, up to a bounded value, the right hand side of the above identity has absolute value $\vert\rho\_t-\rho^\ast\_t\vert$, which is not integrable, so is not a martingale. The case where $\rho$ is not continuous can be handled similarly, but is a bit messier.