If you think about it, $\mathbf{Z}_P$ will almost never be $L^1$-compact, the reason being that it lacks closedness. Suppose that you had $Z^e$ which comes from an equivalent local martingale measure and $Z^a$, which comes from an absolutely continuous but not equivalent local martingale measure. Then $Z^e > 0$. Therefore, consider the convex combinations $\frac{1}{n}Z^e + \frac{n-1}{n}Z^a$. These are equivalent local martingales measures since $Z^e$ is postive, but they converge to $Z^a$ (in $L^1$), which is not equivalent. So, as soon as you have any nonequivalent local martingale measure, you lose closedness.
If you think about it, $\mathbf{Z}_P$ will almost never be $L^1$-compact, the reason being that it lacks closedness. Suppose that you had $Z^e$ which comes from an equivalent local martingale measure and $Z^a$, which comes from an absolutely continuous but not equivalent local martingale measure. Then $Z^e > 0$. Therefore, consider the convex combinations $\frac{1}{n}Z^e + \frac{n-1}{n}Z^a$. These are equivalent local martingales since $Z^e$ is postive, but they converge to $Z^a$ (in $L^1$), which is not equivalent. So, as soon as you have any nonequivalent local martingale measure, you lose closedness.