Under what conditions does $L^{1}(X)$ have a predual?

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I know this question has been asked a million times—but they seem to always be with some special flair. I've looked at many and cannot extract from them an answer to my plain question:

Question: Let $(X,\mu)$ be a topological space endowed with a regular Borel measure. Consider the Banach space $\left(L^{1}(X,\mu),~\|\cdot\|_{1}\right)$. What do $X$ and $\mu$ have to satisfy (sufficient would be okay, but necessary-and-sufficient better), so that $L^{1}(X,\mu)$ has a predual?

Note: I don't care at this stage if the predual is separable.

Note: I am fully aware of papers that exist, that show what a Banach space $Y$ has to satisfy so that $\exists{X,\mu:~}Y^{\prime}\cong L^{1}(X,\mu)$. However I am asking the reverse: Given $(X,\mu)$ when does such a $Y$ exist?

Does $\mu$ have to be finite? Does $X$ have to be separable? Does $\mu$ have to have only finitely many atoms? etc.

Thanks in advance.