Is the $L^1$-space dual to a Banach space Let $(\Omega,\mu)$ be a measure space. It is well known that for $1<p\leq \infty$ one has the duality
$$L^p=(L^{p*})^*,$$
where $1/p+1/p^*=1$.
Question. Is it known that the Banach space $L^1$ is not isomorphic to the dual space of any Banach space?
To avoid trivial cases let us assume that $L^1$ is infinite dimensional. I am particularly interested in the case when the total measure of $\Omega$ is finite.
 A: I would like to mention that for general measure spaces the answer is negative: The space $L_1(0,1)^{**}$ (by, e. g., Proposition  II.5.3 in Lindenstrauss--Tzafriri (1973)) is a dual $L_1$-space. 
Actually for purely atomic (even finite) measures the negative answer is obvious: consider a probability measure on a countable set.
A: OP's question was about being isomorphic to a dual space so we need to observe that $L_1$ lacks the Radon–Nikodym property, which is invariant under isomorphisms, and separable dual spaces have this property.
Also, the $\ell_1$-sum of continuum many copies of $L_1[0,1]$ is isometric to $C[0,1]^*$.
Edit of 31.07.2016: It has been pointed out that my answer is incomplete as I do not treat the case when $L_1(\mu)$ is non-separable for a $\sigma$-finite measure $\mu$. By the Radon–Nikodym theorem, we may assume without loss of generality that $\mu$ is actually finite. 
The argument is then almost exactly the same as in this case the inclusion $L_2(\mu)\subset L_1(\mu)$ has dense range so $L_1(\mu)$ is weakly compactly generated. It is well known that weakly compactly generated dual spaces have the Radon–Nikodym property, a property that $L_1(\mu)$ is clearly lacking (by Maharam's theorem, $L_1(\mu)$ is isometric to $L_1(\{0,1\}^\lambda)$, where $\{0,1\}^\lambda$ is considered with the product fair-coin-toss (Haar) measure and $\lambda$ is the density character of $L_1(\mu)$). 
