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.



Is the $L^1$-space dual to a Banach space
OP's question was about being *isomorphic* to a dual space so we need to observe that [$L\_1$ lacks the Radon–Nikodym property](https://books.google.co.uk/books?id=-YLoYK26rTsC&pg=PA258&dq=the%20Radon-Nikodym%20property&hl=pl&sa=X&ved=0ahUKEwi0rN6n8NzNAhVnJsAKHc00AMkQ6AEIKDAC#v=onepage&q=the%20Radon-Nikodym%20property&f=false), 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](http://link.springer.com/chapter/10.1007/978-1-4419-7515-7_13#page-1). 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 $L\_1(\mu)$). 


