Proving that $L^1(X,M,\mu)$ is not reflexive

Solution 1:

I'll turn my comment into an answer. Suppose that $E_i$ are pairwise disjoint sets which satisfy $\mu(E_i)\in (0,\infty)$ for all $i$ and call $e_i = \mu(E_i)^{-1} \chi_{E_i}$, where $\chi$ denotes the indicator function of a set.

By the Eberlein-Smulian theorem, sequential weak compactness is the same as weak compactness for Banach spaces, so it suffices to argue that the sequence $(e_i : i =1,\dots)$, which lie in the unit ball of $L^1(X,M,\mu)$, cannot have a weakly convergent subsequence. Suppose that a subsequence $e_{n_k}$ converges weakly to some $\tilde{e}$ and define $$ g = \sum_{k=1}^\infty (-1)^k e_{n_k}. $$ Notice that because the $E_i$ are pairwise disjoint, $\|g\|_{L^\infty(X,M,\mu)} = 1.$ Define a bounded linear functional on $L^1(X,M,\mu)$ by $$ \Lambda(f) = \int_{X} g f d\mu $$ so that $|\Lambda(f)| \leq \|f\|_{L^1(X,M,\mu)}$. By construction, we have $\Lambda(e_{n_k}) = (-1)^k$, which does not converge, a contradiction.