Failure of Banach Alaoglu in $L^1$ We know that dual of $L^{\infty}(\mathbb{R})$ is not $L^{1}(\mathbb{R})$ and hence Banach Alaoglu theorem is not applicable for the $(L^1,L^{\infty})$ pairing.
In other words if $\{f_n\}$ is a uniformly bounded sequence in $L^1(\mathbb{R})$ (i.e. $||f_n||_{L^1(\mathbb{R})} \leq C ),$ then there may not exist subsequence $f_{n_k}$ and an $f\in L^1(\mathbb{R})$ such that $\int\limits_{\mathbb{R}} f_{n_k}(x)g(x) dx \rightarrow \int\limits_{\mathbb{R}}f(x)g(x)dx$ for all $g\in L^{\infty}(\mathbb{R}).$
Is there any simple counter example which demonstrates this?
 A: Try $f_n = n I_{[0,1/n]}$, where $I_{[0,1/n]}$ is the indicator function of $[0,1/n]$.
If $g$ is any bounded continuous function, $\int_{\mathbb R} f_n(x) g(x)\; dx \to g(0)$ as $n \to \infty$.  But there is no $f \in L^1$ such that $\int_{\mathbb R} f(x) g(x)\; dx = g(0)$ for all such $g$.
A: The example in my previous comment $\{f_n(x)=\mathbb{1}_{(n,n+1]}(x): n\in\mathbb{N}\}$ is uniformly bounded in $L_1(\mathbb{R},\lambda)$ (here $\lambda$ is Lebesgue's measure on the real line) and has not weakly convergence sequence: $\int \mathbf{1}\, f_n=1$ for all $n$ and for any $g\in L_\infty$ with compact support $\int f_ng\xrightarrow{n\rightarrow\infty}0$.
A deep theorem in analysis states that under some additional conditions on $\Phi:=(f_n:n\in\mathbb{N})$ (uniform integrability) the $\Phi$ is weakly compact (compact under the topology $\sigma(L_\infty,L_1)$.

Theorem (Dunford-Pettis):  Suppose $(\Omega,\mathscr{F},\mu)$ is $\sigma$-finite. A subset $K\subset L_1$ is $\sigma(L_1,L_\infty)$--relatively compact iff $K$ is uniformly integrable.

See Bogachev, V., Measure Theory, Vol I, Springer, 2007, pp.285-286.
The result in many textbooks is stated for probability spaces; the result generalizes easily to $\sigma$-finite spaces.
Notice that $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$ is $\sigma$-finite. The sequence in my counter example $f_n=\mathbb{1}_{(n,n+1]}(x)$ is not uniformly integrable:
$$\inf_{g\in L^+_1}\sup_n\int_{\{f_n>g\}}f_n(x)\,dx=1>0$$
The counterexample given by our stemmed member Robert Israel also fails to be uniformly integrable.
Uniform integrability in a sense prevents for mass to concentrate in sets of measure zero (in Robert's example) or to escape to infinity (in my example)
