Showing $L^{\infty}(\mathbb R)^*\neq L^1(\mathbb R)$ using the weak star compactness of unit ball in $L^{\infty}(\mathbb R)^*$. $\lambda_n(f)=\frac{1}{2n}\int_{-n}^{n}f$ defines  a dual element of  $ L^{\infty}(\mathbb R)$. It is easy to see that $\lambda_n\in L^{\infty}(\mathbb R)^*_1.$
By using the weak-star compactness of the unit ball we can say that  $\{\lambda_n\}_{n\in \mathbb N}$ has a weak-star accumulation point  $\lambda$. Then we will be done if we can show that  $\lambda(f)$ is not given by $\int fg $ for any $g\in L^1(\mathbb R)$.
How can I show that there is no $g\in L^1(\mathbb R)$ such that $\frac{1}{2n}\int_{-n}^{n}f\rightarrow \int\ fg$ for all $f\in L^{\infty}(\mathbb R)$?
Any help would be appreciated. Thanks in advance.
 A: Your sequence $\lambda_n=\frac{1}{2n}\mathbb{1}_{(-n,n)}$, considered as a sequence in $L^*_\infty(\mathbb{R})$, is indeed in the unit ball of $L^*_\infty(\mathbb{R})$ however,  $\sigma(L^*_\infty, L_\infty)$-subsequential compactness does not apply here since $L_\infty(\mathbb{R})$ is not separable.

To show that $L_1(\mathbb{R}) \subsetneq L^*_\infty(\mathbb{R})$ notice that $\lambda(f)=\lim_{n\rightarrow\infty}\lambda_n(f)$ is well define for $f\in C_0\oplus\mathbb{R}$ (or rather the measure-0 mod classes of equivalence in $C_0\oplus\mathbb{R}$ and that  $|\lambda(f)|\leq\|f\|_\infty$ for $f\in C_0\oplus\mathbb{R}$. You can use Hahn-Banach to extend $\lambda$ to all of $L_\infty$ as an element in $L^*_\infty(\mathbb{R})$.
The Banach limit (or sub sequential limit) $\lambda$ is finitely additive but not countably additive. Indeed, $(0,\infty)=\bigcup^\infty_{m=0}(m,m+1]$, $\lambda(\mathbb{1}_{(0,\infty)})=\frac12$ but $\lambda(\mathbb{1}_{(m,m+1]})=0$ for each $m$.
From that you see that $\lambda$ is not even a Radon measure on $\mathbb{R}$ with the Borel structure.
