 Singular functional on L infinity

Point evaluation at zero is bounded on $C([0,1])$ with the essential-supremum norm, and Hahn--Banach extends it to a bounded functional on $L^{\infty }([0,1])$. No $g\in L^1$ represents this extension: continuous functions supported in shrinking neighbourhoods of zero have value one at zero while their integrals against $g$ tend to zero. Hence $(L^{\infty }{)}^{\prime }$ is strictly larger than $L^1$.