- Let $(X,\mu,M)$ be a measure space with $X=[0,1]$, $\mu$=counting measure and $M$ is the $\sigma$-algebra contains all the subset $E$ of $[0,1]$ such that either $E$ or $E^c$(complement) is countable. Since $X$ is not $\sigma$-finite in $\mu$, $(L^1)^* \neq L^{\infty}$. The proof is as follow:
Let $g(x)=x$ which is not measurable. Let $\Lambda f=\int fg d\mu$ for all $f\in L^1$, the integration here makes sense since we can prove $fg$ is measurable. We check that $\|\Lambda \|\leq1$ hence $\Lambda $ is a linear bounded functional. Suppose there is $h \in L^{\infty}$ such that $\Lambda f=\int fh d\mu=\int fg d\mu$, then we can choose some special $f$ to obtain $h=g$ a.e.. Thus we have a contradiction here since $h$ is measurable but $g$ is not.
2 Let $(X,\mu,M)$ be a measure space with $X=[0,1]$, $\mu$=Lebesgue measure. It is well known that $(L^{\infty})^* \neq L^{1}$. The proof is as follow:
Let $C(X)$ be the space containing all continuous functions in $X$, which is proper closed subspace of $L^{\infty}$. Thus, by Hahn-Banach theorem, there is $\Lambda \in (L^{\infty})^*$ such that $\Lambda =0$ on $C(X)$ and $\Lambda \neq 0$ in $L^{\infty}$. Suppose there is $g\in L^1$ such that $\Lambda f= \int fgd\mu $. But $\Lambda f= \int fgd\mu=0 $ for all $f \in C(X)$, we can prove that $g=0$ which implies $\Lambda = 0$ contradicts to $\Lambda \neq 0$ in $L^{\infty}$.
In both of the proof, it seems that they assume that the space $X$ is the dual of $Y$ if and only if they have a Riesz representation. But I am not sure if it is true. The above proof shows that the map $g \mapsto \Lambda$ by $\Phi(g)=\Lambda_g(f)=\int fg d\mu$ is not a isometry (not onto). But it does not mean that there exist no other isometry between the space.
Best Answer
The point of the dual is to use its elements. For instance you don't get anything useful from saying that the dual of $\mathbb C^4$ is isomorphic to $M_2 (\mathbb C) $, but you do from saying that it is isomorphic to $\mathbb C^4$ with the duality given by the inner product.
In the case of $L^1$ and $L^\infty $, "integration against" gives you natural isometric embeddings $$L^\infty\subset (L^1)^*,\ \ \ \ \ L^1\subset (L^\infty)^*. $$The examples you quoted show that in general both embeddings fail to be onto.