$(L^\infty)^*$ is not isomorphic to $L^1$

I am working on this problem in Rudin:

Let $L^\infty=L^\infty(m)$, where $m$ is Lebesgue measure on $I=[0,1]$. Show that there is a bounded linear functional $\lambda\neq 0$ on $L^\infty$ that is $0$ on $C(I)$, and that therefore there is no $g\in L^1(m)$ that satisfies $\lambda f=\int_I fg dm$ for every $f\in L^\infty$. Thus $(L^\infty)^*\neq L^1$.

I've figured out all but the part that saids " $(L^\infty)^*\neq L^1$". Most of the problem seems to be showing that the normal method of showing $(L^p)^* \cong L^q$ doesn't work. However, I don't see how that helps me do the last deduction.

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The original problem isn't about reiterating the other proof and finding a gap in the copycat argument — it asks for something much stronger. It is about constructing a specific counterexample to the inclusion $(L^\infty)^* \subseteq L^1$ (or at least proving its existence). – Erick Wong Feb 12 '13 at 3:53
You know that every $g \in L^1$ gives a linear functional $\lambda_g \in (L^{\infty})^\ast$ by $f \mapsto \int fg$ and that for $g_1 \neq g_2$ you have $\lambda_{g_1} \neq \lambda_{g_2}$. This shows that $L^1 \subseteq (L^\infty)^\ast$. You also know that the only $g \in L^1$ such that $\int fg$ for all $f \in C(I)$ is $g = 0$. Therefore the functional $\lambda$ you have constructed is not of the form $\lambda_g$ for any $g \in L^1$ and hence $L^1 \subsetneqq (L^\infty)^\ast$. – Martin Feb 12 '13 at 6:26

Every $g \in L^1$ yields a linear functional $\lambda_g \in (L^\infty)^\ast$ defined by $\lambda_g(f) = \int fg\,dm$ on $L^\infty$.
The estimate $\lvert \lambda_g(f)\rvert \leq \lVert f \rVert_\infty \lVert g\rVert_1$ shows that $\lVert \lambda_g\rVert \leq \lVert g\rVert_1$, so $\lambda_g$ is continuous. Taking $$f(x) = \begin{cases} 0 & \text{if } g(x) = 0 \cr \frac{\overline{g(x)}}{\lvert g(x)\rvert} & \text{if } g(x) \neq 0 \end{cases}$$ we get $\lambda_g(f) = \lVert g\rVert_1$, so $\lVert \lambda_g\lVert \geq \lVert g \rVert_1$. Thus, the linear map $L^1 \to (L^\infty)^\ast, g \mapsto \lambda_g$ is isometric and hence injective. We can therefore view $L^1 \subseteq (L^\infty)^\ast$ as a subspace.
The point of Rudin's exercise is to establish that $L^1 \subsetneqq (L^\infty)^\ast$, in words $L^1$ is a proper subspace of $(L^\infty)^\ast$.
Indeed, let $0 \neq \lambda \in (L^\infty)^\ast$ be a linear functional such that $\lambda(f) = 0$ for all $f \in C(I)$ which you say you have proved to exist. Then $\lambda$ can't be of the form $\lambda = \lambda_g$ for any $g \in L^1$ since for $g \in L^1$ the vanishing $\int gf\,dm = 0$ for all $f \in C(I)$ implies that $g = 0$ and hence $\lambda_g = 0$. But we have $\lambda \neq 0$ and thus $\lambda \in (L^\infty)^\ast \setminus L^1$.
You're welcome. By the way: $L^1(X,\Sigma,\mu)$ is almost always a proper subspace of $L^\infty(X,\Sigma,\mu)^\ast$ except in very degenerate cases like $X$ a finite set. See When is $L^1 = (L^\infty)^\ast$? for a precise statement. – Martin Feb 12 '13 at 22:10
For $g\in L^1$ such that $\int gf\,dm=0$ for all $f\in C(I)$, then $g=0$, how to show this? – Xiang Yu Jan 5 at 11:59