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The answer isIndeed, Lebesgue measure on $E\subset {\mathbb{R}}^n$ is a sigma-finite measure, and duality of Lp spaces gives $(L^1(E){)}^{\ast }=L^{\infty }(E)$. The restricted Lebesgue measure is a non-atomic measure and $E$ has positive measure, so the Dual of L infinity is larger than L1: the canonical integral representation does not exhaust $(L^{\infty }(E){)}^{\ast }$.

For $1<p<\infty$, the Holder conjugate exponent also lies strictly between one and infinity. Duality of Lp spaces then holds in both directions, in agreement with Reflexivity of Lp spaces. For $p=\infty$, the first required identification already fails.

Let $q$ be the Holder conjugate exponent of $p$. On ${\mathbb{R}}^n$ with Lebesgue measure, the duality of Lp spaces isunder the pairing ${\Lambda }_g(f)=\int fg$. Thus $(L^1{)}^{\ast }=L^{\infty }$, while for $1<p<\infty$ the dual exponent satisfies $1/p+1/q=1$.

A Banach space is reflexive when its canonical map into its bidual is surjective. The spaces $L^p({\mathbb{R}}^n)$ are reflexive exactly forFor these exponents, applying $L^p$ duality twice identifies the bidual with $L^p$. The endpoint spaces $L^1$ and $L^{\infty }$ are not reflexive; for example, disjoint unit cubes give closed copies of ${\ell }^1$ and ${\ell }^{\infty }$, respectively.

Integration gives an isometric embedding $L^1\hookrightarrow (L^{\infty }{)}^{\ast }$, but it is not onto. On the subspacedefine $\Lambda (c+h)=c$, the limit at infinity. Since $|\Lambda (v)|\le \Vert v{\Vert }_{\infty }$, the Hahn-Banach theorem extends $\Lambda$ to all of $L^{\infty }$. If this extension were integration against some $g\in L^1$, its vanishing on every compactly supported bounded function would force $g=0$ almost everywhere, but $\Lambda (1)=1$. ThereforeThis is the Dual of L infinity is larger than L1.