For $1<p<\mathrm{\infty }$, the same representation theorem holds on arbitrary measure spaces

Let $\mathcal{S}$ be the family of sigma-finite measurable subsets of $X$. For each $E\in \mathcal{S}$, define the restricted functional ${\mathrm{\Lambda }}_E([u]):=\mathrm{\Lambda }([\tilde{u}])$ on $L^p(\mu {\mathrm{\mid }}_E)$, where $\tilde{u}$ is extended by $0$ outside $E$. By [L1], there is a unique $g_E\in L^q(\mu {\mathrm{\mid }}_E)$ such that ${\mathrm{\Lambda }}_E([u])={\int }_{E}u{g}_{E}d\mu ([u]\in L^p(\mu {\mathrm{\mid }}_E)),$ and $\mathrm{\parallel }{g}_E{\mathrm{\parallel }}_q=\mathrm{\parallel }{\mathrm{\Lambda }}_E\mathrm{\parallel }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{.}$

The empty set lies in $\mathcal{S}$ and has $\mathrm{\mid }{g}_{\mathrm{\varnothing }}{\mathrm{\parallel }}_q=0$, so the bounded set below is nonempty. Put $s:=\sup \{\mathrm{\parallel }{g}_E{\mathrm{\parallel }}_q^q:E\in \mathcal{S}\}\le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q\mathrm{.}$ For each $n\ge 1$, the definition of the supremum makes the family $\{E\in \mathcal{S}:\mathrm{\parallel }{g}_E{\mathrm{\parallel }}_q^q>s-2^{-n}\}$ nonempty. By [L0], choose one $E_n$ from each family. Because each $E_n$ is sigma-finite, the family of sequences $(E_{n,k}{)}_{k\ge 1}$ of measurable finite-measure subsets of $E_n$ whose union is $E_n$ is nonempty. Using [L0] again, choose one such sequence for every $n$. Let $Z:={\bigcup }_{n=1}^{\mathrm{\infty }}{E}_n\mathrm{.}$ The doubly indexed family $(E_{n,k}{)}_{n,k\ge 1}$ has union $Z$, and a diagonal enumeration of it proves that $Z$ is sigma-finite. By [L1], there is a unique $g_Z\in L^q(\mu {\mathrm{\mid }}_Z)$ representing ${\mathrm{\Lambda }}_Z$. Since $E_n\subseteq Z$ and both $g_Z{\mathrm{\mid }}_{E_n}$ and $g_{E_n}$ represent ${\mathrm{\Lambda }}_{E_n}$, uniqueness in [L1] gives $g_Z=g_{E_n}$ almost everywhere on $E_n$. Therefore $\mathrm{\parallel }{g}_Z{\mathrm{\parallel }}_q^q\ge \mathrm{\parallel }{g}_{E_n}{\mathrm{\parallel }}_q^q>s-2^{-n}(n\ge 1)\mathrm{.}$ Since $Z\in \mathcal{S}$, the definition of $s$ also gives $\mathrm{\parallel }{g}_Z{\mathrm{\parallel }}_q^q\le s$, and hence $\mathrm{\parallel }{g}_Z{\mathrm{\parallel }}_q^q=s$.

Let $g$ be the zero extension of a measurable representative of $g_Z$ from $Z$ to $X$. Since $q<\mathrm{\infty }$, step 2.1 gives ${\int }_X\mathrm{\mid }g{\mathrm{\mid }}^{q}d\mu ={\int }_Z\mathrm{\mid }{g}_Z{\mathrm{\mid }}^{q}d\mu =s<\mathrm{\infty },$ so $g\in L^q(\mu )$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_q^q=s$. Now let $E\in \mathcal{S}$. Then $Z\cup E$ is sigma-finite, so [L1] supplies $g_{Z\cup E}$. Uniqueness on $Z$ and on $E$ gives $g_{Z\cup E}=g_Z\text{ a.e. on }Z,g_{Z\cup E}=g_E\text{ a.e. on }E\mathrm{.}$ Therefore $\mathrm{\parallel }{g}_{Z\cup E}{\mathrm{\parallel }}_q^q={\int }_Z\mathrm{\mid }{g}_Z{\mathrm{\mid }}^{q}d\mu +{\int }_{E\setminus Z}\mathrm{\mid }{g}_E{\mathrm{\mid }}^{q}d\mu \mathrm{.}$ The left side is at most $s$ by definition of $s$, while the first term is $s$ by step 2.1. Hence ${\int }_{E\setminus Z}\mathrm{\mid }{g}_E{\mathrm{\mid }}^{q}d\mu =0,$ so $g_E=0$ almost everywhere on $E\setminus Z$ and $g_E=g$ almost everywhere on $E\cap Z$.

Let $[f]\in L^p(\mu )$. By [L2], choose a sigma-finite measurable set $S\subseteq X$ such that $f=0$ almost everywhere on $X\setminus S$. Applying step 1.1 to $S$ gives $g_S\in L^q(\mu {\mathrm{\mid }}_S)$ with $\mathrm{\Lambda }([f])={\mathrm{\Lambda }}_S([f{\mathrm{\mid }}_S])={\int }_{S}f{g}_{S}d\mu \mathrm{.}$ Step 3.1 gives $g_S=g$ almost everywhere on $S$, so $\mathrm{\Lambda }([f])={\int }_{S}fgd\mu ={\int }_{X}fgd\mu \mathrm{.}$ Thus $g$ represents $\mathrm{\Lambda }$ on all of $L^p(\mu )$.

If $h\in L^q(\mu )$ also represents $\mathrm{\Lambda }$, then $h-g\in L^q(\mu )$ and $q<\mathrm{\infty }$. By [L2], choose a sigma-finite measurable set $S$ such that $h-g=0$ almost everywhere on $X\setminus S$. The restrictions of $h$ and $g$ to $S$ represent the same bounded functional on the sigma-finite space $S$, so [L1] gives $h=g$ almost everywhere on $S$. Together with the choice of $S$, this proves $h=g$ almost everywhere on $X$. Finally [L3] gives $\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }=\mathrm{\parallel }g{\mathrm{\parallel }}_q\mathrm{.}$ So the representing class is unique and has the correct norm.