The Radon-Nikodym density of a bounded $L^p$ functional belongs to $L^q$

Assume first $1<p<\mathrm{\infty }$. Choose a representative $u$ of $g$ and, for $k\ge 1$, define $u_k:=u{\mathbf{1}}_{\{\mathrm{\mid }u\mathrm{\mid }\le k\}},f_k:=\mathrm{\mid }{u}_k{\mathrm{\mid }}^{q-1}\mathrm{sgn}u\mathrm{.}$ Each $f_k$ is bounded, so [L1] gives $\mathrm{\Lambda }([f_k])=\int f_{k}gd\mu =\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \mathrm{.}$ Also $\mathrm{\mid }{f}_k{\mathrm{\mid }}^p=\mathrm{\mid }{u}_k{\mathrm{\mid }}^q$ by [L2], hence $\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu =\mathrm{\mid }\mathrm{\Lambda }([f_k])\mathrm{\mid }\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }\mathrm{\parallel }[f_k]{\mathrm{\parallel }}_p=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }{(\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu )}^{1\mathrm{/}p}\mathrm{.}$ Therefore $\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q(k\ge 1)\mathrm{.}$

Assume instead $p=1$, so $q=\mathrm{\infty }$. Put $M:=\mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$ and let $u$ be a representative of $g$. Fix $t>M$ and set $E_t:=\{\mathrm{\mid }u\mathrm{\mid }>t\}$. If $\mu (E_t)>0$, then $E_t$ has finite measure because $\mu (X)<\mathrm{\infty }$, and $f_t:=\mathrm{sgn}u{\mathbf{1}}_{E_t}$ is bounded with $\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1=\mu (E_t)$. By [L1], $\mathrm{\mid }\mathrm{\Lambda }([f_t])\mathrm{\mid }={\int }_{E_t}\mathrm{\mid }u\mathrm{\mid }d\mu >t\mu (E_t)=t\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1,$ contradicting the bound $\mathrm{\mid }\mathrm{\Lambda }([f_t])\mathrm{\mid }\le M\mathrm{\parallel }[f_t]{\mathrm{\parallel }}_1$. Hence $\mu (E_t)=0$ for every $t>M$, and [L4] gives $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M$.

Step 1.1 and monotone convergence prove the strict-exponent case, while step 1.2 proves the endpoint case. Indeed, from step 1.1 the sequence $\mathrm{\mid }{u}_k{\mathrm{\mid }}^q$ increases pointwise to $\mathrm{\mid }u{\mathrm{\mid }}^q$, so [L3] gives $\int \mathrm{\mid }u{\mathrm{\mid }}^{q}d\mu ={\lim }_{k\to \mathrm{\infty }}\int \mathrm{\mid }{u}_k{\mathrm{\mid }}^{q}d\mu \le \mathrm{\parallel }\mathrm{\Lambda }{\mathrm{\parallel }}^q\mathrm{.}$ Thus in every case $g\in L^q(\mu )$ and $\mathrm{\parallel }g{\mathrm{\parallel }}_q\le \mathrm{\parallel }\mathrm{\Lambda }\mathrm{\parallel }$.

Let $[f]\in L^p(\mu )$ and choose a representative $w$. For each $n\ge 1$, set $w_n:=w{\mathbf{1}}_{\{\mathrm{\mid }w\mathrm{\mid }\le n\}}\mathrm{.}$ Then each $w_n$ is bounded and $\mathrm{\mid }{w}_n-w{\mathrm{\mid }}^p\le \mathrm{\mid }w{\mathrm{\mid }}^p$ with $\mathrm{\mid }{w}_n-w{\mathrm{\mid }}^p\to 0$ pointwise, so dominated convergence gives $\mathrm{\parallel }[w_n]-[f]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{w}_n-w{\mathrm{\mid }}^{p}d\mu ⟶0.$ By step 2.1, $g\in L^q(\mu )$, so [L5] shows that $I_g([h]):=\int hgd\mu$ is a bounded linear functional on $L^p(\mu )$. Since [L1] gives $\mathrm{\Lambda }([w_n])=I_g([w_n])$ for every $n$, continuity of both functionals yields $\mathrm{\Lambda }([f])=I_g([f])=\int wgd\mu \mathrm{.}$ Thus the Radon-Nikodym density represents $\mathrm{\Lambda }$ on all of $L^p(\mu )$.