On a finite-measure space, a bounded $L^p$ functional is integration against its Radon-Nikodym density

By [L1] the induced measure is finite signed and absolutely continuous. By [L6] its total variation is finite; the constant exhaustion $X_n=X$ is common and finite for it and $\mu$. Under AC, [L2] therefore applies: choose $g\in L^1(\mu )$ such that $\nu (E)={\int }_{E}gd\mu (E\in \mathcal{A})\mathrm{.}$ Since $\nu (E)=\mathrm{\Lambda }([{\mathbf{1}}_E])$, this means $\mathrm{\Lambda }([{\mathbf{1}}_E])=\int {\mathbf{1}}_{E}gd\mu (E\in \mathcal{A})\mathrm{.}$

Let $u$ be bounded and measurable, with $\mathrm{\mid }u\mathrm{\mid }\le M$. By [L3], choose simple functions $(s_n)$ such that $\mathrm{\mid }{s}_n\mathrm{\mid }\le M$ for every $n$ and $s_n\to u$ pointwise. Since $\mu (X)<\mathrm{\infty }$, $\mathrm{\mid }{s}_n-u{\mathrm{\mid }}^p\le (2M{)}^p{\mathbf{1}}_X,$ and the majorant is integrable. Therefore [L4] gives $\mathrm{\parallel }[s_n]-[u]{\mathrm{\parallel }}_p^p=\int \mathrm{\mid }{s}_n-u{\mathrm{\mid }}^{p}d\mu ⟶0,$ so continuity of $\mathrm{\Lambda }$ yields $\mathrm{\Lambda }([s_n])⟶\mathrm{\Lambda }([u])\mathrm{.}$

For each simple approximant $s_n$, step 1.1 and linearity give $\mathrm{\Lambda }([s_n])=\int s_{n}gd\mu \mathrm{.}$ Thus the formula already holds for every simple function, and in particular for the chosen sequence $(s_n)$.

Because $g\in L^1(\mu )$ and $\mathrm{\mid }(s_n-u)g\mathrm{\mid }\le 2M\mathrm{\mid }g\mathrm{\mid }$ with $2M\mathrm{\mid }g\mathrm{\mid }\in L^1(\mu )$, [L4] gives $\int s_{n}gd\mu ⟶\int ugd\mu \mathrm{.}$ Combining this with step 1.2 and step 2.1 yields $\mathrm{\Lambda }([u])=\int ugd\mu \mathrm{.}$ So the Radon-Nikodym density represents $\mathrm{\Lambda }$ on every bounded measurable representative.