= Solution
For a measurable set $E\subseteq U$, define
$$
\nu(E)=\Lambda(\mathbf1_E),
$$
where $\mathbf1_E$ is its <indicator function>. This is finite because $U$ has <finite measure> and hence $\mathbf1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E=\bigcup_{k\geq1}E_k$, then
$$
\left\|\mathbf1_E-
\sum_{k=1}^N\mathbf1_{E_k}\right\|_p^p
=|E\setminus\bigcup_{k=1}^NE_k|\longrightarrow0.
$$
The continuity of the <positive linear functional> $\Lambda$ therefore gives
$$
\nu(E)=\sum_{k=1}^{\infty}\nu(E_k),
$$
so $\nu$ is a <measure> by <countable additivity>. Moreover, a set of zero <Lebesgue measure> has zero indicator in $L^p$, so $\nu$ is <absolute continuity of measures>[absolutely continuous] with respect to Lebesgue measure.
The <Radon-Nikodym theorem> now supplies a measurable function $\omega\geq0$ such that
$$
\nu(E)=\int_E\omega.
$$
Consequently, first for nonnegative simple functions and then, by <approximation by nonnegative simple functions>, for every nonnegative $f\in L^p(U)$,
$$
\int_U f\omega=\Lambda(f)
\leq\|\Lambda\|\,\|f\|_p.
$$
Applying the stated characterization of the <Lp norm> to $\omega$ shows that $\omega\in L^q(U)$ and $\|\omega\|_q\leq\|\Lambda\|$. The <Holder inequality> makes $f\mapsto\int_Uf\omega$ continuous on $L^p(U)$, while bounded simple functions form a <dense subset>; hence
$$
\boxed{\Lambda(f)=\int_U f\omega\quad(f\in L^p(U)).}
$$
This is the <positive functional representation on Lp>.
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