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For a nonnegative measurable $f$, defineEach $f_n$ takes only finitely many values, its level sets are measurable, and $f_n(x)\to f(x)$ pointwise. Thus measurable nonnegative functions are pointwise limits of simple functions.

If $f$ is integrable, the same functions satisfy $0\le f_n\le f$ and converge pointwise. The dominated convergence theorem givesThey are integrable because they are bounded above by $f$.

If $0\le f\le M$, instead setThis has finitely many values andeverywhere, so $g_n\to f$ uniformly.

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.