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For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

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.