Definition of the Nonnegative Integral

From Measurability to Integration

In the previous tutorial, nonnegative measurable functions were allowed to be unbounded or to take the value $+\infty$. Measurability supplies useful structure, such as measurable level sets and finite-valued truncations, but it does not assign a numerical size to a function. The next step is to define that size using simple functions that lie below it.

A nonnegative simple function has a well-defined integral from the earlier tutorial on the Integral of a Simple Function. We use those integrals as lower estimates. The integral of a general nonnegative measurable function is the least upper bound of all such estimates. This definition permits the value $+\infty$, and it does not require the function itself to be bounded or finite everywhere.

Definition: Let $(X,\mathcal{F},\mu)$ be a measure space, and let $f:X\to[0,\infty]$ be extended nonnegative measurable. Define $$ \int_X f\,d\mu = \sup\left\{ \int_X s\,d\mu: s\text{ is a nonnegative measurable simple function and }s\leq f\text{ on }X \right\}. $$ The supremum is taken in $[0,\infty]$. The simple function $s=0$ is always among the functions being considered, so the set of candidate integrals is nonempty.

The condition $s\leq f$ is pointwise: it must hold at every $x\in X$, including points where $f(x)=+\infty$. A candidate $s$ is a finite-valued simple function, even though $f$ may be extended-valued. Its integral may nevertheless be infinite if one of its positive level sets has infinite measure.

The supremum is not necessarily achieved by a candidate. The definition says that no candidate integral exceeds $\int_X f\,d\mu$, and that candidate integrals come arbitrarily close to it when it is finite. If the supremum is infinite, candidate integrals are unbounded. These facts follow from the least-upper-bound property, and are useful ways to interpret the definition.

Compatibility with the Simple-Function Integral

The new definition must give the same answer when the function being integrated is already simple. This is important: the definition extends the earlier one rather than replacing it with a different convention.

Theorem (Agreement with the Simple-Function Integral): If $f:X\to[0,\infty)$ is a measurable simple function, then its integral as defined by the supremum of lower simple functions equals its previously defined simple-function integral.

Proof. Since $f$ is itself a nonnegative measurable simple function and $f\leq f$, it is an admissible candidate in the defining supremum. Thus the supremum is at least the previously defined value $\int_X f\,d\mu$.

Now let $s$ be any admissible candidate, so $s$ is a nonnegative measurable simple function and $s\leq f$. By the Monotonicity of the Lebesgue Integral established earlier in the course, $\int_X s\,d\mu\leq\int_X f\,d\mu$. Every candidate integral is therefore at most the previously defined integral of $f$, so their supremum is at most that value. The two inequalities prove equality. $\square$

In particular, the finite-value formula from the Integral of a Simple Function tutorial can still be used whenever the integrand is simple. The definition by supremum matters when the integrand is not simple: in that case, it collects the integral values of all simple lower estimates.

Indicators and Simple Integrands

An indicator function is the simplest useful test of the definition. For a measurable set $A$, the function $\mathbf{1}_A$ is simple, so its integral is already covered by the agreement theorem. This yields a direct connection between integration and measure.

Theorem (Integral of an Indicator): If $A\in\mathcal{F}$, then $$ \int_X \mathbf{1}_A\,d\mu=\mu(A), $$ where either side may be $+\infty$.

Proof. The function $\mathbf{1}_A$ is a nonnegative measurable simple function. By the agreement theorem, its integral under the supremum definition is its simple-function integral. The simple-function formula gives $\int_X\mathbf{1}_A\,d\mu=1\cdot\mu(A)+0\cdot\mu(X\setminus A)=\mu(A)$. This holds whether $\mu(A)$ is finite or infinite. $\square$

The convention in this formula is that a zero value contributes zero, including when the corresponding set has infinite measure. There is no ambiguous product $0\cdot\infty$ to evaluate: the zero level contributes nothing to the simple-function integral.

Worked Example: Integrating an Indicator on an Infinite-Measure Set

Take $X=\mathbb{R}$ with Lebesgue measure and $A=[3,\infty)$. The indicator $\mathbf{1}_A$ equals $1$ on $A$ and $0$ outside $A$. Since $\lambda(A)=\infty$, the Integral of an Indicator theorem gives $$ \int_{\mathbb{R}}\mathbf{1}_{[3,\infty)}\,d\lambda =\lambda([3,\infty)) =\infty. $$ The function is bounded and simple, but its integral is infinite because it is positive on a set of infinite measure.

This example shows that boundedness of the function alone does not guarantee a finite integral. The measure of the region where the function is positive also matters.

Worked Example: Integrating a Two-Level Simple Function

Let $X=[0,5]$ with Lebesgue measure, $A=[0,1]$, and $B=(1,5]$. These sets are disjoint, have union $X$, and have measures $\lambda(A)=1$ and $\lambda(B)=4$. Define $s=6\mathbf{1}_A+2\mathbf{1}_B$. Then $s$ equals $6$ on $A$ and $2$ on $B$, so the simple-function formula gives $$ \int_X s\,d\lambda =6\lambda(A)+2\lambda(B) =6\cdot1+2\cdot4 =14. $$ By the Agreement with the Simple-Function Integral theorem, the supremum definition gives exactly the same value.

The value $14$ also makes clear why the integral is not simply the largest value of the function times the measure of its support: the two regions contribute according to their own function values and measures.

The Supremum as an Approximation Principle

Although a supremum need not be attained, it provides a precise approximation statement. This is often the practical way to use the definition: propose simple lower estimates, calculate their integrals, and show that those values approach a target or become arbitrarily large.

Theorem (Simple-Function Approximation of the Supremum): Let $f:X\to[0,\infty]$ be extended nonnegative measurable. If $I=\int_X f\,d\mu<\infty$, then for every $\varepsilon>0$ there is a nonnegative measurable simple function $s\leq f$ such that $$ I-\varepsilon<\int_X s\,d\mu\leq I. $$ If $\int_X f\,d\mu=\infty$, then for every finite $M\geq0$ there is such an $s$ with $\int_X s\,d\mu>M$.

Proof. Let $S$ be the nonempty set of candidate integrals in the definition, so $I=\sup S$. Every element of $S$ is at most $I$, which gives the upper bound in the finite case. If $I-\varepsilon$ were an upper bound for $S$, then $\sup S\leq I-\varepsilon<I$, a contradiction. Thus some element of $S$ is greater than $I-\varepsilon$, giving the required candidate $s$. In the infinite case, if every element of $S$ were at most $M$, then $M$ would be a finite upper bound and $\sup S\leq M$, contradicting $\sup S=\infty$. Hence a candidate with integral greater than $M$ exists. $\square$

The theorem does not say that the simple functions must approximate $f$ uniformly, or even that one candidate works for every desired accuracy. It says that the integrals of admissible simple functions approximate the supremum in the precise sense stated. The Increasing Simple Approximation theorem from earlier in the course supplies a particularly useful sequence of simple functions below a finite-valued nonnegative measurable function; the Monotone Convergence Theorem, in the next tutorial, will explain how integrals behave along increasing sequences of this kind.

Worked Example: An Infinite Integral on the Counting Measure Space

Let $X=\mathbb{N}$, equip it with the power-set sigma-algebra and counting measure, and define $f(n)=1/n$. This is a nonnegative measurable function. For each positive integer $N$, define $$ s_N(n)= \begin{cases} 1/n,&1\leq n\leq N,\\ 0,&n>N. \end{cases} $$ Each $s_N$ has finite range, is measurable, and satisfies $0\leq s_N\leq f$. Its integral is a finite sum: $$ \int_{\mathbb{N}}s_N\,d\mu=\sum_{n=1}^{N}\frac{1}{n}. $$

These sums are unbounded. For each integer $k\geq1$, the terms with $2^{k-1}<n\leq2^k$ number $2^{k-1}$, and each is at least $1/2^k$. Their sum is therefore at least $2^{k-1}/2^k=1/2$. As more such blocks are included, the partial sums grow without bound. The Simple-Function Approximation of the Supremum theorem, or directly the definition, now gives $$ \int_{\mathbb{N}}f\,d\mu=\infty. $$ No single finite-support candidate is responsible for this infinite value; the candidate integrals have no finite upper bound.

Nonnegative Homogeneity

The definition also gives a useful scaling rule. Scaling a function by a nonnegative constant scales every simple lower estimate by the same constant. The zero constant requires separate attention, since the integral of the zero function is zero even when the measure space has infinite measure.

Theorem (Nonnegative Homogeneity of the Integral): If $f:X\to[0,\infty]$ is extended nonnegative measurable and $c\geq0$ is finite, then $$ \int_X cf\,d\mu=c\int_X f\,d\mu, $$ with $c\cdot\infty=\infty$ for $c>0$.

Proof. If $c=0$, then $cf=0$ everywhere, so $\int_X cf\,d\mu=0=0\cdot\int_X f\,d\mu$. Suppose $c>0$. If $s$ is a nonnegative measurable simple function with $s\leq f$, then $cs$ is a nonnegative measurable simple function and $cs\leq cf$. The simple-function homogeneity established earlier gives $\int_X cs\,d\mu=c\int_X s\,d\mu$. Conversely, if $t$ is a nonnegative measurable simple function with $t\leq cf$, then $t/c$ is a nonnegative measurable simple function and $t/c\leq f$. Thus the candidates below $cf$ are exactly the functions $cs$ for candidates $s$ below $f$.

Taking suprema therefore gives $\int_X cf\,d\mu=c\sup\{\int_X s\,d\mu:0\leq s\leq f,\ s\text{ simple}\}=c\int_X f\,d\mu$. Multiplication by $c>0$ preserves suprema of nonnegative sets, including an unbounded set, so this reasoning also covers the case $\int_X f\,d\mu=\infty$. $\square$

Worked Example: Scaling a Function with Infinite Integral

For the function $f(n)=1/n$ on $\mathbb{N}$ with counting measure, the preceding example showed that $\int_{\mathbb{N}}f\,d\mu=\infty$. For any fixed $c>0$, nonnegative homogeneity gives $$ \int_{\mathbb{N}}\frac{c}{n}\,d\mu =c\int_{\mathbb{N}}\frac{1}{n}\,d\mu =c\cdot\infty =\infty. $$ For $c=0$, the scaled function is identically zero, and its integral is $0$, not $\infty$. This is why the zero case is stated separately in the theorem.

What the Definition Does—and Does Not—Say

The integral is built from measurable simple functions lying below the integrand. This lower-bound requirement is essential: arbitrary simple functions need not give estimates for $f$, and simple functions lying above $f$ are not the candidates in this definition. The integral can be finite or infinite, even when $f$ is finite at every point. Conversely, an infinite value of $f$ at some points does not by itself determine the integral; the measure and the rest of the function also matter.

For a measurable set $E$, one can speak of integrating over $E$ by considering the function restricted to that region, or equivalently by using $f\mathbf{1}_E$ on $X$. The indicator theorem shows how this convention recovers $\mu(E)$ when the function is $1$ on $E$. Later results will develop further properties, including how integrals behave under limits. At this stage, the central object is the supremum itself and the simple lower estimates that define it.

Key takeaway: The nonnegative integral of an extended nonnegative measurable function is the supremum of the integrals of all measurable simple functions lying below it. This agrees with the simple-function integral, recovers measure on indicators, and may equal $+\infty$.

Check Your Understanding

Use the supremum definition and the results proved here to check your understanding.

Why is the set of candidate simple-function integrals in the definition always nonempty?
Explain why the supremum definition agrees with the earlier simple-function integral when the integrand is simple.
What is $\int_X\mathbf{1}_A\,d\mu$ when $A$ is measurable and has infinite measure?
For an infinite integral, what does the definition guarantee about candidate simple-function integrals?
Why must the proof of nonnegative homogeneity handle $c=0$ separately?