The Nonnegative Integral as a Supremum

The Integral Defined from Below

The nonnegative Lebesgue integral is built by measuring how much integral can be captured by simple functions that lie below a given function. This definition applies whether the underlying measure is finite or infinite, and whether the function itself is bounded or takes the value $+\infty$. In this tutorial, $(X,\mathcal{M},\mu)$ is an arbitrary measure space: no finiteness, sigma-finiteness, or completeness assumption is imposed. Functions in the definition are measurable and take values in $[0,\infty]$; inequalities between functions are pointwise unless explicitly stated otherwise.

A bounded nonnegative simple function is an $\mathcal{M}$-measurable function with finite range in $[0,\infty)$. It can be written using finitely many disjoint measurable sets as $\varphi=\sum_{\ell=1}^r b_\ell\chi_{F_\ell}$, where $b_\ell\ge0$. Its integral is $\int\varphi\,d\mu=\sum_{\ell=1}^r b_\ell\mu(F_\ell)$, with $0\cdot\infty=0$. This finite-sum formula for simple functions is part of the earlier material on integration of nonnegative simple functions.

Definition: If $f:X\to[0,\infty]$ is $\mathcal{M}$-measurable, its nonnegative integral is $$ \int f\,d\mu = \sup\left\{\int\varphi\,d\mu:\ \varphi\text{ is a bounded nonnegative simple function and }0\le\varphi\le f\right\}. $$ The supremum is taken in $[0,\infty]$, so the integral is allowed to be infinite. The zero function is always an admissible minorant.

The condition $\varphi\le f$ is pointwise. In particular, a bounded simple minorant must vanish wherever $f$ vanishes. The supremum need not be achieved by one of the minorants for an arbitrary measurable $f$. For bounded simple $f$, however, $f$ itself is an admissible minorant, and the supremum can be computed exactly.

Indicators

Proposition: For every $E\in\mathcal{M}$, the indicator $\chi_E$ satisfies $$ \int\chi_E\,d\mu=\mu(E), $$ including when $\mu(E)=\infty$.

Proof. The function $\chi_E$ is bounded, nonnegative, and simple, so it is itself an admissible minorant of $\chi_E$. Its simple-function integral is $\mu(E)$. Thus the supremum in the definition is at least $\mu(E)$. For the reverse bound, let $\varphi$ be any admissible minorant and write its level-set representation as $\varphi=\sum_{\ell=1}^r b_\ell\chi_{F_\ell}$, with the $F_\ell$ disjoint and $b_\ell\ge0$. Since $\varphi\le\chi_E$, it vanishes on $E^c$; every level set with positive value is therefore contained in $E$, and its value is at most $1$. Hence $$ \int\varphi\,d\mu =\sum_{\ell=1}^r b_\ell\mu(F_\ell) \le \sum_{\ell:\,b_\ell>0}\mu(F_\ell) \le\mu(E). $$ These inequalities also hold if $\mu(E)=\infty$. Taking the supremum over admissible $\varphi$ gives the upper bound, and hence equality. $\square$

Worked Example: The Indicator of a Finite Interval

Take $X=\mathbb{R}$, $\mathcal{M}=\mathcal{L}$, and $\mu=m$, Lebesgue measure. Let $E=(2,5]$. The indicator $\chi_E$ is bounded and measurable, so it is an admissible minorant of itself. Its simple-function integral is $1\cdot m((2,5])$. The interval has length $5-2=3$, so $m(E)=3$. The proposition gives $\int\chi_E\,dm=3$.

Check: The interval endpoints do not change its Lebesgue measure: $m((2,5])=m([2,5])=5-2=3$. The computed integral agrees with this measure.

If the set has infinite measure, the proposition still gives a definite extended-real value. The integral is $+\infty$; it is not necessary for any one finite-valued minorant to have infinite integral, although in the indicator case $\chi_E$ itself already does.

Worked Example: An Indicator of Infinite Measure

On $(\mathbb{R},\mathcal{L},m)$, let $E=(0,\infty)$. The indicator proposition gives $\int\chi_E\,dm=m((0,\infty))=\infty$. To see the unbounded supremum directly using finite values, for each positive integer $n$ let $\varphi_n=\chi_{(0,n)}$. Then $0\le\varphi_n\le\chi_E$, and the interval $(0,n)$ has measure $n$, so $\int\varphi_n\,dm=n$. For every real $K>0$, choosing an integer $n>K$ gives an admissible minorant with integral greater than $K$; therefore the supremum is infinite.

Check: Each $\varphi_n$ has integral $1\cdot m((0,n))=1\cdot n=n$. These values have no finite upper bound, in agreement with $m(E)=\infty$.

Finite-Valued Nonnegative Functions

Here “finite-valued” means that the function takes only finitely many finite real values; in particular, it does not take $+\infty$. Its positive level sets are measurable and disjoint. The function is zero outside their union, so its integral formula is determined entirely by those level sets. Empty level sets and zero values cause no difficulty and may be omitted.

Proposition: Let $f:X\to[0,\infty)$ be a measurable function with finite range. Write its distinct positive values as $a_1,\ldots,a_k$, and let $E_j=\{x\in X:f(x)=a_j\}$. Then the $E_j$ are pairwise disjoint and measurable, and $$ \int f\,d\mu=\sum_{j=1}^k a_j\mu(E_j). $$ The sum is zero if $k=0$, and it may equal $+\infty$ if some $E_j$ has infinite measure and $a_j>0$.

Proof. Because $f$ has finite range, it is the simple function $\sum_{j=1}^k a_j\chi_{E_j}$; it is zero outside the union of the $E_j$. It is bounded and measurable, so it is an admissible minorant of itself. The simple-function integral formula therefore gives $$ \int f\,d\mu\ge\int f\,d\mu\text{ as a minorant's integral }=\sum_{j=1}^k a_j\mu(E_j). $$ For clarity, this establishes that the supremum is at least the displayed weighted sum.

For the upper bound, fix any admissible bounded nonnegative simple function $\varphi\le f$. Write $\varphi=\sum_{\ell=1}^r b_\ell\chi_{F_\ell}$, where $b_\ell\ge0$ and the $F_\ell$ are disjoint measurable level sets. Partition each $E_j$ by these levels: its pieces are $E_j\cap F_\ell$, together with any portion of $E_j$ where $\varphi=0$. On each piece $E_j\cap F_\ell$, the inequality $\varphi\le f$ gives $b_\ell\le a_j$. Thus, if $\mu(E_j)<\infty$, $$ \int_{E_j}\varphi\,d\mu =\sum_{\ell=1}^r b_\ell\mu(E_j\cap F_\ell) \le a_j\sum_{\ell=1}^r\mu(E_j\cap F_\ell) \le a_j\mu(E_j). $$ The zero-level portion contributes zero. If $\mu(E_j)=\infty$ and $a_j>0$, the required inequality $\int_{E_j}\varphi\,d\mu\le a_j\mu(E_j)$ holds because its right side is $+\infty$. If $a_j=0$, then $E_j$ is not among the positive level sets; on the zero set of $f$, $\varphi=0$. Also, $\varphi=0$ outside $\bigcup_{j=1}^kE_j$, since $\varphi\le f=0$ there. Summing over the finitely many positive level sets gives $$ \int\varphi\,d\mu =\sum_{j=1}^k\int_{E_j}\varphi\,d\mu \le\sum_{j=1}^k a_j\mu(E_j). $$ Since this holds for every admissible $\varphi$, taking the supremum proves the upper bound. Together with the lower bound, it proves the formula. If $k=0$, then $f=0$ everywhere, every admissible $\varphi$ is zero, and both sides are zero. $\square$

Worked Example: A Finite-Valued Function on an Interval

On $X=[0,2]$ with Lebesgue measure, define $f=2$ on $E_1=[0,1)$, $f=\tfrac12$ on $E_2=[1,2]$, and $f=0$ elsewhere in $X$. The sets are disjoint, with $m(E_1)=1-0=1$ and $m(E_2)=2-1=1$. The finite-valued formula gives $$ \int_X f\,dm =2\cdot1+\frac12\cdot1 =2+\frac12 =\frac52. $$ The function $f$ itself is an admissible minorant, so this value is attained in the supremum. For any other admissible simple $\varphi$, its values on $E_1$ are at most $2$, its values on $E_2$ are at most $\tfrac12$, and it is zero outside $E_1\cup E_2$. The partition argument in the proposition bounds its integral by $2m(E_1)+\tfrac12m(E_2)=\tfrac52$.

Check: The two disjoint pieces contribute $2\cdot1=2$ and $\tfrac12\cdot1=\tfrac12$; their sum is $\tfrac52$, as given by the formula.

Worked Example: A Finite Weighted Atomic Space

Let $X=\{a,b,c\}$, let $\mathcal{M}$ be all subsets of $X$, and assign the point masses $\mu(\{a\})=2$, $\mu(\{b\})=\tfrac13$, and $\mu(\{c\})=4$. Define $f(a)=0$, $f(b)=3$, and $f(c)=1$. The positive level sets are $E_1=\{b\}$, with value $a_1=3$, and $E_2=\{c\}$, with value $a_2=1$. Therefore $$ \int f\,d\mu =3\cdot\frac13+1\cdot4 =1+4 =5. $$ The point $a$ contributes zero because $f(a)=0$, even though its mass is $2$. In the supremum definition, $f$ is an admissible minorant, while every admissible $\varphi$ has $\varphi(a)=0$, $\varphi(b)\le3$, and $\varphi(c)\le1$. Hence its integral is at most $3\cdot\tfrac13+1\cdot4=5$.

Check: Directly summing value times point mass gives $0\cdot2+3\cdot\tfrac13+1\cdot4=0+1+4=5$, the same value.

A Common Upper-Bound Error

When computing a supremum, the upper bound must apply to every admissible minorant and must be the proposed finite weighted sum. It is not enough to compare each minorant with the integral whose value is being defined.

Common mistake: For $f=2\chi_E$, one might write “$\varphi\le f$, so $\int\varphi\,d\mu\le\int f\,d\mu$; taking the supremum proves $\int f\,d\mu\le2\mu(E)$.” The first inequality only bounds the supremum by $\int f\,d\mu$, the unknown quantity being defined. It does not establish the claimed bound by $2\mu(E)$.

The correction is to work with the level sets of $\varphi$. If $\varphi=\sum_{\ell=1}^r b_\ell\chi_{F_\ell}$ is admissible, then it vanishes outside $E$, and $b_\ell\le2$ on each $F_\ell\subseteq E$ with positive value. Consequently, $$ \int\varphi\,d\mu =\sum_{\ell:\,b_\ell>0}b_\ell\mu(F_\ell) \le 2\sum_{\ell:\,b_\ell>0}\mu(F_\ell) \le2\mu(E). $$ This estimate is valid whether $\mu(E)$ is finite or infinite. Since $2\chi_E$ itself is admissible and has simple-function integral $2\mu(E)$, the supremum is exactly $2\mu(E)$.

The same logic is essential for a function with several positive levels: partition each level set of $f$ by the finitely many levels of an arbitrary minorant, bound the minorant's value by the corresponding value of $f$, and then sum. This is what supplies the needed upper bound without assuming the answer in advance. The definition also explains why the integral can be infinite: even if each admissible simple function has finite range, its integral can be arbitrarily large, or one of its positive level sets can already have infinite measure.

Check Your Understanding

In each question, identify the admissible minorants and justify both bounds on the supremum.

For a measurable $E$ with $\mu(E)=0$, compute $\int\chi_E\,d\mu$, and explain why every admissible minorant has integral zero.
Suppose $f=4$ on a measurable set $A$, $f=1$ on a disjoint measurable set $B$, and $f=0$ elsewhere. State the finite-valued formula, including the case $\mu(A)=\infty$.
On $\mathbb{R}$ with Lebesgue measure, use admissible indicators of bounded intervals to show that $\int\chi_{(3,\infty)}\,dm=\infty$.
Why does the statement $\int\varphi\,d\mu\le\int f\,d\mu$ for every $\varphi\le f$ not by itself prove a proposed explicit upper bound for the supremum?
If a finite-valued $f$ is identically zero, what is the positive-level-set sum, and what are all its admissible nonnegative simple minorants?