Integrating Finite Simple Functions

Finite Simple Functions and Their Level Sets

A finite simple function records only finitely many values, each on a measurable part of the space. Its integral is therefore a finite weighted sum of measures. The essential care is to use disjoint level sets: if a function is written as a sum of indicators of overlapping sets, its coefficients do not in general describe its values separately on those sets.

Throughout, $(X,\mathcal{M},\mu)$ is an arbitrary measure space; no finiteness, completeness, or sigma-finiteness assumption is imposed. A finite nonnegative simple function is an $\mathcal{M}$-measurable function $\varphi:X\to[0,\infty)$ with finite range. Its integral may nevertheless be $+\infty$ if a positive level set has infinite measure. We use $0\cdot\infty=0$ when a zero level is included in a sum.

Definition: If the distinct positive values of $\varphi$ are $c_1,\ldots,c_k$, its positive level sets are $E_i=\{x\in X:\varphi(x)=c_i\}$. They are measurable and pairwise disjoint. The integral of $\varphi$ is $$ \int\varphi\,d\mu=\sum_{i=1}^{k}c_i\,\mu(E_i). $$ The sum is zero if $\varphi$ has no positive values. The zero level set $\{\varphi=0\}$ contributes zero and need not be included.

This is the simple-function integral: when a simple function is expressed as a sum of nonnegative constants times indicators of disjoint measurable sets, the integral is the sum of each constant times the measure of its set. The level-set form makes explicit that the sets in the sum are disjoint and that the values are the actual values of the function. In particular, a positive value on a set of infinite measure contributes $+\infty$.

Worked Example: Integrating by Measurable Levels

On $X=[-1,2]$ with its Lebesgue sigma-algebra and Lebesgue measure $m$, let $\varphi=3$ on $[-1,0)$, $\varphi=1$ on $[0,2]$, and $\varphi=0$ elsewhere in $X$. Find $\int\varphi\,dm$.

The level sets are $E_3=[-1,0)$ and $E_1=[0,2]$. They are measurable and disjoint: the first excludes $0$, where the second begins. Their measures are $m(E_3)=0-(-1)=1$ and $m(E_1)=2-0=2$. Therefore

\int\varphi\,dm=3m(E_3)+1m(E_1)=3\cdot1+1\cdot2=5.

Check: On all of $X$, the same function is $\varphi=\chi_X+2\chi_{[-1,0)}$. This is an overlapping-indicator representation, but linearity gives $\int\varphi\,dm=m(X)+2m([-1,0))=3+2=5$, agreeing with the level-set computation.

Why the Level-Set Formula Is Well Defined

A single function can have several finite indicator representations. The level-set definition assigns an integral to the function itself, not to an arbitrary choice of representation.

Proposition: For a finite nonnegative simple function on any measure space $(X,\mathcal{M},\mu)$, the level sets $\{\varphi=c\}$ for its positive values are measurable and disjoint, and their weighted-measure sum is independent of the representation of $\varphi$.

Proof. Measurability follows because $\varphi$ is measurable and $\{c\}$ is a Borel subset of $\mathbb{R}$, so $\{\varphi=c\}\in\mathcal{M}$. If $c\ne d$, no $x$ can satisfy both $\varphi(x)=c$ and $\varphi(x)=d$; thus the level sets are disjoint. Their values and sets are determined by $\varphi$, so any two representations of the same function give the same positive level sets and the same sum $\sum_{c>0}c\,\mu(\{\varphi=c\})$. This remains true if a level set has infinite measure, since its positive coefficient makes its contribution $+\infty$. The zero level contributes zero under every representation. Hence the formula is representation-independent. $\square$

For example, suppose $\varphi=\sum_{j=1}^r a_j\chi_{A_j}$, where the $A_j\in\mathcal{M}$ are pairwise disjoint and the $a_j\ge0$. If several coefficients have the same positive value $c$, then $\{\varphi=c\}$ is the union of exactly those $A_j$ with $a_j=c$. Finite additivity of $\mu$ shows that the level-set formula gives $\sum_j a_j\mu(A_j)$, with zero-coefficient terms contributing zero. Thus it agrees with the disjoint-set definition of the simple-function integral.

Worked Example: A Finite Atomic Measure

Let $X=\{a,b,c,d\}$, let $\mathcal{M}$ be the set of all subsets of $X$, and define $\mu(E)=\sum_{x\in E}w_x$, where $w_a=2$, $w_b=\tfrac12$, $w_c=3$, and $w_d=4$. Let $\varphi(a)=\varphi(c)=2$, $\varphi(b)=5$, and $\varphi(d)=0$.

The positive level sets are $E_2=\{a,c\}$ and $E_5=\{b\}$. Their measures are $\mu(E_2)=2+3=5$ and $\mu(E_5)=\tfrac12$. Hence

\int\varphi\,d\mu=2\cdot5+5\cdot\tfrac12=10+\tfrac52=\tfrac{25}{2}.

Check: Directly summing the value at each point times its atomic weight gives $2\cdot2+5\cdot\tfrac12+2\cdot3+0\cdot4=4+\tfrac52+6+0=\tfrac{25}{2}$. The $d$-term is zero, even though $w_d=4$.

Comparing Integrals on a Common Partition

To compare two simple functions, partition the space according to both of their values. Intersections of their level sets form finitely many measurable pieces on which each function is constant. This puts the pointwise comparison and the integral comparison on the same sets.

Theorem: Let $\varphi,\psi:X\to[0,\infty)$ be finite simple measurable functions on a measure space $(X,\mathcal{M},\mu)$. If $\varphi(x)\le\psi(x)$ for every $x\in X$, then $\int\varphi\,d\mu\le\int\psi\,d\mu$, with $+\infty$ allowed as an integral value.

Proof. Let $A_1,\ldots,A_r$ be the positive level sets of $\varphi$, and let $B_1,\ldots,B_s$ be the positive level sets of $\psi$. Add the respective zero level sets to each family, so each family partitions $X$; the zero level sets may be empty. The intersections $C_{ij}=A_i\cap B_j$ are measurable, pairwise disjoint, and their union is $X$. Write $a_i\ge0$ and $b_j\ge0$ for the values of $\varphi$ and $\psi$ on the corresponding sets, including value zero. Pointwise comparison gives $a_i\le b_j$ whenever $C_{ij}\ne\varnothing$. Since $\mu(C_{ij})\ge0$, we have $a_i\mu(C_{ij})\le b_j\mu(C_{ij})$, also when the measure is infinite, using $0\cdot\infty=0$. Summing these inequalities over the finite partition gives $\int\varphi\,d\mu\le\int\psi\,d\mu$. If one of the sums is infinite, the same termwise inequalities ensure the asserted order in $[0,\infty]$. $\square$

A particularly useful case is increasing both the value and the supporting set. For $E\in\mathcal{M}$, the function $a\chi_E$ has integral $a\mu(E)$ when $a>0$, and has integral zero when $a=0$. If $E\subseteq F$ and $0\le a\le b<\infty$, then $a\chi_E\le b\chi_F$ everywhere: on $E$ this follows from $a\le b$, on $F\setminus E$ it follows from $0\le b$, and outside $F$ both sides are zero. The theorem therefore gives $\int a\chi_E\,d\mu\le\int b\chi_F\,d\mu$. No finiteness assumption on $\mu(E)$ or $\mu(F)$ is needed.

Worked Example: Increasing Values and Support

On $X=[0,4]$ with Lebesgue measure, define $\varphi=2$ on $[0,1)$, $\varphi=4$ on $[2,3)$, and zero elsewhere. Define $\psi=3$ on $[0,2)$, $\psi=5$ on $[2,4]$. Verify that $\varphi\le\psi$, and compare their integrals.

The intervals $[0,1)$, $[1,2)$, $[2,3)$, and $[3,4]$ partition $X$. On them, respectively, $(\varphi,\psi)=(2,3),(0,3),(4,5),(0,5)$. Thus $\varphi\le\psi$ at every point, including $4$. The level sets of $\varphi$ each have measure $1$, so $\int\varphi\,dm=2\cdot1+4\cdot1=6$. The level set of $\psi$ at $3$ is $[0,2)$, of measure $2$; its level set at $5$ is $[2,4]$, also of measure $2$. Consequently

\int\psi\,dm=3\cdot2+5\cdot2=6+10=16, \qquad 6\le16.

Check: The difference $\psi-\varphi$ is $1,3,1,5$ on the four listed intervals, each of measure $1$. Its integral is $1+3+1+5=10$, and $16-6=10$, as required by additivity for these nonnegative simple functions.

A Common Error: Overlapping Indicators

The expression $\sum_j a_j\chi_{A_j}$ adds the coefficients at points lying in multiple sets. It is not generally a function that takes value $a_j$ on $A_j$ independently of the other sets. Compute the resulting level sets before applying the weighted-measure formula.

Worked Example: Correcting an Overlap Calculation

On $X=[0,2]$ with Lebesgue measure, let $A=[0,1]$, $B=[1,2]$, and $\theta=\chi_A+\chi_B$. A common incorrect calculation is $\int\theta\,dm=m(A)+m(B)=2$, on the assumption that the two sets are disjoint.

In fact, $A\cap B=\{1\}$, so the sets overlap. At $x=1$, $\theta(1)=\chi_A(1)+\chi_B(1)=1+1=2$. The level set where $\theta=1$ is $[0,1)\cup(1,2]$, of measure $1+1=2$; the level set where $\theta=2$ is $\{1\}$, of measure $0$. The level-set calculation is therefore

\int\theta\,dm=1\cdot2+2\cdot0=2.

The numerical answer happens to match the incorrect calculation because the overlap is a null set, but the pointwise description and the presumed disjointness were wrong. To see why overlap can change the answer, take $A=B=[0,1]$ instead. Then $\theta=2$ on $[0,1]$, so $\int\theta\,dm=2\cdot1=2$, whereas treating the sets as a disjoint partition would give $1+1=2$ without correctly identifying the level. If the coefficients differ, the error can also change the computed value: for $\eta=\chi_{[0,1]}+2\chi_{[0,1]}$, the actual function is $3$ on $[0,1]$ and its integral is $3$, not $1\cdot1+2\cdot1$ interpreted as contributions on disjoint levels. The safe method is always to determine the function’s actual values first.

The level-set formula is a finite calculation, but it is also the right foundation for more elaborate integral arguments: it makes the contribution of each value explicit and supports comparisons without ambiguity about how the function is represented. For finite simple functions, pointwise order is enough to compare integrals; no convergence theorem is being used here.

Check Your Understanding

For each question, identify the measure space and use disjoint measurable level sets before calculating.

On $[0,3]$ with Lebesgue measure, a function equals $2$ on $[0,1)$, $4$ on $[1,2]$, and $0$ elsewhere. Compute its integral.
On a measure space, explain why two distinct positive level sets of a function are disjoint.
If $E\subseteq F$ are measurable and $0\le a\le b<\infty$, prove pointwise that $a\chi_E\le b\chi_F$, then state the resulting integral inequality.
On $[0,1]$ with Lebesgue measure, compute the integral of $3\chi_{[0,1]}+2\chi_{[0,1/2]}$ by first finding its actual levels.
Can a finite nonnegative simple function have infinite integral? Give a precise condition on one of its level sets.