Weighted Partitions and Exact Integral Calculations

From Partition Data to an Integral

A simple function is constant on each cell of a finite measurable partition. Its integral is therefore calculated by weighting each cell’s measure by the value of the function there. The details matter when cells have measure zero or infinite measure: a zero coefficient contributes nothing, even on an infinite-measure cell, while a positive coefficient on such a cell contributes infinity.

Throughout, let $(X,\mathcal{M},\mu)$ be an arbitrary measure space. No assumption that $\mu$ is finite or sigma-finite is needed. A finite measurable partition of $X$ is a finite collection of pairwise disjoint sets in $\mathcal{M}$ whose union is $X$; empty cells are permitted. Consider a function $f:X\to[0,\infty)$ that is constant with a finite nonnegative value $a_j$ on each cell $E_j$ of such a partition. Its integral is an extended nonnegative real number.

Recall the level-set definition for a nonnegative simple function: if $c_1,\ldots,c_r$ are its distinct positive values, then

\int f\,d\mu=\sum_{k=1}^{r}c_k\,\mu(\{x\in X:f(x)=c_k\}).

This is a finite sum in $[0,\infty]$. The zero level set is not included: points where $f=0$ contribute zero, whether that set has measure zero, finite measure, or infinite measure. We will use the convention that a zero coefficient contributes $0$, including on a cell of infinite measure; we do not interpret the expression $0\cdot\infty$ as an ordinary product.

Theorem (Weighted partition formula): Let $E_1,\ldots,E_N$ be a finite measurable partition of $X$ in the measure space $(X,\mathcal{M},\mu)$, and let $a_1,\ldots,a_N$ be finite nonnegative real numbers. If $f=a_j$ on $E_j$, then $$ \int f\,d\mu=\sum_{\substack{1\le j\le N\\a_j>0}}a_j\,\mu(E_j). $$ The sum is allowed to equal $+\infty$. Cells with $a_j=0$ are omitted, regardless of their measure.

Proof. Let $c_1,\ldots,c_r$ be the distinct positive values among $a_1,\ldots,a_N$. For each $k$, the level set $\{x:f(x)=c_k\}$ is the union of exactly those partition cells $E_j$ for which $a_j=c_k$. These cells are disjoint and measurable. Finite additivity of the measure gives $\mu(\{f=c_k\})=\sum_{j:a_j=c_k}\mu(E_j)$, with the value $+\infty$ allowed. Substituting these level-set measures into the definition of the integral gives $\int f\,d\mu=\sum_{k=1}^{r}c_k\sum_{j:a_j=c_k}\mu(E_j)$. Each positive coefficient $a_j$ appears once in this regrouping, with its cell measure, so this equals the stated sum. Cells with coefficient zero do not belong to any positive level set and contribute nothing. If there are no positive coefficients, $f=0$ everywhere and both sides are zero. $\square$

The formula is insensitive to whether partition cells with the same value are listed separately: those cells are grouped into one level set. It also makes the infinite cases precise. A positive finite coefficient multiplied by an infinite cell measure gives an infinite contribution. A zero coefficient does not.

Worked Example: Repeated Values on Finite-Measure Cells

Let $X=[0,4]$ with Lebesgue measure $m$. Take the partition $E_1=[0,1)$, $E_2=[1,3)$, and $E_3=[3,4]$. Define $f=2$ on $E_1$, $f=\tfrac12$ on $E_2$, and $f=2$ on $E_3$. Compute $\int_X f\,dm$.

The cell measures are $m(E_1)=1-0=1$, $m(E_2)=3-1=2$, and $m(E_3)=4-3=1$. Applying the weighted partition formula, the first cell contributes $2\cdot1=2$, the second contributes $\tfrac12\cdot2=1$, and the third contributes $2\cdot1=2$. Therefore

\int_X f\,dm=2+1+2=5.

Check by level sets: the set where $f=2$ is $E_1\cup E_3$, of measure $1+1=2$, and the set where $f=\tfrac12$ is $E_2$, of measure $2$. Thus the level-set calculation is $2\cdot2+\tfrac12\cdot2=4+1=5$, agreeing with the cell-by-cell calculation.

Worked Example: A Null Cell and an Infinite Cell with Value Zero

Let $X=\mathbb{R}$ with Lebesgue measure $m$, and partition $X$ into $E_1=\{0\}$, $E_2=(0,1]$, and $E_3=\mathbb{R}\setminus[0,1]$. Set $f=7$ on $E_1$, $f=3$ on $E_2$, and $f=0$ on $E_3$. Determine the integral, accounting for every cell.

The singleton cell has measure $m(E_1)=0$. The interval cell has measure $m(E_2)=1-0=1$. The third cell has infinite measure because it contains the disjoint intervals $[-n,-n+1)$ for every positive integer $n$; each has measure $1$, so the measure of their union, and hence of $E_3$, is infinite. The weighted partition calculation is $7\cdot0=0$ from $E_1$, $3\cdot1=3$ from $E_2$, and $0$ from $E_3$, since its coefficient is zero. Thus

\int_{\mathbb{R}} f\,dm=0+3+0=3.

Check by level sets: the positive level sets are $\{f=7\}=E_1$, with measure $0$, and $\{f=3\}=E_2$, with measure $1$. Hence $\int f\,dm=7\cdot0+3\cdot1=3$. The infinite-measure zero level set is not part of the positive-value sum.

Worked Example: A Positive Value on an Infinite-Measure Cell

On $X=\mathbb{R}$ with Lebesgue measure $m$, partition $X$ into $E_1=[0,1]$ and $E_2=\mathbb{R}\setminus[0,1]$. Let $f=0$ on $E_1$ and $f=2$ on $E_2$. Compute $\int f\,dm$.

The first cell has measure $m(E_1)=1-0=1$. The second has infinite measure: it contains the disjoint intervals $[-n,-n+1)$ for every positive integer $n$, each of measure $1$. Its coefficient is $2>0$, so the weighted partition formula gives a contribution $2\cdot\infty=\infty$. The first cell contributes zero because its coefficient is zero. Therefore $\int f\,dm=\infty$.

Check by level sets: $\{f=2\}=E_2$ has infinite measure, so the defining positive-level-set sum is $2\,m(E_2)=\infty$. The value agrees with the partition calculation.

When Is the Integral Finite?

Proposition: Under the hypotheses of the weighted partition formula, $\int f\,d\mu$ is finite if and only if every cell $E_j$ with $a_j>0$ has finite measure. No condition on the measures of cells with $a_j=0$ is required.

Proof. Suppose first that every cell with positive coefficient has finite measure. There are only finitely many cells, and every coefficient is finite, so each product $a_j\mu(E_j)$ in the weighted sum is finite. A finite sum of finite nonnegative numbers is finite; hence $\int f\,d\mu<\infty$. Conversely, suppose there is a cell $E_j$ with $a_j>0$ and $\mu(E_j)=\infty$. Since $a_j$ is positive and finite, its contribution $a_j\mu(E_j)$ is $+\infty$. The weighted sum is then $+\infty$, so the integral is not finite. This proves both directions. $\square$

This criterion is useful when the partition data are given in a table: inspect only the cells with strictly positive function values. Infinite measure on a zero-valued cell is irrelevant, while infinite measure on any positive-valued cell determines the integral to be infinite. A cell of measure zero is likewise harmless for any finite coefficient.

A Common Mistake: Treating Zero Times Infinity as a Product

Suppose a partition includes a cell $E$ with $\mu(E)=\infty$, and the function has value zero there. A tempting but incorrect calculation writes the contribution as $0\cdot\infty$ and then calls it undefined, or assigns it infinity. The weighted formula does neither: zero-coefficient cells are omitted. Equivalently, the positive level sets in the definition of the integral exclude the zero level set altogether. In the second worked example, the cell $E_3$ has infinite measure but value zero; the integral is still exactly $3$, not undefined or infinite.

Another useful check is to group all cells with the same positive value before multiplying. If a value $c>0$ occurs on several disjoint cells, their combined level set has measure equal to the sum of those cell measures, with infinity allowed. This agrees with adding their separate weighted contributions and prevents a repeated value from being counted only once.

Check Your Understanding

For each question, identify the positive-valued cells before deciding whether the integral is finite.

A finite measurable partition has cell measures $0$, $4$, and $\infty$, and respective values $6$, $0$, and $0$. What is the integral?
On a partition with cell measures $2$, $3$, and $0$, the respective function values are $1$, $0$, and $5$. Compute the integral by the weighted formula and by positive level sets.
State and prove the condition under which a finite-valued nonnegative simple function on a finite measurable partition has finite integral.
Can a zero-valued cell of infinite measure force the integral to be infinite? Explain using the positive-level-set definition.
If two disjoint cells both carry the value $4$, explain how the level-set calculation combines their measures.

Weighted Partitions and Exact Integral Calculations

What You'll Learn