Finite Weighted Atomic Spaces

From Point Values to Integrals

On a finite space, a nonnegative simple function is specified by finitely many point values, and a measure is specified by the weight assigned to each point. The integral can therefore be calculated without approximation: multiply each function value by the weight of its point and add. Care is needed when weights repeat, when a function has the same value at several points, or when a point has weight zero. In particular, repeated weights belong to distinct points unless we deliberately group their contributions.

Throughout, let $X=\{x_1,\ldots,x_N\}$ be a finite set, possibly empty, and let $\mathcal{M}=\mathcal{P}(X)$, the collection of all subsets of $X$. Prescribe finite weights $w_i\ge0$, one for each $x_i$. This is a finite, complete measure space: every subset is measurable, and every subset of a null set is measurable. Every real-valued function on $X$ is measurable and simple. Unless stated otherwise, functions in this tutorial are real-valued and nonnegative, and integral equalities hold exactly, not merely almost everywhere.

Definition: The finite weighted measure $\mu$ on $(X,\mathcal{M})$ is given by $$ \mu(E)=\sum_{\{i:\,x_i\in E\}} w_i\qquad(E\subseteq X). $$ An empty sum is zero. In particular, $\mu(\{x_i\})=w_i$, and $\mu(X)=\sum_{i=1}^Nw_i<\infty$. A point with $w_i>0$ has positive measure; a point with $w_i=0$ is a null point.

To see why this defines a measure, consider a countable collection of pairwise disjoint subsets of $X$. Each point of their union belongs to exactly one set in the collection. Its weight is therefore counted exactly once when the measures of those sets are summed. Since $X$ is finite, only finitely many sets in the collection can be nonempty, so this sum equals the measure of the union. Completeness follows because every subset of $X$, including every subset of a null set, belongs to $\mathcal{M}$.

A simple function $\varphi:X\to[0,\infty)$ can be described by a finite measurable partition $E_1,\ldots,E_k$ of $X$ and constants $c_j\ge0$, with $\varphi=c_j$ on $E_j$. Its Lebesgue integral is defined by summing the value on each piece times the measure of that piece. When $X$ is empty, this partition has no nonempty pieces, and the integral is zero.

Theorem: For the finite weighted measure space $(X,\mathcal{M},\mu)$ above and any nonnegative simple function $\varphi:X\to[0,\infty)$, $$ \int_X\varphi\,d\mu=\sum_{i=1}^Nw_i\varphi(x_i). $$ The sum is zero when $X$ is empty.

Proof. Choose a finite measurable partition $E_1,\ldots,E_k$ of $X$ on which $\varphi$ takes the respective values $c_1,\ldots,c_k$. By the definition of the integral of a simple function and the definition of $\mu$, $$ \int_X\varphi\,d\mu =\sum_{j=1}^k c_j\mu(E_j) =\sum_{j=1}^k c_j\sum_{\{i:\,x_i\in E_j\}}w_i. $$ Every point $x_i$ belongs to exactly one set of the partition, namely the set on which $\varphi(x_i)$ is its value. Because both sums are finite, regrouping terms gives $\sum_{i=1}^Nw_i\varphi(x_i)$. If $X$ is empty, both sides are empty sums and equal zero. $\square$

Two Equivalent Ways to Organize the Sum

The pointwise formula is usually the quickest calculation. Sometimes it is shorter to group points at which the function has the same value. For $c\ge0$, the set $\{x\in X:\varphi(x)=c\}$ is called the level set of $\varphi$ at $c$. Its measure is the total weight of all points in that level set. The next formula says that the two ways of organizing the finite sum agree.

Proposition: Let $c_1,\ldots,c_k$ be the distinct values of a nonnegative simple function $\varphi$ on $(X,\mathcal{M},\mu)$. Then $$ \int_X\varphi\,d\mu =\sum_{j=1}^k c_j\,\mu(\{x\in X:\varphi(x)=c_j\}). $$

Proof. The level sets $F_j=\{x\in X:\varphi(x)=c_j\}$ are disjoint and their union is $X$. By the definition of $\mu$, $\mu(F_j)=\sum_{\{i:\,x_i\in F_j\}}w_i$. Thus $$ \sum_{j=1}^k c_j\mu(F_j) =\sum_{j=1}^k c_j\sum_{\{i:\,x_i\in F_j\}}w_i. $$ Each $x_i$ belongs to exactly one level set, and on that set $c_j=\varphi(x_i)$. Regrouping this finite sum yields $\sum_iw_i\varphi(x_i)$, which equals the integral by the theorem. If $X$ is empty, there are no level sets with points in them, and the identity is again the empty-sum identity. $\square$

There are two distinct kinds of repetition to keep track of. Points may have the same weight but different function values; they still contribute separately in the pointwise formula. Alternatively, points may have the same function value; then the level-set formula combines their weights. Both are ordinary finite-sum regroupings, and neither discards a positive-weight point.

Worked Example: Repeated Weights and a Null Point

Let $X=\{x_1,x_2,x_3,x_4,x_5\}$, with weights $(w_1,w_2,w_3,w_4,w_5)=(2,1,2,0,3)$, and let $\varphi$ have values $(4,4,1,7,0)$ at these points. Calculate $\int_X\varphi\,d\mu$ in both ways.

Point by point, multiply each value by the weight at the same point: $$ \int_X\varphi\,d\mu =2\cdot4+1\cdot4+2\cdot1+0\cdot7+3\cdot0 =8+4+2+0+0 =14. $$ The equal weights $w_1=w_3=2$ do not make $x_1$ and $x_3$ the same point: their contributions are $8$ and $2$, respectively. The point $x_4$ has weight zero, so its value $7$ contributes zero.

For a second calculation, the value $4$ occurs at $x_1,x_2$, whose total weight is $2+1=3$; the value $1$ occurs at $x_3$, of weight $2$; the value $7$ occurs at $x_4$, of weight $0$; and the value $0$ occurs at $x_5$, of weight $3$. Hence $$ 4\cdot3+1\cdot2+7\cdot0+0\cdot3=12+2+0+0=14. $$ Check: both computations give $14$. The total space weight is $2+1+2+0+3=8$; the integral is not obtained by multiplying that total weight by one of the function values, since the function is not constant.

Worked Example: Grouping by Function Value

Let $X=\{y_1,y_2,y_3,y_4\}$ have weights $(\tfrac12,\tfrac32,0,\tfrac12)$. Suppose $\psi(y_1)=0$, $\psi(y_2)=3$, $\psi(y_3)=5$, and $\psi(y_4)=3$. Compute the integral and identify which points affect it.

The direct calculation is $$ \int_X\psi\,d\mu =\tfrac12\cdot0+\tfrac32\cdot3+0\cdot5+\tfrac12\cdot3 =0+\tfrac92+0+\tfrac32 =6. $$ In the level-set calculation, the value $0$ has total weight $\tfrac12$; the value $3$ has total weight $\tfrac32+\tfrac12=2$; and the value $5$ has total weight $0$. Therefore $$ 0\cdot\tfrac12+3\cdot2+5\cdot0=0+6+0=6. $$ Check: the weight of the level set where $\psi=3$ is exactly $2$, so its contribution is $3\cdot2=6$, agreeing with the sum of the two pointwise contributions $\tfrac92+\tfrac32=6$. The value at $y_3$ does not contribute because $y_3$ is a null point.

Worked Example: A Function Given by Indicators

Let $X=\{z_1,z_2,z_3,z_4\}$ have weights $(3,0,1,2)$. Define $A=\{z_1,z_2\}$, $B=\{z_3,z_4\}$, and $\theta=2\chi_A+5\chi_B$. Calculate $\int_X\theta\,d\mu$ directly and by using the measures of $A$ and $B$.

Because $A$ and $B$ partition $X$, the values of $\theta$ at $z_1,z_2,z_3,z_4$ are respectively $2,2,5,5$. The pointwise sum is $$ \int_X\theta\,d\mu =3\cdot2+0\cdot2+1\cdot5+2\cdot5 =6+0+5+10 =21. $$ For the second calculation, $\mu(A)=3+0=3$ and $\mu(B)=1+2=3$. Thus $$ \int_X\theta\,d\mu =2\mu(A)+5\mu(B) =2\cdot3+5\cdot3 =6+15 =21. $$ Check: the two sets have equal measure $3$, but their function values differ; the integral is $2\cdot3+5\cdot3=21$, not a constant value times the total measure $6$. The zero-weight point $z_2$ is included in $A$ but adds nothing to $\mu(A)$ or to the integral.

Zero Weights and Almost-Everywhere Equality

A function can differ at a null point without changing its integral. On this finite space, the exact criterion is particularly simple: only values at points of positive weight matter. Recall that two functions are equal $\mu$-almost everywhere if the set on which they differ has measure zero.

Proposition: For a nonnegative simple function $\varphi$ on $(X,\mathcal{M},\mu)$, $$ \int_X\varphi\,d\mu=0 \quad\Longleftrightarrow\quad \varphi(x_i)=0\ \text{for every }i\text{ such that }w_i>0. $$ Equivalently, $\varphi=0$ $\mu$-almost everywhere.

Proof. By the pointwise integral formula, $\int_X\varphi\,d\mu=\sum_iw_i\varphi(x_i)$, a finite sum of nonnegative terms. If $\varphi(x_i)=0$ whenever $w_i>0$, then each term is zero: it is zero because $w_i=0$ or because $\varphi(x_i)=0$. Hence the integral is zero. Conversely, suppose the integral is zero. If some $i$ had $w_i>0$ and $\varphi(x_i)>0$, then the nonnegative sum would be at least $w_i\varphi(x_i)>0$, a contradiction. Thus every positive-weight point has value zero. Finally, the set of points where $\varphi\ne0$ has measure zero exactly when it contains no positive-weight point, because the measure of any subset is the sum of its point weights. This proves the almost-everywhere formulation as well. $\square$

This criterion also explains why values at zero-weight points can be changed without affecting an integral: any set containing only such points has measure zero. It does not say that all points are irrelevant when a weight is small; a small positive weight still contributes its weight times the function value.

A Common Counting Error

A frequent mistake is to count a repeated weight only once, as though a weight belonged to a value rather than to its point. In Worked Example 1, the value $4$ occurs at two points, with weights $2$ and $1$. A mistaken calculation might take the weight $2$ only once and write $4\cdot2+1\cdot2=10$, omitting the contribution $4\cdot1=4$ from $x_2$. The correct contribution from the level set $\{\varphi=4\}$ is $$ 4\,\mu(\{\varphi=4\})=4(2+1)=12, $$ not $4\cdot2=8$. The other level-set contributions are $1\cdot2=2$, $7\cdot0=0$, and $0\cdot3=0$, giving $12+2=14$. The correction is to sum all point weights in the level set, including repeated weights and any zero weights, before multiplying by the shared function value.

The finite weighted formula matters because it turns the definition of a nonnegative simple-function integral into a concrete computation. It also separates the roles of values and weights: the function supplies the value at each point, while the measure supplies that point’s weight. One may calculate point by point or regroup by level sets, but each point’s contribution must be accounted for exactly once. On finite spaces, this bookkeeping is the whole calculation.

Check Your Understanding

In each question, state which points have positive weight and use the finite weighted formula or the level-set formula as appropriate.

For weights $(1,0,2)$ and function values $(3,8,3)$, compute the integral both point by point and by grouping equal values.
Suppose a finite weighted space has weights $(0,0,0)$. What is the measure of $X$, and what is the integral of any nonnegative real-valued function on $X$?
Give the condition on the point values of a nonnegative simple function that is equivalent to its integral being zero.
If two distinct points have the same positive weight but different function values, explain why their contributions cannot be replaced by a single point contribution without preserving both values.
For a subset $A\subseteq X$, express $\int_X c\chi_A\,d\mu$ in terms of $c$ and $\mu(A)$, and evaluate it when $c=0$.

What You'll Learn