Why the Simple-Function Integral Is Well Defined

Why a Definition Needs a Well-Defined Value

A nonnegative simple function can be written in more than one way as a finite sum of constants times indicators of disjoint measurable sets. The integral is defined by adding each constant multiplied by the measure of its set. For this to define an integral of the function, rather than of a particular written expression, the resulting value must be the same for every such representation.

Let \((X,\mathcal{M},\mu)\) be any measure space. No assumption of finiteness, sigma-finiteness, or completeness is imposed. A nonnegative simple function here means a measurable function \(s:X\to[0,\infty)\) with finite range. Its coefficients are finite real numbers. Its values are defined everywhere, not just almost everywhere. The integral may nevertheless be infinite when one of the sets in a representation has infinite measure.

Definition: A finite disjoint measurable representation of \(s\) is an expression \(s=\sum_{i=1}^{r}a_i\chi_{E_i}\), where \(a_i\ge0\), \(E_i\in\mathcal{M}\), and the sets \(E_i\) are pairwise disjoint. The sum is pointwise on \(X\); outside \(\bigcup_{i=1}^{r}E_i\), it has value zero. The weighted-sum value of this representation is \(\sum_{i=1}^{r}a_i\mu(E_i)\). We define a zero-coefficient term to contribute zero even if \(\mu(E_i)=\infty\); in particular, \(0\cdot\infty\) is not treated as an arithmetic product.

The empty representation, with \(r=0\), represents the zero function and has weighted-sum value zero. Empty sets can also appear in a representation; their measure is zero, so their terms contribute zero. These conventions cover the zero function and degenerate terms without requiring any exception in the result below.

Finite Refinements Preserve a Weighted Sum

The key calculation is that splitting a measurable set into finitely many disjoint measurable pieces preserves its contribution. This uses only finite additivity of the measure and ordinary multiplication by a finite nonnegative coefficient.

Lemma: Let \(E\in\mathcal{M}\), let \(c\in[0,\infty)\) be finite, and let \(A_1,\ldots,A_q\in\mathcal{M}\) be pairwise disjoint sets whose union is \(E\). Then, with the zero-term convention above, \(c\mu(E)=\sum_{\ell=1}^{q}c\mu(A_\ell)\), including when \(\mu(E)=\infty\).

Proof. If \(c=0\), both sides are zero by the stated convention, regardless of whether the measures are finite or infinite. Suppose \(c>0\). Finite additivity of \(\mu\) gives \(\mu(E)=\sum_{\ell=1}^{q}\mu(A_\ell)\), where the sum is in \([0,\infty]\). If \(\mu(E)<\infty\), multiplying this finite equality by \(c\) gives the result. If \(\mu(E)=\infty\), at least one of the finitely many \(\mu(A_\ell)\) must be infinite; otherwise their finite sum would be finite. Thus the left side is infinity and the right side also has an infinite term multiplied by \(c>0\), so it is infinity. \(\square\)

This lemma is the precise reason a representation can be subdivided without changing its weighted sum. It also handles infinite values of the integral: no subtraction of infinities is used, and the argument compares finite sums of nonnegative extended values.

The Weighted-Sum Definition Is Independent of Representation

Theorem: Let \((X,\mathcal{M},\mu)\) be a measure space, and let \(s:X\to[0,\infty)\) be a nonnegative simple function. If \(s=\sum_{i=1}^{r}a_i\chi_{E_i}=\sum_{j=1}^{t}b_j\chi_{F_j}\) are two finite disjoint measurable representations, then \(\sum_{i=1}^{r}a_i\mu(E_i)=\sum_{j=1}^{t}b_j\mu(F_j)\). The equality is in \([0,\infty]\), so both sides may be infinite.

Proof. The equality of representations is pointwise. The sets \(E_i\cap F_j\), as \(j\) ranges from \(1\) to \(t\), are pairwise disjoint for each fixed \(i\). The part of \(E_i\) not covered by the \(F_j\) is \(R_i=E_i\setminus\bigcup_{j=1}^{t}F_j\). These sets are measurable, are disjoint from every \(E_i\cap F_j\), and together with those intersections form a finite disjoint partition of \(E_i\).

On \(R_i\), the second representation has value zero. Since the two representations define the same function at every point, \(s=a_i\) on \(R_i\) and also \(s=0\) there. Consequently \(a_i=0\) if \(R_i\) is nonempty. If \(R_i\) is empty, its measure is zero. In either case, \(a_i\mu(R_i)\) contributes zero under our convention. Applying the lemma to this partition of each \(E_i\) gives \(\sum_i a_i\mu(E_i)=\sum_i\sum_j a_i\mu(E_i\cap F_j)\).

Interchanging these two finite sums is valid for nonnegative extended real numbers. Now fix \(i,j\). If \(E_i\cap F_j\) is nonempty, then on that intersection \(s=a_i\) by the first representation and \(s=b_j\) by the second. Pointwise equality implies \(a_i=b_j\), so the corresponding terms \(a_i\mu(E_i\cap F_j)\) and \(b_j\mu(E_i\cap F_j)\) agree, including if this measure is infinite. If \(E_i\cap F_j\) is empty, its measure is zero and both terms are zero. Therefore \(\sum_i\sum_j a_i\mu(E_i\cap F_j)=\sum_j\sum_i b_j\mu(E_i\cap F_j)\).

For each \(F_j\), the part not covered by the \(E_i\) has value zero in the first representation, and hence \(b_j=0\) there if it is nonempty. Applying the lemma to the resulting partition of each \(F_j\) shows that \(\sum_j\sum_i b_j\mu(E_i\cap F_j)=\sum_j b_j\mu(F_j)\). Combining the equalities proves the assertion. If either representation has no terms, it represents the zero function; the same reasoning shows that every coefficient on a nonempty set in the other representation must be zero, so its weighted sum is zero as well. \(\square\)

The theorem justifies writing \(\int s\,d\mu\) for this common value. This notation is well-defined because every allowed representation yields exactly the same element of \([0,\infty]\). The theorem is about representations of the same function everywhere. It does not assert here that changing a function on a null set preserves its integral.

Worked Representations

Worked Example: Two Partitions of an Interval

Take \(X=[0,3]\), \(\mathcal{M}\) to be the Lebesgue-measurable subsets of \(X\), and \(\mu\) to be Lebesgue measure restricted to \(X\). Define \(s=2\) on \([0,1)\), \(s=5\) on \([1,3]\), and \(s=0\) nowhere else in \(X\). One representation is \(s=2\chi_{[0,1)}+5\chi_{[1,3]}\). The intervals are disjoint: the point \(1\) belongs to the second interval, not the first. Their measures are \(1\) and \(2\), so this representation has weighted sum \(2\cdot1+5\cdot2=2+10=12\).

Refine the second interval at \(2\). A second disjoint representation is \(s=2\chi_{[0,1)}+5\chi_{[1,2)}+5\chi_{[2,3]}\). The measures of these three sets are respectively \(1\), \(1\), and \(1\). Its weighted sum is \(2\cdot1+5\cdot1+5\cdot1=2+5+5=12\).

Check: the values and sets agree pointwise: the value is \(2\) on \([0,1)\), \(5\) on \([1,2)\) and \([2,3]\), and there are no points of \(X\) left uncovered. The refined representation therefore computes the same function, and both weighted sums equal \(12\).

Worked Example: A Representation with Infinite Integral

Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Let \(s(x)=4\) for \(x\in[0,\infty)\), and \(s(x)=0\) for \(x\in(-\infty,0)\). One representation is \(s=4\chi_{[0,\infty)}\). Since \(m([0,\infty))=\infty\), its weighted sum is \(4\cdot\infty=\infty\).

For a different representation, split the positive half-line into the disjoint sets \([0,1)\) and \([1,\infty)\): \(s=4\chi_{[0,1)}+4\chi_{[1,\infty)}\). Here \(m([0,1))=1\), and \(m([1,\infty))=\infty\). The weighted sum is \(4\cdot1+4\cdot\infty=4+\infty=\infty\).

Check: the point \(1\) is included in \([1,\infty)\), while \(0\) is included in \([0,1)\). Thus the two sets cover \([0,\infty)\) without overlap. Both expressions give value \(4\) on that half-line and zero on its complement, and the two sums are both infinity.

Worked Example: Zero Coefficient on a Set of Infinite Measure

Let \(X=\mathbb{R}\) with Lebesgue measure, and define \(s=2\) on \([0,1]\) and \(s=0\) elsewhere. The representation \(s=2\chi_{[0,1]}\) has weighted sum \(2\cdot1=2\).

Another representation is \(s=2\chi_{[0,1)}+2\chi_{\{1\}}+0\chi_{(1,\infty)}\). The three sets are pairwise disjoint, and their union is \([0,\infty)\); the function is zero on the part of \(\mathbb{R}\) not covered by them. Their measures are \(1\), \(0\), and \(\infty\), respectively. By the zero-term convention, the weighted sum is \(2\cdot1+2\cdot0+0=2+0+0=2\).

Check: the singleton \(\{1\}\) has Lebesgue measure zero, and the last set has infinite measure but coefficient zero. The second representation equals \(2\) at every point of \([0,1]\), including the endpoint \(1\), and equals zero elsewhere. Its value \(2\) agrees with the first representation.

A Common Mistake and Its Correction

A weighted sum is not the sum of the coefficients alone. For example, on \([0,1]\) with Lebesgue measure, the function \(\chi_{[0,1]}\) can be represented as \(\chi_{[0,1]}\), giving \(1\cdot m([0,1])=1\). It can also be represented as \(\chi_{[0,1/2)}+\chi_{[1/2,1]}\). A mistaken calculation might add the two coefficients and report \(1+1=2\). That omits the measures of the sets.

The correct calculation uses both coefficients and measures: \(1\cdot m([0,1/2))+1\cdot m([1/2,1])=1\cdot\frac12+1\cdot\frac12=1\). The endpoint \(1/2\) belongs to the second interval only, so this is a disjoint partition. A related error is to treat \(0\cdot\infty\) as an undefined product in a representation with a zero coefficient. The weighted-sum definition assigns that term zero: a zero-valued part of a function contributes nothing, even if its underlying set has infinite measure.

Why the Result Matters

The integral of a nonnegative simple function is the starting point for defining integrals of more general nonnegative measurable functions. The representation theorem ensures that the starting value is intrinsic to the function: refining pieces, joining pieces with the same value, or choosing a different finite disjoint partition cannot alter it. The proof needs only pointwise equality, measurability of the sets, finite additivity of \(\mu\), and finite nonnegative sums. In particular, it does not require that the measure space be finite or sigma-finite, and it allows the common value to be \(+\infty\).

When checking a proposed representation, verify both that the sets are pairwise disjoint and that the expression gives the specified function at every point. Then compute each coefficient-times-measure term, treating zero coefficients by the convention above. Those checks distinguish the function being integrated from the way it happens to be written.

Check Your Understanding

For each question, keep track of the sets, their disjointness, and whether any measure is infinite.

On a measure space, a set \(E\) is partitioned into finitely many disjoint measurable sets \(A_1,\ldots,A_q\). Explain why \(c\mu(E)=\sum_{\ell=1}^{q}c\mu(A_\ell)\) also holds when \(c=0\) or \(\mu(E)=\infty\).
On \([0,2]\) with Lebesgue measure, compare the weighted sums for \(3\chi_{[0,1)}+3\chi_{[1,2]}\) and \(3\chi_{[0,2]}\).
Why does pointwise equality of two representations force their coefficients to agree on every nonempty intersection of a set from the first representation with a set from the second?
On \(\mathbb{R}\) with Lebesgue measure, what contribution does \(0\chi_{[0,\infty)}\) make, and why is it not computed as an undefined product?
Does the theorem as proved require two functions to be equal only almost everywhere, or everywhere? Identify where the proof uses the answer.