Properties of the Simple-Function Integral

Algebraic Rules for Simple-Function Integrals

The previous tutorial defined the integral of a simple function by using its values on measurable level sets. That definition gives a finite sum, but the terms in the sum may be infinite when a positive level set has infinite measure. This tutorial establishes the algebraic rules that remain valid in that setting, and then identifies the additional finiteness condition needed for signed linearity.

Throughout, let $(X,\mathcal{F},\mu)$ be a measure space. We use the finite partition formula and the definition of the signed integral from the previous tutorial. For nonnegative simple functions, integrals take values in $[0,+\infty]$. Addition and multiplication by a positive finite constant are well-defined for such extended values. Multiplication by zero will be treated separately, so that no expression of the form $0\cdot(+\infty)$ is needed.

Additivity and Homogeneity for Nonnegative Simple Functions

If $s$ and $t$ are nonnegative simple functions, then $s+t$ is also a nonnegative simple function. The key is to use a common finite measurable partition on which both functions are constant. On each cell, adding the functions adds their values; the finite partition formula then turns this pointwise calculation into a calculation of integrals.

Theorem (Additivity and Nonnegative Homogeneity): Let $s,t:X\to[0,\infty)$ be measurable simple functions and let $c\geq 0$ be finite. Then $$ \int_X(s+t)\,d\mu=\int_Xs\,d\mu+\int_Xt\,d\mu $$ and $$ \int_X(cs)\,d\mu=c\int_Xs\,d\mu. $$ In the second equality, if $c=0$, the right-hand side is defined to be $0$, including when $\int_Xs\,d\mu=+\infty$.

Proof. Choose a finite measurable partition $C_1,\ldots,C_r$ of $X$ on which both $s$ and $t$ are constant. Such a partition can be obtained by intersecting their finite level-set partitions. Write $s=a_i$ and $t=b_i$ on $C_i$, where $a_i,b_i\geq 0$. By the finite partition formula, $$ \int_X(s+t)\,d\mu=\sum_{i=1}^{r}(a_i+b_i)\mu(C_i), \quad \int_Xs\,d\mu+\int_Xt\,d\mu =\sum_{i=1}^{r}a_i\mu(C_i)+\sum_{i=1}^{r}b_i\mu(C_i). $$ For each cell, $(a_i+b_i)\mu(C_i)=a_i\mu(C_i)+b_i\mu(C_i)$. If $\mu(C_i)=+\infty$, this identity still holds: when both coefficients are zero, both sides are zero; otherwise both sides are $+\infty$. Regrouping the finitely many nonnegative terms proves additivity. If $c=0$, then $cs$ is identically zero and its integral is zero, as required. If $c>0$, on $C_i$ the function $cs$ has value $ca_i$, so the partition formula gives $$ \int_X(cs)\,d\mu=\sum_{i=1}^{r}(ca_i)\mu(C_i) =c\sum_{i=1}^{r}a_i\mu(C_i) =c\int_Xs\,d\mu. $$ For $c>0$, multiplying an infinite nonnegative value by $c$ gives $+\infty$, consistent with the finite partition calculation. This proves both identities. $\square$

The theorem includes cases in which one or more integrals are infinite. In particular, additivity does not require the functions to have finite integrals. The restriction $c\geq 0$ matters: a negative scalar produces a signed function, whose integral requires separate care.

Worked Example: Additivity on a Finite Measure Space

Let $X=\{p,q,r\}$, let $\mathcal{F}$ contain every subset of $X$, and define a measure by assigning masses $\mu(\{p\})=2$, $\mu(\{q\})=1$, and $\mu(\{r\})=3$. Define $s$ to have values $2,0,1$ at $p,q,r$, respectively, and define $t$ to have values $1,4,0$ there. The singleton partition gives

\int_Xs\,d\mu=2(2)+0(1)+1(3)=7, \qquad \int_Xt\,d\mu=1(2)+4(1)+0(3)=6.

The values of $s+t$ at $p,q,r$ are $3,4,1$. Therefore

\int_X(s+t)\,d\mu=3(2)+4(1)+1(3)=13 =7+6 =\int_Xs\,d\mu+\int_Xt\,d\mu.

This calculation uses the same cells for both functions, so each cell contributes the sum of the two weighted values.

Worked Example: Scaling on a Space of Infinite Measure

Give $\mathbb{N}$ counting measure and let $s(n)=2$ for every $n$. Its only positive level set is all of $\mathbb{N}$, which has infinite counting measure. Thus $$ \int_{\mathbb{N}}s\,d\mu=2\cdot(+\infty)=+\infty. $$ For a positive scalar, such as $c=3$, the scaled function has value $6$ everywhere and

\int_{\mathbb{N}}3s\,d\mu=6\cdot(+\infty)=+\infty =3\int_{\mathbb{N}}s\,d\mu.

For $c=0$, however, $0s$ is the zero function, so $\int_{\mathbb{N}}0s\,d\mu=0$. The zero case is not evaluated by multiplying zero by the infinite integral; it is determined directly from the function $0s$.

Linearity for Signed Simple Functions

For signed functions, additivity and homogeneity are safe when the integrals involved are finite. A useful condition ensuring this is integrability in absolute value. For a measurable simple function $f$, the function $|f|$ is a nonnegative simple function, and its integral is always defined as a nonnegative extended real number.

Definition: A real-valued measurable simple function $f$ is integrable if $$ \int_X |f|\,d\mu<\infty. $$ In that case its signed integral is finite.

The finiteness criterion from the previous tutorial says that $\int_X|f|\,d\mu$ is finite exactly when each positive level set of $|f|$ has finite measure. Equivalently, each nonzero level set of $f$ has finite measure. This makes finite weighted-sum calculations possible for signed functions.

Theorem (Linearity for Integrable Simple Functions): Let $f$ and $g$ be integrable real-valued measurable simple functions, and let $\alpha,\beta\in\mathbb{R}$. Then $\alpha f+\beta g$ is integrable and $$ \int_X(\alpha f+\beta g)\,d\mu =\alpha\int_Xf\,d\mu+\beta\int_Xg\,d\mu. $$

Proof. Take a finite measurable partition $C_1,\ldots,C_r$ on which both $f$ and $g$ are constant. Write their respective values on $C_i$ as $u_i$ and $v_i$. If $u_i\neq 0$, then $C_i$ is contained in the level set $\{x:f(x)=u_i\}$, which has finite measure because $f$ is integrable. If $v_i\neq 0$, the same reasoning shows that $\mu(C_i)<\infty$. Thus every cell on which at least one of $f,g$ is nonzero has finite measure. The union of these finitely many cells has finite measure, and $\alpha f+\beta g$ vanishes outside that union. Since $\alpha f+\beta g$ takes only finitely many finite values, its absolute integral is finite.

On a cell where both $u_i$ and $v_i$ are zero, all three functions $f,g,\alpha f+\beta g$ contribute zero. On every other cell the measure is finite, and the finite signed formula gives $$ \int_X(\alpha f+\beta g)\,d\mu =\sum_{i=1}^{r}(\alpha u_i+\beta v_i)\mu(C_i). $$ The terms are finite, so the sum can be regrouped: $$ \sum_{i=1}^{r}(\alpha u_i+\beta v_i)\mu(C_i) =\alpha\sum_{i=1}^{r}u_i\mu(C_i) +\beta\sum_{i=1}^{r}v_i\mu(C_i) =\alpha\int_Xf\,d\mu+\beta\int_Xg\,d\mu. $$ This proves the claimed linearity. $\square$

Worked Example: A Signed Linear Combination

On $X=[0,3]$ with Lebesgue measure, let $f$ equal $2$ on $[0,1)$, $-1$ on $[1,3]$, and let $g$ equal $1$ on $[0,2]$, $-2$ on $(2,3]$. All nonzero level sets have finite measure, so both functions are integrable. Their integrals are

\int_Xf\,d\lambda=2(1)+(-1)(2)=0, \qquad \int_Xg\,d\lambda=1(2)+(-2)(1)=0.

Take $\alpha=2$ and $\beta=-1$. On $[0,1)$, $2f-g=4-1=3$; on $[1,2]$, $2f-g=-2-1=-3$; and on $(2,3]$, $2f-g=-2-(-2)=0$. Hence

\int_X(2f-g)\,d\lambda=3(1)+(-3)(1)+0(1)=0 =2\int_Xf\,d\lambda-\int_Xg\,d\lambda.

The partition includes every region where either function changes value. In particular, the two nonzero contributions cancel only after both have been computed as finite numbers.

Bounds Involving the Absolute Value

Linearity also leads to a basic estimate. For integrable simple functions, the absolute value of the integral cannot exceed the integral of the absolute value. The corresponding triangle inequality for absolute integrals follows from the pointwise inequality $|f+g|\leq |f|+|g|$, together with the Monotonicity of the Lebesgue Integral established earlier in this course and the additivity for nonnegative simple functions proved above.

Theorem (Absolute-Value Bounds): If $f$ and $g$ are integrable real-valued measurable simple functions, then $$ \left|\int_X f\,d\mu\right|\leq\int_X|f|\,d\mu $$ and $$ \int_X|f+g|\,d\mu\leq\int_X|f|\,d\mu+\int_X|g|\,d\mu. $$

Pointwise, $|f+g|\leq |f|+|g|$. Both sides are nonnegative measurable simple functions, so the Monotonicity of the Lebesgue Integral gives $$ \int_X|f+g|\,d\mu\leq\int_X(|f|+|g|)\,d\mu =\int_X|f|\,d\mu+\int_X|g|\,d\mu. $$ This proves the second inequality. $\square$

Worked Example: Cancellation and the Absolute-Value Bound

On $X=[0,2]$ with Lebesgue measure, let $h$ equal $5$ on $[0,1)$ and $-3$ on $[1,2]$. Then

\int_Xh\,d\lambda=5(1)+(-3)(1)=2, \qquad \int_X|h|\,d\lambda=5(1)+3(1)=8.

Why Unrestricted Signed Linearity Fails

The finite-integral hypothesis in the signed linearity theorem cannot simply be dropped. On $\mathbb{R}$ with Lebesgue measure, let $f(x)=1$ and $g(x)=-1$ for every $x$. The signed integrals are defined as extended values: $\int_{\mathbb{R}}f\,d\lambda=+\infty$ and $\int_{\mathbb{R}}g\,d\lambda=-\infty$. But $f+g=0$, so its integral is $0$. The proposed right-hand side $(+\infty)+(-\infty)$ is undefined, not zero.

The issue is not that the pointwise identity $f+g=0$ fails. Rather, the two separate infinite contributions cannot be combined by the arithmetic rules for finite integrals. This is why the signed integral was defined through the positive and negative parts, and why the linearity theorem above is stated for integrable functions.

For nonnegative simple functions.
Use additivity and nonnegative homogeneity, including infinite integrals; treat the zero scalar directly.

For signed simple functions.
Check that the absolute integrals are finite before applying linearity or rearranging terms.

For estimates.
Use pointwise absolute-value inequalities and monotonicity to bound integrals.

These rules are fundamental because they allow integrals of simple functions to behave like finite weighted sums when the relevant quantities are finite, while preserving the correct extended-value behavior for nonnegative functions. The distinction between those two settings is essential: nonnegative additivity permits infinity, but signed cancellation requires finite contributions.

Key takeaway: Nonnegative simple-function integrals are additive and homogeneous for nonnegative scalars, even when infinite. Signed linearity and absolute-value bounds apply when the functions are integrable, so no undefined cancellation of infinities occurs.

Check Your Understanding

Use common partitions and the finiteness conditions to check your understanding.

Why does additivity hold for nonnegative simple functions even if one of their integrals is infinite?
How should the homogeneity identity be interpreted when the scalar is zero and the integral is infinite?
What condition on a signed simple function ensures that its integral is finite?
Why can the finite signed linearity theorem regroup terms over a common partition?
Give the pointwise inequality used to prove the triangle inequality for absolute integrals.
Why is unrestricted signed linearity invalid when one integral is positive infinite and the other is negative infinite?