Pointwise Order and the Integral
A basic test of any definition of integration is whether it respects order: if one nonnegative function never exceeds another, its integral should not exceed the other's integral. For the Lebesgue integral of nonnegative functions, this follows directly from its definition as a supremum of integrals of simple functions lying below the function. The key is to compare the two families of admissible simple functions, not to assume that either supremum is attained.
Throughout, \((X,\mathcal{M},\mu)\) is an arbitrary measure space. No assumption that \(\mu\) is finite, sigma-finite or complete is made. Unless a particular example specifies otherwise, functions are measurable and take values in \([0,\infty]\), and pointwise inequalities and equalities are meant everywhere on \(X\). A nonnegative simple function is a measurable function with finite range in \([0,\infty)\). For such a function \(\varphi\), its integral is the sum of each value times the measure of the set on which that value is taken, with \(0\cdot\infty=0\).
The zero function is always a simple minorant, so this family is nonempty even when \(\mu\) is the zero measure or \(f\) vanishes everywhere. The supremum need not be achieved by any one simple minorant. Monotonicity requires only that every minorant available for the smaller function is also available for the larger one.
The Order Principle for Suprema
Proof. By the definition of supremum, \(\sup B\) is an upper bound for every element of \(B\). Since \(A\subseteq B\), it is therefore an upper bound for every element of \(A\). The least upper bound of \(A\) cannot exceed this upper bound, so \(\sup A\le\sup B\). This reasoning also applies if either supremum is \(+\infty\). \(\square\)
Monotonicity of the Nonnegative Integral
Proof. Let \(\mathcal{S}_f\) be the family of nonnegative simple functions \(\varphi\) satisfying \(\varphi\le f\) everywhere, and define \(\mathcal{S}_g\) in the same way. If \(\varphi\in\mathcal{S}_f\), then \(0\le\varphi(x)\le f(x)\le g(x)\) for every \(x\), so \(\varphi\in\mathcal{S}_g\). Thus \(\mathcal{S}_f\subseteq\mathcal{S}_g\). Taking the integrals of these simple functions gives an inclusion of their sets of integral values. By the definition of the two nonnegative integrals and the order principle for suprema, $$ \int f\,d\mu =\sup_{\varphi\in\mathcal{S}_f}\int\varphi\,d\mu \le \sup_{\varphi\in\mathcal{S}_g}\int\varphi\,d\mu =\int g\,d\mu. $$ All suprema here are taken in \([0,\infty]\), so the argument covers zero and infinite values as well as finite ones. \(\square\)
Proof. The equality \(f=g\) gives both \(f\le g\) and \(g\le f\) everywhere. Applying the theorem once in each direction yields \(\int f\,d\mu\le\int g\,d\mu\) and \(\int g\,d\mu\le\int f\,d\mu\). Hence the two extended nonnegative values are equal. \(\square\)
Worked Comparisons
Worked Example: A Weighted Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be the collection of all subsets of \(X\), and define a measure by \(\mu(\{a\})=2\), \(\mu(\{b\})=\tfrac12\), and \(\mu(\{c\})=3\). Define simple functions by \(f(a)=1\), \(f(b)=0\), \(f(c)=2\), and \(g(a)=2\), \(g(b)=1\), \(g(c)=2\). At \(a,b,c\), respectively, the comparisons are \(1\le2\), \(0\le1\), and \(2\le2\). Thus \(f\le g\) everywhere.
The singleton sets form a partition, so the simple-function integral formula gives $$ \int f\,d\mu =1\cdot2+0\cdot\tfrac12+2\cdot3 =2+0+6 =8, $$ and $$ \int g\,d\mu =2\cdot2+1\cdot\tfrac12+2\cdot3 =4+\tfrac12+6 =\tfrac{21}{2}. $$ Therefore \(8\le\tfrac{21}{2}\). Check: the difference between the computed values is \(\tfrac{21}{2}-8=\tfrac52\). The pointwise increases are \(1\) at \(a\), \(1\) at \(b\), and \(0\) at \(c\); weighting those increases gives \(1\cdot2+1\cdot\tfrac12+0\cdot3=\tfrac52\), in agreement with the comparison.
Worked Example: Indicators of Nested Intervals
Take \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{L}\), and \(\mu=m\), Lebesgue measure. Let \(f=\chi_{(0,1)}\) and \(g=\chi_{(0,2)}\). Since \((0,1)\subset(0,2)\), the indicator functions satisfy \(f(x)\le g(x)\) for every real \(x\). In particular, the endpoints are handled by the open intervals: both functions vanish at \(0\), \(f(1)=0\) while \(g(1)=1\), and both vanish at \(2\).
These functions are simple. The interval \((0,1)\) has Lebesgue measure \(1\), and \((0,2)\) has Lebesgue measure \(2\). Hence $$ \int f\,dm=1\cdot m((0,1))=1\cdot1=1, \qquad \int g\,dm=1\cdot m((0,2))=1\cdot2=2. $$ Thus the theorem gives the exact comparison \(1\le2\). Check: the interval lengths used are \(1-0=1\) and \(2-0=2\), so the calculated integrals agree with the lengths of the sets where the respective indicators equal \(1\).
Worked Example: Strict Pointwise Inequality but Equal Integrals
Let \(X=\mathbb{R}\), \(\mathcal{M}=\mathcal{B}_{\mathbb{R}}\), and take the Dirac measure \(\delta_0\), defined by \(\delta_0(E)=1\) if \(0\in E\) and \(\delta_0(E)=0\) if \(0\notin E\). Set \(f=\chi_{\{0\}}\) and \(g=\chi_{\{0,1\}}\). Both sets are Borel, and \(f\le g\) everywhere. At \(x=1\), the inequality is strict: \(f(1)=0<1=g(1)\).
Using the simple-function integral formula, \(\{0\}\) has \(\delta_0\)-measure \(1\), while \(\{1\}\) has \(\delta_0\)-measure \(0\). Consequently, $$ \int f\,d\delta_0=1\cdot\delta_0(\{0\})=1, \qquad \int g\,d\delta_0=1\cdot\delta_0(\{0,1\})=1\cdot1=1. $$ So strict pointwise inequality does not force strict inequality of integrals. Check: \(0\in\{0,1\}\), hence \(\delta_0(\{0,1\})=1\), exactly as used in the second calculation. The strict difference at \(1\) does not change either computed value.
Worked Example: The Comparison Can Have Infinite Values
On \(X=\mathbb{N}=\{1,2,\ldots\}\) with the counting measure, every subset is measurable and the measure of a set is its number of elements, possibly \(+\infty\). Let \(f(n)=1\) and \(g(n)=2\) for every \(n\in\mathbb{N}\). These are nonnegative simple functions and \(f\le g\) pointwise. Their level sets are both all of \(\mathbb{N}\), whose counting measure is \(+\infty\). Therefore $$ \int f\,d\mu=1\cdot\mu(\mathbb{N})=1\cdot\infty=\infty, \qquad \int g\,d\mu=2\cdot\mu(\mathbb{N})=2\cdot\infty=\infty. $$ The conclusion \(\infty\le\infty\) is valid in the extended order on \([0,\infty]\). Check: for each finite subset \(\{1,\ldots,N\}\), the corresponding contributions are \(N\) and \(2N\); as \(N\) increases without bound, both full integrals are infinite.
What Monotonicity Does—and Does Not—Say
The theorem is useful because it transfers a pointwise estimate into an integral estimate without requiring a formula for either integral. Once \(f\le g\) is established everywhere, no approximation sequence, limiting argument or interchange of limits and integrals is needed: the minorant family for \(f\) is contained in the minorant family for \(g\). This reasoning is valid even when the integral of \(g\) is infinite.
A common mistake is to reverse that inclusion. If \(f\le g\), then a simple function below \(f\) is certainly below \(g\); a simple function below \(g\) need not be below \(f\). For example, on \(\mathbb{R}\) with Lebesgue measure, take \(f=\chi_{(0,1)}\) and \(g=\chi_{(0,2)}\). The simple function \(\chi_{(0,2)}\) is a minorant of \(g\), but it is not a minorant of \(f\), since it equals \(1\) at \(x=\tfrac32\) while \(f(\tfrac32)=0\). Thus the correct inclusion is \(\mathcal{S}_f\subseteq\mathcal{S}_g\), not the reverse. Reversing it would incorrectly suggest \(\int g\,dm\le\int f\,dm\), contrary to the computed values \(2\) and \(1\).
Another important distinction is between \(f\le g\) and \(f<g\). The theorem guarantees a non-strict inequality between the integrals. The Dirac example shows that even if \(f<g\) at a point, the integrals may be equal. Monotonicity alone makes no claim that pointwise strictness forces a strict integral inequality. Finally, the order hypothesis here is pointwise. The result as stated does not silently replace it with an almost-everywhere comparison; that is a separate formulation requiring its own justification.
Check Your Understanding
For each question, identify the measure space and verify the pointwise order before comparing the integrals.
- For measurable \(f,g:X\to[0,\infty]\), explain why every simple minorant of \(f\) is a simple minorant of \(g\) when \(f\le g\) everywhere.
- On \(\mathbb{R}\) with Lebesgue measure, compare \(\chi_{(1,3)}\) and \(\chi_{(0,4)}\), and compute both integrals.
- On \(\mathbb{N}\) with counting measure, let \(f=\chi_{\{2,4\}}\) and \(g=\chi_{\{1,2,4\}}\). Verify the pointwise order and compute both integrals.
- Give an example of \(f\le g\) everywhere with \(f<g\) at some point but \(\int f\,d\mu=\int g\,d\mu\).
- Does the proof require \(\mu(X)<\infty\)? Identify the precise step that answers this question.