Order and the Lebesgue Integral
Monotonicity lets us compare integrals without evaluating them. The Monotonicity of the Lebesgue Integral theorem, established in “The Lebesgue Integral,” says that if nonnegative measurable functions \(f\) and \(g\) satisfy \(f\leq g\) everywhere, then \(\int_X f\,d\mu\leq\int_X g\,d\mu\). This tutorial develops two useful refinements: the comparison remains valid when the order holds only almost everywhere, and it becomes strict under appropriate additional hypotheses.
Throughout, \((X,\mathcal{F},\mu)\) is a measure space. Integrals of nonnegative measurable functions are allowed to equal \(+\infty\). We use the Zero Integral Criterion and finite additivity of the nonnegative integral from earlier in this course. In particular, a nonnegative measurable function has integral zero exactly when it vanishes almost everywhere.
The phrase “almost everywhere” permits exceptions on a null set, but it does not permit exceptions on a set of positive measure. The following result explains why null exceptions do not affect comparisons of nonnegative integrals.
Order Holding Almost Everywhere
Proof. Choose a measurable null set \(N\) such that \(f\leq g\) on \(X\setminus N\). The functions \(f\mathbf{1}_N\) and \(g\mathbf{1}_N\) vanish outside \(N\), so each is zero almost everywhere. By the Zero Integral Criterion, $$ \int_X f\mathbf{1}_N\,d\mu=0 \qquad\text{and}\qquad \int_X g\mathbf{1}_N\,d\mu=0. $$ Decompose each function into its parts on \(N\) and \(X\setminus N\). Finite additivity of the nonnegative integral gives $$ \int_X f\,d\mu = \int_X f\mathbf{1}_N\,d\mu+\int_X f\mathbf{1}_{X\setminus N}\,d\mu = \int_X f\mathbf{1}_{X\setminus N}\,d\mu, $$ and likewise $$ \int_X g\,d\mu = \int_X g\mathbf{1}_{X\setminus N}\,d\mu. $$ On all of \(X\), \(f\mathbf{1}_{X\setminus N}\leq g\mathbf{1}_{X\setminus N}\): on \(X\setminus N\) this is the assumed order, and on \(N\) both functions are zero. The Monotonicity of the Lebesgue Integral theorem therefore gives $$ \int_X f\mathbf{1}_{X\setminus N}\,d\mu \leq \int_X g\mathbf{1}_{X\setminus N}\,d\mu. $$ The two integral identities prove the claim. This reasoning also applies if either integral is infinite.
Worked Example: Changing a Function at One Point
Take \(X=[0,1]\) with Lebesgue measure, and define \(f(x)=1\) for every \(x\). Define \(g(0)=0\) and \(g(x)=2\) for \(0<x\leq1\). Both functions are measurable and nonnegative. At \(x=0\), \(f(0)>g(0)\), but the exceptional set \(\{0\}\) has measure zero. Thus \(f\leq g\) almost everywhere, and the theorem gives \(\int_0^1 f\,dx\leq\int_0^1 g\,dx\). Direct calculation confirms this: $$ \int_0^1 f\,dx=1, \qquad \int_0^1 g\,dx=0\cdot\lambda(\{0\})+2\cdot\lambda((0,1])=2. $$ The pointwise failure at zero does not affect the integral comparison.
When the Inequality Is Strict
Pointwise order alone gives a weak inequality. To obtain a strict inequality, the functions must differ on a set of positive measure, and the integral of the smaller function must be finite. This finite-integral condition prevents the comparison from becoming \(+\infty<+\infty\), which is not a meaningful strict inequality.
Proof. Set \(h=g-f\). Since \(f\) and \(g\) are finite-valued measurable functions and \(f\leq g\), \(h\) is a finite-valued nonnegative measurable function. It is positive precisely on the set where \(f<g\). That set has positive measure, so \(h\) is not zero almost everywhere. The Zero Integral Criterion implies that \(\int_X h\,d\mu\neq0\), and therefore $$ \int_X h\,d\mu>0, $$ where the integral may be infinite. Since \(g=f+h\), finite additivity of the nonnegative integral gives $$ \int_X g\,d\mu=\int_X f\,d\mu+\int_X h\,d\mu. $$ The first term on the right is finite, and the second is strictly positive. Their sum is strictly greater than \(\int_X f\,d\mu\), including when \(\int_X h\,d\mu=+\infty\). This proves the assertion.
Worked Example: A Strict Comparison on an Interval
On \(X=[0,2]\) with Lebesgue measure, let \(f(x)=x\) and \(g(x)=x+1\). They are finite-valued and measurable, with \(f(x)<g(x)\) at every point. Also, $$ \int_0^2 f(x)\,dx=\int_0^2 x\,dx =\left[\frac{x^2}{2}\right]_0^2=2<\infty. $$ The difference is \(g-f=1\), so $$ \int_0^2(g-f)\,dx=\int_0^2 1\,dx=2. $$ Consequently, $$ \int_0^2g(x)\,dx = \int_0^2f(x)\,dx+\int_0^2(g-f)(x)\,dx = 2+2=4. $$ The strict inequality follows both from the theorem and from the direct calculations.
The positive-measure condition is essential: a strict pointwise inequality on a null set need not change an integral. The finite-integral condition is also essential if one wants to conclude a strict inequality between the two integral values.
Worked Example: Strict Inequality Only on a Null Set
On \([0,1]\) with Lebesgue measure, set \(f(x)=1\) everywhere and \(g(x)=1+\mathbf{1}_{\{0\}}(x)\). Then \(f\leq g\) everywhere and \(f<g\) at \(x=0\). But the strict inequality holds only on the null set \(\{0\}\), and $$ \int_0^1 f(x)\,dx=1, \qquad \int_0^1 g(x)\,dx = \int_0^1 1\,dx+\int_0^1\mathbf{1}_{\{0\}}\,dx = 1+\lambda(\{0\})=1. $$ Thus pointwise strictness somewhere does not suffice; the strict inequality must hold on a set of positive measure.
Worked Example: Why Finiteness Matters
On \(X=\mathbb{R}\) with Lebesgue measure, let \(f(x)=1\) and \(g(x)=2\). These functions satisfy \(f<g\) everywhere, so the set of strict inequality has positive measure. Nevertheless, $$ \int_{\mathbb{R}} f\,d\lambda=+\infty \qquad\text{and}\qquad \int_{\mathbb{R}} g\,d\lambda=+\infty. $$ There is no strict inequality between these extended integral values. Here the difference \(g-f=1\) also has infinite integral, and adding it to the already infinite integral of \(f\) still gives \(+\infty\). This is why the strict monotonicity theorem assumes \(\int_X f\,d\mu<\infty\).
Bounding an Integral by Pointwise Bounds
Monotonicity also turns pointwise estimates into integral estimates. For example, suppose \(E\in\mathcal{F}\) has finite measure and \(f\) is measurable with \(0\leq a\leq f\leq b\) on \(E\), where \(a\) and \(b\) are finite constants. Applying monotonicity to \(a\mathbf{1}_E\leq f\mathbf{1}_E\leq b\mathbf{1}_E\), and using the Integral of an Indicator theorem and nonnegative homogeneity, gives $$ a\mu(E)\leq\int_E f\,d\mu\leq b\mu(E). $$ These bounds remain useful even when the exact integral is difficult to compute.
Worked Example: Estimating an Integral from Bounds
Let \(E=[0,2]\) with Lebesgue measure, and let \(f(x)=2+x/2\). For \(0\leq x\leq2\), $$ 2\leq 2+\frac{x}{2}\leq3. $$ Since \(\lambda(E)=2\), the bounds imply $$ 4=2\lambda(E)\leq\int_E f\,d\lambda\leq3\lambda(E)=6. $$ In this case the integral can also be calculated: $$ \int_0^2\left(2+\frac{x}{2}\right)\,dx = \left[2x+\frac{x^2}{4}\right]_0^2 = 4+1=5. $$ The calculated value lies between the two bounds.
What the Order Hypotheses Do—and Do Not—Say
For nonnegative measurable functions, pointwise order gives an integral inequality, and almost-everywhere order gives the same conclusion because changing values on a null set does not change a nonnegative integral. Strict inequality requires more: a positive-measure set where the functions differ and a finite integral for the smaller function. If the functions are integrable real-valued functions rather than nonnegative ones, the same strict-comparison idea can be applied to their nonnegative difference when the order holds; the difference is integrable and its integral is positive whenever it is positive on a set of positive measure.
A useful habit is to check the hypotheses before drawing a conclusion. “\(f<g\) somewhere” does not imply strict inequality between the integrals, because that may happen only on a null set. Even “\(f<g\) almost everywhere” does not force strict inequality when both integrals are infinite. In contrast, pointwise bounds on a finite-measure set directly yield numerical upper and lower bounds for the integral over that set.
Check Your Understanding
Use the order results and the examples above to answer these questions.
- Why does \(f\leq g\) almost everywhere imply the same integral inequality as pointwise order?
- In the strict monotonicity theorem, how does the Zero Integral Criterion show that \(\int_X(g-f)\,d\mu>0\)?
- Give the two distinct reasons that the strict comparison theorem’s hypotheses cannot be dropped, as illustrated by the examples.
- If \(0\leq f\leq 4\) on a measurable set \(E\) with \(\mu(E)=3\), what integral bounds follow?
- On \([0,1]\), why do \(f=1\) and \(g=1+\mathbf{1}_{\{0\}}\) have equal integrals despite \(f<g\) at one point?