Order Preservation of the Integral

Pointwise Order and Integral Comparison

The Linearity of the Integral tutorial established how to integrate sums and scalar multiples. Another fundamental tool is order preservation: a pointwise inequality between integrable functions gives the same inequality between their integrals. This tutorial develops consequences of that result, especially the distinction between a weak integral inequality and a strict one.

We will use the Order Preservation for the Riemann Integral Theorem, established in Properties of the Riemann Integral, without reproving it. In particular, if \(p\) and \(q\) are Riemann integrable on \([a,b]\) and \(p(x)\leq q(x)\) throughout that interval, then \(\int_a^b p(x)\,dx\leq\int_a^b q(x)\,dx\). The key question here is what additional information can make this comparison quantitative or strict.

A Uniform Gap Forces a Quantitative Integral Gap

Knowing only that \(f\leq g\) gives an ordering of the integrals, but it does not say how far apart they are. If the difference \(g-f\) is at least a fixed positive amount on an interval of positive length, then its integral must account for at least that much area there.

Proof. By linearity, \(h=g-f\) is Riemann integrable. The assumptions give \(h(x)\geq0\) on \([a,b]\) and \(h(x)\geq\delta\) on \([u,v]\). Define a step function \(s:[a,b]\to\mathbb{R}\) by setting \(s(x)=\delta\) on \([u,v]\) and \(s(x)=0\) outside \([u,v]\). This is a step function, so it is Riemann integrable. At every point of \([u,v]\), \(h(x)\geq s(x)=\delta\); outside that interval, \(h(x)\geq0=s(x)\). Therefore \(s(x)\leq h(x)\) throughout \([a,b]\).

By order preservation, \(\int_a^b s(x)\,dx\leq\int_a^b h(x)\,dx\). The formula for the integral of a step function gives \(\int_a^b s(x)\,dx=\delta(v-u)\). By linearity, \(\int_a^b h(x)\,dx=\int_a^b g(x)\,dx-\int_a^b f(x)\,dx\). Combining these equalities and the inequality proves the result. \(\square\)

The interval must have positive length: if the gap occurs only at one point, this argument supplies no positive lower bound. The bound also depends on both ingredients, the size \(\delta\) of the gap and the length \(v-u\) of the interval where it holds.

When a Strict Pointwise Inequality Gives a Strict Integral Inequality

For continuous functions, even one strict inequality is enough to produce a strict inequality between the integrals, provided the weak inequality holds everywhere. Continuity is what turns a positive difference at one point into a positive difference throughout a short interval.

Proof. Set \(h=g-f\). Then \(h\) is continuous, \(h(x)\geq0\) throughout \([a,b]\), and \(h(x_0)>0\). Write \(d=h(x_0)\), so \(d>0\). By continuity at \(x_0\), there is an \(\eta>0\) such that \(h(x)>d/2\) whenever \(x\in[a,b]\) and \(|x-x_0|<\eta\).

Because \(a<b\), we can choose \(u,v\in[a,b]\) with \(u<v\) such that every point of \([u,v]\) lies within distance \(\eta\) of \(x_0\). If \(x_0\) is an endpoint, choose a sufficiently short interval extending into \([a,b]\); if it is an interior point, choose a sufficiently short interval around it. Thus \(h(x)\geq d/2\) on \([u,v]\). The Integral Bound from a Uniform Gap, applied to \(f\) and \(g\) with \(\delta=d/2\), yields

The final inequality is strict because \(d>0\) and \(v-u>0\). Rearranging proves the theorem. \(\square\)

One immediate consequence is an equality test for continuous functions. If \(f\) and \(g\) are continuous, \(f\leq g\) everywhere, and their integrals are equal, then \(f=g\) everywhere. Indeed, if they differed at any point, the strict integral comparison theorem would make their integrals unequal. This conclusion depends on continuity; it does not hold for all Riemann integrable functions.

Why Strictness Can Fail Without Continuity

For arbitrary Riemann integrable functions, \(f(x)<g(x)\) at a point does not imply that their integrals are strictly ordered. A single point has no interval length, so changing a function only there need not change its integral. The earlier Finite Point Changes Do Not Affect the Integral Theorem makes this precise.

It is also important to distinguish pointwise order from the sign of an integral. A function may take both positive and negative values and still have a positive, negative, or zero integral, depending on the balance of its contributions. Order preservation compares two functions; it does not say that the integral of every function that is positive somewhere must be positive. For a dependable strict lower bound, establish a positive gap on an interval of positive length.

Using Order Preservation to Bound Integrals

In applications, the comparison functions need not be polynomials. If an integrable function \(f\) is trapped between integrable functions \(p\) and \(q\), order preservation gives

This is useful when \(p\) and \(q\) have simpler integrals than \(f\). If the bounds are constants, the Bounds for the Integral Theorem gives the familiar estimate \(m(b-a)\leq\int_a^b f(x)\,dx\leq M(b-a)\) whenever \(m\leq f(x)\leq M\) throughout the interval. A comparison may be sharp or quite rough; its value is that every bound follows from an inequality verified pointwise.

When a strict conclusion is needed, check the hypotheses carefully. A weak pointwise comparison always gives a weak comparison of integrals for integrable functions. A strict pointwise inequality at one isolated location need not improve it. A uniform positive gap over an interval of positive length does improve it, and continuity can supply such an interval from a strict inequality at a single point.