Properties of the Riemann Integral

From Integrability to Useful Integral Properties

The previous tutorials established when a bounded function is Riemann integrable and how to certify integrability with Darboux sums. Once integrability is known, the integral can be used to compare functions and to break calculations into smaller intervals. Here we establish several such properties without assuming the linearity of the integral, which is the subject of the next tutorial.

We will use two earlier results. First, for an integrable function, its integral agrees with both its Darboux lower integral and its Darboux upper integral. Second, the upper and lower integrals are monotone: pointwise order between bounded functions gives the same order between their upper integrals and between their lower integrals. These facts let us prove order preservation for the Riemann integral directly.

Order Is Preserved by Integration

Proof. By monotonicity of the Darboux lower integrals, $$ \underline{\int_a^b} f\leq\underline{\int_a^b} g. $$ Because both functions are Riemann integrable, each lower integral equals the corresponding Riemann integral. Substituting those equalities gives the claimed inequality. \(\square\)

Taking \(f\) to be the zero function shows that a nonnegative integrable function has a nonnegative integral. More generally, if \(m\leq f(x)\leq M\) on the whole interval, order preservation compares the integral of \(f\) with the integrals of the constant functions \(m\) and \(M\). This recovers the integral bounds established earlier.

Splitting an Interval

A second basic property is additivity across adjacent intervals. This is distinct from linearity: it concerns splitting the domain of integration, not integrating a sum of functions. The additivity of the Darboux lower and upper integrals across adjacent intervals was established earlier. Together with integrability of restrictions, it gives the corresponding property for Riemann integrals.

Proof. By the theorem on restriction to subintervals, \(f\) is integrable on \([a,c]\) and \([c,b]\). The additivity theorem for Darboux lower integrals gives $$ \underline{\int_a^b}f = \underline{\int_a^c}f+\underline{\int_c^b}f. $$ Each lower integral in this equality equals the Riemann integral on its interval, because \(f\) is integrable on all three intervals. Replacing the lower integrals by those Riemann integrals proves the result. \(\square\)

Repeated application allows any finite subdivision of the interval. Values at the shared endpoints do not cause a problem: each restriction uses the original function, and the additivity theorem for the Darboux integrals already accounts for the endpoints.

Absolute Values and the Integral

Absolute values introduce two related questions. If \(f\) is integrable, is \(|f|\) also integrable? And how large can the integral of \(f\) be compared with the integral of \(|f|\)? Both answers follow from a basic pointwise inequality and control of Darboux gaps.

Proof. Since \(f\) is integrable, it is bounded, so \(|f|\) is bounded. For any two real numbers \(s,t\), the triangle inequality gives $$ \bigl||s|-|t|\bigr|\leq|s-t|. $$ Consequently, on each subinterval of any partition, the oscillation of \(|f|\) is at most the oscillation of \(f\). The Oscillation Formula for the Darboux Gap therefore gives $$ 0\leq U(|f|,P)-L(|f|,P) \leq U(f,P)-L(f,P). $$ For every \(\varepsilon>0\), the Darboux Criterion supplies a partition \(P\) with \(U(f,P)-L(f,P)<\varepsilon\). The displayed inequality then gives \(U(|f|,P)-L(|f|,P)<\varepsilon\). The Darboux Criterion proves that \(|f|\) is integrable.

Write \(I=\int_a^b f(x)\,dx\) and \(J=\int_a^b |f(x)|\,dx\). Fix \(\delta>0\), and choose a partition \(P\) with Darboux gap for \(f\) less than \(\delta\). The gap for \(|f|\) is also less than \(\delta\). Choose one tag in each cell of \(P\), and let \(S_f\) and \(S_{|f|}\) be the resulting tagged sums. The integral lies between the lower and upper sums for every partition, and every tagged sum lies between those same sums. Hence $$ |I-S_f|<\delta \quad\text{and}\quad |J-S_{|f|}|<\delta. $$ At each tag, \(|f(x)|\) is the absolute value of \(f(x)\), so the triangle inequality for finite sums gives \(|S_f|\leq S_{|f|}\). It follows that $$ |I| \leq |S_f|+\delta \leq S_{|f|}+\delta \leq J+2\delta. $$ This holds for every \(\delta>0\). If \(|I|>J\), choosing \(2\delta<|I|-J\) would contradict the last inequality. Therefore \(|I|\leq J\), as required. \(\square\)

What a Zero Integral Does—and Does Not—Tell You

For a continuous nonnegative function, a zero integral forces the function to vanish everywhere. Continuity matters: without it, a nonnegative integrable function can be nonzero at a point and still have integral zero.

Proof. Suppose instead that \(f(c)>0\) for some \(c\in[a,b]\), and set \(q=f(c)\). By continuity at \(c\), there is a neighborhood of \(c\), relative to \([a,b]\), on which \(f(x)>q/2\). Since \(a<b\), we can choose \(u,v\in[a,b]\) with \(u<v\), both in that neighborhood. Then \(f(x)\geq q/2\) on \([u,v]\). Order preservation gives $$ \int_u^v f(x)\,dx \geq \int_u^v \frac{q}{2}\,dx =\frac{q}{2}(v-u)>0. $$ By repeated interval additivity, the integral on \([a,b]\) is the sum of the integrals over \([a,u]\), \([u,v]\), and \([v,b]\), omitting any piece of zero length if an endpoint coincides. Each remaining piece has nonnegative integral by order preservation, since \(f\geq0\). Therefore \(\int_a^b f(x)\,dx>0\), contradicting the hypothesis. Thus no such \(c\) exists, and \(f\) is identically zero. \(\square\)

The continuity hypothesis cannot simply be dropped. The function that equals \(1\) at \(x=1/2\) and \(0\) everywhere else on \([0,1]\) is Riemann integrable by the theorem on finite point changes, and its integral is the same as that of the zero function: \(0\). Yet the function is not identically zero. An integral records accumulated area in the Riemann sense; it does not detect every isolated change in function values.