Riemann Integrability

When Do Upper and Lower Integrals Agree?

For a bounded function, the Darboux lower integral and upper integral are always defined, and the lower one never exceeds the upper one. Riemann integrability is the special case in which there is no gap between them. This definition turns the upper and lower integrals developed in the previous tutorials into a test for whether a function has a single well-defined integral.

Throughout, let $a<b$, and let $f:[a,b]\to\mathbb{R}$ be bounded. We use $\underline{\int_a^b}f$ and $\overline{\int_a^b}f$ for its Darboux lower and upper integrals. The definition below is the Darboux form of integrability; its equivalence with the tagged-sum definition introduced earlier in the course will be proved later in the course (Equivalence of Darboux and Riemann Definitions).

Definition (Riemann Integrability): A bounded function $f:[a,b]\to\mathbb{R}$ is Riemann integrable on $[a,b]$ if $$ \underline{\int_a^b}f=\overline{\int_a^b}f. $$ Their common value is the Riemann integral of $f$ on $[a,b]$, written $\int_a^b f(x)\,dx$.

The equality is important: a lower integral is built from the best possible lower-sum approximations, while an upper integral comes from the best possible upper-sum approximations. When they agree, both approaches determine the same number. The definition requires boundedness; unbounded functions on closed intervals are not Riemann integrable in this sense.

A First Calculation: The Square Function

To use the definition, it is enough in some examples to calculate lower and upper sums along a useful family of partitions. On $[0,1]$, take the uniform partition with $n$ subintervals, whose points are $x_i=i/n$. The function $f(x)=x^2$ is increasing there, so the infimum and supremum on each subinterval are its values at the left and right endpoints, respectively.

Worked Example: Integrating the Square Function

For the uniform partition $P_n$, the lower and upper sums are

L(f,P_n)=\frac{1}{n^3}\sum_{i=1}^{n}(i-1)^2 =\frac{(n-1)n(2n-1)}{6n^3}, \qquad U(f,P_n)=\frac{1}{n^3}\sum_{i=1}^{n}i^2 =\frac{n(n+1)(2n+1)}{6n^3}.

Here we used $\sum_{i=1}^n i^2=n(n+1)(2n+1)/6$, and the left-endpoint sum is the same formula with $n-1$ in place of $n$. Subtracting gives

U(f,P_n)-L(f,P_n) =\frac{n(n+1)(2n+1)-(n-1)n(2n-1)}{6n^3} =\frac{6n^2}{6n^3} =\frac1n.

For every partition, the lower integral is at least its lower sum and the upper integral is at most its upper sum. Also, the lower integral does not exceed the upper integral. Thus, for every positive integer $n$,

0\leq \overline{\int_0^1}f-\underline{\int_0^1}f \leq U(f,P_n)-L(f,P_n) =\frac1n.

The only nonnegative real number that is at most $1/n$ for every positive integer $n$ is zero. The lower and upper integrals therefore agree. Moreover, each of the displayed lower and upper sums tends to $1/3$, so the common integral is $1/3$:

\int_0^1 x^2\,dx=\frac13.

This calculation uses a particular sequence of partitions to establish equality of the two integrals. The next tutorial develops the general criterion that explains when small differences between upper and lower sums suffice.

Integrability Is Preserved by Linear Operations

The addition inequalities from Upper and Lower Integrals immediately suggest a useful result. When the upper and lower integrals of each function already agree, the one-sided bounds for their sum force agreement for the sum as well.

Theorem (Linearity of Riemann Integrability): If $f$ and $g$ are Riemann integrable on $[a,b]$, and $c\in\mathbb{R}$, then $f+g$ and $cf$ are Riemann integrable on $[a,b]$, and $$ \int_a^b(f+g)(x)\,dx =\int_a^b f(x)\,dx+\int_a^b g(x)\,dx, \qquad \int_a^b(cf)(x)\,dx =c\int_a^b f(x)\,dx. $$

Proof. Let $I=\int_a^b f(x)\,dx$ and $J=\int_a^b g(x)\,dx$. By integrability, the lower and upper integrals of $f$ are both $I$, and those of $g$ are both $J$. The Superadditivity and Subadditivity theorem gives

\underline{\int_a^b}(f+g)\geq I+J, \qquad \overline{\int_a^b}(f+g)\leq I+J.

For every bounded function, its lower integral is at most its upper integral. Applying that fact to $f+g$ yields

I+J \leq \underline{\int_a^b}(f+g) \leq \overline{\int_a^b}(f+g) \leq I+J.

All these quantities must be equal, proving that $f+g$ is integrable and has integral $I+J$. For scalar multiplication, the Scaling Upper and Lower Integrals theorem shows that when $c\geq0$, both integrals of $cf$ equal $cI$. When $c<0$, they equal $cI$ as well: the lower integral of $cf$ is $c$ times the upper integral of $f$, and the upper integral of $cf$ is $c$ times the lower integral of $f$. Since both integrals of $f$ equal $I$, these values agree. This proves the result for every real $c$. $\square$

Worked Example: Integrating a Linear Combination

The square-function calculation gives $\int_0^1 x^2\,dx=1/3$, and the constant-function calculation gives $\int_0^1 1\,dx=1$. Linearity therefore shows that $h(x)=3x^2-4$ is integrable on $[0,1]$, with

\int_0^1(3x^2-4)\,dx =3\int_0^1x^2\,dx-4\int_0^1 1\,dx =3\left(\frac13\right)-4(1) =-3.

The integrand is negative on much of the interval, which is no obstacle: integrability concerns agreement of upper and lower integrals, not nonnegativity of the function.

Restricting an Integrable Function to a Subinterval

Integrability also behaves well when an interval is split. The Additivity Across Adjacent Intervals theorem applies to upper and lower integrals whether or not the function is integrable. That fact lets us deduce integrability on each piece from integrability on the whole interval.

Theorem (Restriction to Subintervals): Suppose $f$ is bounded on $[a,b]$ and $a<c<b$. If $f$ is Riemann integrable on $[a,b]$, then it is Riemann integrable on both $[a,c]$ and $[c,b]$. Conversely, if it is integrable on both of those intervals, then it is integrable on $[a,b]$.

Proof. Define the nonnegative gap on any interval $[u,v]$ by $\overline{\int_u^v}f-\underline{\int_u^v}f$. It is nonnegative because the Darboux lower integral never exceeds the upper integral. By additivity across adjacent intervals, the gap on $[a,b]$ is the sum of the two gaps:

\overline{\int_a^b}f-\underline{\int_a^b}f = \left(\overline{\int_a^c}f-\underline{\int_a^c}f\right) + \left(\overline{\int_c^b}f-\underline{\int_c^b}f\right).

If $f$ is integrable on $[a,b]$, the left side is zero. The two terms on the right are nonnegative, so each must be zero. Thus the upper and lower integrals agree on each subinterval, proving integrability there. Conversely, if both subinterval gaps are zero, the same identity makes the gap on $[a,b]$ zero. Hence $f$ is integrable on the whole interval. $\square$

Worked Example: Restricting the Square Function

Since $x^2$ is integrable on $[0,1]$, the restriction theorem shows that it is integrable on $[0,1/2]$ and $[1/2,1]$. Its value on the first interval can also be calculated directly. For the uniform partition of $[0,1/2]$ into $n$ pieces, the endpoints are $i/(2n)$, and the upper-minus-lower sum is

\sum_{i=1}^n \left(\left(\frac{i}{2n}\right)^2-\left(\frac{i-1}{2n}\right)^2\right)\frac{1}{2n} = \frac{1}{8n^3}\sum_{i=1}^n(2i-1) = \frac{1}{8n}.

This tends to zero. The lower and upper sums both tend to $1/24$: for example, the upper sum is $\sum_{i=1}^n i^2/(8n^3)=(n+1)(2n+1)/(48n^2)$, which tends to $1/24$. Therefore $\int_0^{1/2}x^2\,dx=1/24$. Additivity and the already calculated integral on $[0,1]$ give the value on the other piece:

\int_{1/2}^1x^2\,dx =\int_0^1x^2\,dx-\int_0^{1/2}x^2\,dx =\frac13-\frac1{24} =\frac7{24}.

Changing Finitely Many Values

A bounded function can be integrable even if its values at a few points differ from nearby values. The following result makes that precise. Its proof first checks that changing a function at one point contributes zero to its integral.

Theorem (Finite Point Changes Do Not Affect the Integral): Let $f$ be Riemann integrable on $[a,b]$. If $g$ is bounded and differs from $f$ at only finitely many points, then $g$ is Riemann integrable and $$ \int_a^b g(x)\,dx=\int_a^b f(x)\,dx. $$

Proof. For $c\in[a,b]$, define $\phi_c(x)=1$ if $x=c$ and $\phi_c(x)=0$ otherwise. We show that $\phi_c$ is integrable with integral zero. Its lower sum on every partition is zero: every subinterval of positive length contains points other than $c$, where $\phi_c$ is zero, so the infimum on that subinterval is zero.

For an interior point $a<c<b$, choose $\delta>0$ small enough that $c-\delta\geq a$ and $c+\delta\leq b$. Take a partition containing $c-\delta,c,c+\delta$. Only the two subintervals touching $c$ can have positive supremum, and their total length is $2\delta$. Thus the upper sum is $2\delta$. If $c=a$ or $c=b$, include the adjacent point at distance $\delta$; the upper sum is then at most $\delta$. Because $\delta$ can be arbitrarily small, the upper integral is zero. The lower integral is also zero, so $\phi_c$ is integrable with integral zero.

Now suppose $g$ differs from $f$ only at distinct points $c_1,\ldots,c_m$. Set $d_j=g(c_j)-f(c_j)$. At every point of $[a,b]$,

g=f+\sum_{j=1}^m d_j\phi_{c_j}.

Each $\phi_{c_j}$ is integrable with integral zero. The Linearity of Riemann Integrability theorem, applied a finite number of times, implies that the sum on the right is integrable with integral zero. Adding it to $f$ proves that $g$ is integrable and that its integral equals the integral of $f$. $\square$

Worked Example: Changing the Square Function at One Point

Let $g(x)=x^2$ for $x\ne 1/2$, and let $g(1/2)=10$, on $[0,1]$. The function differs from $x^2$ at only one point. Since $x^2$ is integrable and has integral $1/3$, the finite point changes theorem gives

\int_0^1g(x)\,dx=\int_0^1x^2\,dx=\frac13.

The altered value can substantially change upper and lower sums for partitions that include $1/2$. It does not, however, change the integral: the contribution of a single point can be confined to intervals of arbitrarily small total length.

What Integrability Does—and Does Not—Say

These results let us build new integrable functions from known ones, split an integral into simpler intervals, and disregard finitely many exceptional values when calculating. They are useful because integrability can often be established once for a basic function and then transferred through these operations.

A common pitfall is to confuse a small gap for one partition with equality of the upper and lower integrals. One partition gives an upper bound on that gap, as in the square-function example; the definition itself requires the integrals to agree. The Darboux Criterion in the next tutorial gives the general connection between arbitrarily small upper-minus-lower sums and integrability.

Check Your Understanding

Use the definition and results in this tutorial to answer the following questions.

What equality of Darboux integrals defines Riemann integrability?
Why do the addition inequalities imply that the sum of two integrable functions is integrable?
If a function is integrable on a whole interval, why must it be integrable on each of two adjacent subintervals?
Why does changing a bounded integrable function at finitely many points leave its integral unchanged?
In the square-function example, what role does the upper-minus-lower sum for a uniform partition play?

What You'll Learn