Upper and Lower Integrals

How Upper and Lower Integrals Behave

The previous tutorial defined the Darboux lower integral as a supremum of lower sums and the Darboux upper integral as an infimum of upper sums. Those definitions give two finite values for every bounded function, but they do not by themselves say how the values change when the function or interval changes. We now establish useful comparison and algebra rules. They let us estimate integrals without returning to the definitions for every partition.

Throughout, let $a<b$, and let all functions under discussion be bounded on $[a,b]$. We write $\underline{\int_a^b} f$ and $\overline{\int_a^b} f$ for the Darboux lower and upper integrals. As established earlier, the lower integral never exceeds the upper integral. Unless the two are equal, neither value is called the Riemann integral.

Pointwise Order Gives Integral Order

Theorem (Monotonicity of Upper and Lower Integrals): Suppose $f(x)\leq g(x)$ for every $x\in[a,b]$. Then $$ \underline{\int_a^b}f\leq\underline{\int_a^b}g, \qquad \overline{\int_a^b}f\leq\overline{\int_a^b}g. $$

Proof. Fix any partition $P=\{x_0,\ldots,x_n\}$. On each subinterval $[x_{i-1},x_i]$, the pointwise inequality implies $\inf f\leq\inf g$ and $\sup f\leq\sup g$, where these infima and suprema are taken on that subinterval. Since its length $x_i-x_{i-1}$ is positive, multiplying by that length preserves the inequalities. Adding over the subintervals gives

L(f,P)\leq L(g,P), \qquad U(f,P)\leq U(g,P).

Taking suprema of the lower sums gives the lower-integral inequality. Taking infima of the upper sums also preserves the order: because $U(f,P)\leq U(g,P)$ for every $P$, the infimum of the first collection is at most the infimum of the second. Thus the upper-integral inequality follows as well. $\square$

In particular, if $m\leq f(x)\leq M$ throughout the interval, monotonicity and the constant-function calculation from the previous tutorial give bounds for both integrals. This is useful even when their equality has not been established.

Worked Example: Comparing Two Affine Functions

On $[0,1]$, let $f(x)=x$ and $g(x)=x+2$. Since $x\leq x+2$ for every $x\in[0,1]$, monotonicity gives

\underline{\int_0^1}x\,dx\leq\underline{\int_0^1}(x+2)\,dx, \qquad \overline{\int_0^1}x\,dx\leq\overline{\int_0^1}(x+2)\,dx.

For any partition $P$, adding $2$ to the function adds $2$ to each subinterval infimum and supremum. The total added amount in either sum is $2\sum_{i=1}^n(x_i-x_{i-1})=2$. Thus $L(g,P)=L(f,P)+2$ and $U(g,P)=U(f,P)+2$. Taking the supremum and infimum, respectively, shifts both Darboux integrals by $2$. Since the previous tutorial established that both integrals of $x$ on $[0,1]$ equal $1/2$, we obtain

\underline{\int_0^1}(x+2)\,dx =\overline{\int_0^1}(x+2)\,dx =\frac52.

The comparison theorem gives the order directly; the partition calculation identifies the exact values in this particular case.

Multiplication by a Constant

Theorem (Scaling Upper and Lower Integrals): Let $c\in\mathbb{R}$. If $c\geq0$, then $$ \underline{\int_a^b}cf=c\,\underline{\int_a^b}f, \qquad \overline{\int_a^b}cf=c\,\overline{\int_a^b}f. $$ If $c<0$, then $$ \underline{\int_a^b}cf=c\,\overline{\int_a^b}f, \qquad \overline{\int_a^b}cf=c\,\underline{\int_a^b}f. $$

Proof. On each subinterval, multiplying all function values by $c\geq0$ multiplies both the infimum and the supremum by $c$. Hence $L(cf,P)=cL(f,P)$ and $U(cf,P)=cU(f,P)$. Taking the supremum in the first identity and the infimum in the second proves the formulas when $c>0$. When $c=0$, every sum for $cf$ is zero, so both integrals are zero, as the formulas state.

If $c<0$, multiplication reverses order: the infimum of $cf$ on a subinterval is $c$ times the supremum of $f$, and its supremum is $c$ times the infimum of $f$. Therefore $L(cf,P)=cU(f,P)$ and $U(cf,P)=cL(f,P)$. As $P$ varies, the supremum of $cU(f,P)$ is $c$ times the infimum of $U(f,P)$, because multiplication by a negative number reverses order. The infimum of $cL(f,P)$ is $c$ times the supremum of $L(f,P)$. These are exactly the asserted formulas. $\square$

Worked Example: Negating the Identity Function

Let $f(x)=x$ on $[0,1]$. The lower and upper integrals of $f$ are both $1/2$. Taking $c=-1$ in the scaling theorem swaps the roles of the upper and lower integrals and multiplies their values by $-1$. Thus

\underline{\int_0^1}(-x)\,dx =-\overline{\int_0^1}x\,dx=-\frac12, \qquad \overline{\int_0^1}(-x)\,dx =-\underline{\int_0^1}x\,dx=-\frac12.

The swap is necessary: on each subinterval, the smallest value of $-x$ comes from the largest value of $x$. A negative factor does not preserve the distinction between infima and suprema.

Adding Functions: One-Sided Inequalities

Theorem (Superadditivity and Subadditivity): If $f$ and $g$ are bounded on $[a,b]$, then $$ \underline{\int_a^b}(f+g) \geq \underline{\int_a^b}f+\underline{\int_a^b}g, \qquad \overline{\int_a^b}(f+g) \leq \overline{\int_a^b}f+\overline{\int_a^b}g. $$

Proof. On any subinterval $J$, every value of $f+g$ is at least $\inf_J f+\inf_J g$, so $\inf_J(f+g)\geq\inf_J f+\inf_J g$. Similarly, every value of $f+g$ is at most $\sup_J f+\sup_J g$, giving $\sup_J(f+g)\leq\sup_J f+\sup_J g$. Multiplying by subinterval lengths and adding yields, for every partition $P$,

L(f+g,P)\geq L(f,P)+L(g,P), \qquad U(f+g,P)\leq U(f,P)+U(g,P).

To pass from these inequalities to the integrals, fix $\varepsilon>0$. By the Approximation by Partition Sums theorem, choose partitions $P_f,P_g$ such that $L(f,P_f)>\underline{\int_a^b}f-\varepsilon/2$ and $L(g,P_g)>\underline{\int_a^b}g-\varepsilon/2$. Take a common refinement $R$. Refinement does not decrease lower sums, so $L(f,R)\geq L(f,P_f)$ and $L(g,R)\geq L(g,P_g)$. Consequently,

\underline{\int_a^b}(f+g) \geq L(f+g,R) \geq L(f,R)+L(g,R) > \underline{\int_a^b}f+\underline{\int_a^b}g-\varepsilon.

Since this holds for every $\varepsilon>0$, the lower-integral inequality follows. For the upper inequality, choose $P_f,P_g$ with upper sums less than their respective upper integrals plus $\varepsilon/2$. On a common refinement $R$, upper sums do not increase. Therefore

\overline{\int_a^b}(f+g) \leq U(f+g,R) \leq U(f,R)+U(g,R) < \overline{\int_a^b}f+\overline{\int_a^b}g+\varepsilon.

Letting $\varepsilon$ be arbitrary proves the upper-integral inequality. $\square$

These are inequalities rather than unconditional equalities. The lower integral of a sum is bounded below by the sum of the lower integrals, while the upper integral is bounded above by the sum of the upper integrals. If $f$ and $g$ are both Riemann integrable, their lower and upper integrals coincide individually; the two inequalities then force the corresponding values for $f+g$ to coincide at the sum. The next tutorial develops the integrability criterion that makes this observation useful.

Worked Example: Adding a Function and Its Negative

On $[0,1]$, let $f(x)=x$ and $g(x)=1-x$. For any partition $P$, the lower sum of $g$ equals the upper sum of $f$ subtracted from $1$, because on each subinterval the infimum of $1-x$ is $1$ minus the supremum of $x$. More explicitly, if $\Delta_i=x_i-x_{i-1}$, then

L(g,P)=\sum_{i=1}^n(1-x_i)\Delta_i =\sum_{i=1}^n\Delta_i-U(f,P) =1-U(f,P).

Likewise, $U(g,P)=1-L(f,P)$. Since both integrals of $f(x)=x$ are $1/2$, taking the supremum of the first identity and the infimum of the second gives lower and upper integrals of $g$ equal to $1/2$. Also $f(x)+g(x)=1$ at every point, so both integrals of the sum equal $1$. In this example the addition inequalities are equalities:

\underline{\int_0^1}(f+g)\,dx =\frac12+\frac12=1, \qquad \overline{\int_0^1}(f+g)\,dx =\frac12+\frac12=1.

Splitting an Interval

Theorem (Additivity Across Adjacent Intervals): If $a<c<b$ and $f$ is bounded on $[a,b]$, then $$ \underline{\int_a^b}f = \underline{\int_a^c}f+\underline{\int_c^b}f, \qquad \overline{\int_a^b}f = \overline{\int_a^c}f+\overline{\int_c^b}f. $$

Proof. First consider lower integrals. Given any partition $P$ of $[a,b]$, add $c$ to its partition points if necessary, producing a refinement $R$ that splits at $c$. By the Upper and Lower Sums Under Refinement theorem, $L(f,P)\leq L(f,R)$. The sum $L(f,R)$ is the sum of a lower partition sum on $[a,c]$ and one on $[c,b]$; each is at most the lower integral on its own interval. Hence every $L(f,P)$ is at most the sum of the two lower integrals. Taking the supremum over $P$ proves $\underline{\int_a^b}f\leq\underline{\int_a^c}f+\underline{\int_c^b}f$.

For the reverse inequality, use Approximation by Partition Sums to choose partitions on $[a,c]$ and $[c,b]$ whose lower sums are within $\varepsilon/2$ of their respective lower integrals. Joining the partitions at $c$ gives a partition of $[a,b]$ whose lower sum is the sum of those two lower sums. Thus the lower integral on $[a,b]$ is greater than the sum of the component lower integrals minus $\varepsilon$. Since $\varepsilon>0$ is arbitrary, the reverse inequality follows.

For upper integrals, insert $c$ into any partition $P$. Refinement does not increase upper sums, so $U(f,P)\geq U(f,R)$. The refined upper sum is a sum of upper sums on the two component intervals, each at least the upper integral there. Taking the infimum over $P$ gives $\overline{\int_a^b}f\geq\overline{\int_a^c}f+\overline{\int_c^b}f$. Conversely, choose component partitions with upper sums within $\varepsilon/2$ of their upper integrals and join them. The resulting upper sum on $[a,b]$ is less than the sum of the component upper integrals plus $\varepsilon$. Taking the infimum and then letting $\varepsilon$ be arbitrary gives the reverse inequality. This proves both identities. $\square$

Worked Example: Splitting the Integral of the Identity Function

Consider $f(x)=x$ on $[0,2]$, split at $1$. The same telescoping calculation used for the identity function on $[0,1]$, now applied on any interval $[u,v]$, gives for a partition with subinterval lengths $\Delta_i$

L(f,P)=\frac{v^2-u^2}{2}-\frac12\sum_i\Delta_i^2, \qquad U(f,P)=\frac{v^2-u^2}{2}+\frac12\sum_i\Delta_i^2.

Uniform partitions make $\sum_i\Delta_i^2$ arbitrarily small, while every such sum is nonnegative. Therefore the lower and upper integrals on $[u,v]$ both equal $(v^2-u^2)/2$. Applying this on each interval gives

\underline{\int_0^1}x\,dx =\overline{\int_0^1}x\,dx=\frac12, \qquad \underline{\int_1^2}x\,dx =\overline{\int_1^2}x\,dx=\frac32.

The additivity theorem then gives lower and upper integrals on $[0,2]$ equal to $1/2+3/2=2$, which also agrees with the direct value $(2^2-0^2)/2=2$.

Using the Rules Carefully

Monotonicity, scaling, addition inequalities, and interval additivity are often used together. Pointwise bounds can enclose both integrals; negative scaling requires swapping upper and lower values; and splitting at a convenient point does not change either integral except by expressing it as a sum of values on the two pieces. These rules apply to bounded functions whether or not their upper and lower integrals are equal.

A common pitfall is to treat the addition inequalities as equalities before integrability has been shown. The proof only guarantees superadditivity for lower integrals and subadditivity for upper integrals. Another is to multiply an inequality for upper integrals by a negative constant without reversing its direction. The scaling theorem handles this explicitly: a negative multiple interchanges the roles of upper and lower integrals.

The next tutorial uses the gap between upper and lower integrals to state when a bounded function is Riemann integrable. The properties here will then provide efficient ways to establish integrability and calculate the resulting value.

Check Your Understanding

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

If $f\leq g$ everywhere on an interval, what inequalities follow for their lower and upper integrals?
Why does multiplication by a negative constant interchange the upper- and lower-integral formulas?
Which direction does the addition inequality take for lower integrals, and which direction for upper integrals?
What role does a common refinement play in proving the addition inequalities?
How does inserting a point $c$ into a partition help prove additivity across $[a,c]$ and $[c,b]$?

What You'll Learn