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Riemann Integration · Tutorial 473 of 1000

Integrability of Monotone Functions

Learn how monotonicity controls Darboux sums and gives a direct integrability test on a closed interval.

Advanced 8 min read

What You'll Learn

  • Distinguish nondecreasing, nonincreasing, and monotone functions
  • Prove that a monotone function is bounded on a closed bounded interval
  • Compute the exact Darboux gap for a uniform partition
  • Use the Darboux Criterion to establish integrability
  • Recognize why endpoint jumps do not prevent integrability

Monotonicity Controls Variation

For a step function, the previous tutorial controlled the Darboux gap by isolating finitely many breakpoints and making the cells beside them short. A monotone function may have infinitely many changes in value, so that strategy does not apply directly. Instead, monotonicity gives a global control: as we move across the interval, the function can only move in one direction. On a uniform partition, the oscillations of successive cells therefore add by telescoping rather than accumulating without bound.

The relevant setting is a closed bounded interval \([a,b]\), with \(a<b\). We first define the direction of monotonicity precisely. The endpoint values will be especially useful: they bound all the other values and determine the total change appearing in the Darboux-gap estimate.

Definition: A function \(f:[a,b]\to\mathbb{R}\) is nondecreasing if \(f(x)\leq f(y)\) whenever \(a\leq x<y\leq b\). It is nonincreasing if \(f(x)\geq f(y)\) whenever \(a\leq x<y\leq b\). The function is monotone if it is either nondecreasing or nonincreasing.

“Monotone” here allows equality: a constant function is both nondecreasing and nonincreasing. The definition also places no continuity requirement on \(f\). In particular, jumps are allowed, provided their direction is consistent with the specified order.

Boundedness Follows from the Endpoint Values

Before applying the Darboux Criterion, we need the function to be bounded. For a monotone function this follows immediately from its direction of change, including at the endpoints.

Proposition (Boundedness of a Monotone Function): If \(f:[a,b]\to\mathbb{R}\) is monotone, then \(f\) is bounded on \([a,b]\). If \(f\) is nondecreasing, then \(f(a)\leq f(x)\leq f(b)\) for every \(x\in[a,b]\). If \(f\) is nonincreasing, then \(f(b)\leq f(x)\leq f(a)\) for every \(x\in[a,b]\).

Proof. Suppose first that \(f\) is nondecreasing. For any \(x\in[a,b]\), the definition applied to \(a\leq x\) gives \(f(a)\leq f(x)\) when \(x>a\); when \(x=a\), the same inequality is equality. Applying the definition to \(x\leq b\) gives \(f(x)\leq f(b)\), with equality allowed at \(x=b\). Hence \(f(a)\leq f(x)\leq f(b)\) throughout the interval, so \(f\) is bounded.

If \(f\) is nonincreasing, applying its defining inequality to \(a\leq x\) and to \(x\leq b\) gives \(f(a)\geq f(x)\geq f(b)\). Thus \(f(b)\leq f(x)\leq f(a)\), which again proves boundedness. \(\square\)

This proposition uses the closed interval in an essential way: the endpoint values are defined and provide the bounds. On an open interval, monotonicity alone need not give boundedness. For example, \(f(x)=1/x\) is nonincreasing on \((0,1)\) but is not bounded there. That example does not contradict the proposition, which concerns a function defined on the entire closed interval.

The Exact Gap on a Uniform Partition

Let \(n\) be a positive integer and divide \([a,b]\) into \(n\) equal subintervals. Write \(h=(b-a)/n\) for their common length and \(x_i=a+ih\) for the partition points, where \(i=0,\ldots,n\). If \(f\) is nondecreasing, its infimum and supremum on the \(i\)th cell \([x_{i-1},x_i]\) are \(f(x_{i-1})\) and \(f(x_i)\), respectively. These extrema are attained because the endpoints belong to the cell. For a nonincreasing function, the roles of the two endpoint values are reversed.

Proposition (Uniform-Partition Gap for a Monotone Function): Let \(f:[a,b]\to\mathbb{R}\) be monotone, and let \(P_n=\{x_0,\ldots,x_n\}\) be the uniform partition just defined. Then $$ U(f,P_n)-L(f,P_n)=\frac{b-a}{n}\,|f(b)-f(a)|. $$

Proof. Suppose first that \(f\) is nondecreasing. On the cell \([x_{i-1},x_i]\), the supremum minus the infimum is \(f(x_i)-f(x_{i-1})\). The definition of the upper and lower sums therefore gives

$$ U(f,P_n)-L(f,P_n) =h\sum_{i=1}^{n}\bigl(f(x_i)-f(x_{i-1})\bigr). $$

Every interior value \(f(x_i)\) in this sum occurs once with a positive sign and once with a negative sign. The sum telescopes to \(f(x_n)-f(x_0)=f(b)-f(a)\), which is nonnegative. Thus the gap is \(h(f(b)-f(a))\). Since \(h=(b-a)/n\), this is the stated formula in the nondecreasing case.

If \(f\) is nonincreasing, the supremum minus the infimum on the \(i\)th cell is \(f(x_{i-1})-f(x_i)\). Consequently,

$$ U(f,P_n)-L(f,P_n) =h\sum_{i=1}^{n}\bigl(f(x_{i-1})-f(x_i)\bigr) =h\bigl(f(a)-f(b)\bigr). $$

Here \(f(a)-f(b)\geq0\), so this also equals \(\frac{b-a}{n}|f(b)-f(a)|\). If \(f\) is constant, both calculations give zero, as does the stated formula. \(\square\)

The estimate is exact, not merely an upper bound. Subdividing the interval reduces the common cell length, while the sum of the cell oscillations remains precisely the total change from one endpoint to the other. This is the central mechanism for monotone functions: potentially many local changes are controlled by one finite endpoint difference.

Integrability of Monotone Functions

Theorem (Integrability of Monotone Functions): Every monotone function \(f:[a,b]\to\mathbb{R}\), where \(a<b\), is Riemann integrable.

Proof. By the Boundedness of a Monotone Function, \(f\) is bounded on \([a,b]\). Given \(\varepsilon>0\), choose a positive integer \(n\) large enough that

$$ \frac{b-a}{n}|f(b)-f(a)|<\varepsilon. $$

Such an \(n\) exists: if \(|f(b)-f(a)|=0\), the left-hand side is zero for every positive integer \(n\); otherwise, the left-hand side tends to zero as \(n\) increases. For the uniform partition \(P_n\), the Uniform-Partition Gap for a Monotone Function now gives \(U(f,P_n)-L(f,P_n)<\varepsilon\). Since this holds for every \(\varepsilon>0\), the Darboux Criterion implies that \(f\) is Riemann integrable. \(\square\)

This proof establishes integrability; it does not in general calculate the value of the integral. The estimate compares upper and lower sums closely enough to guarantee that they determine a single integral, but finding that value may require a separate argument.

Worked Example: A Square-Root Function

Define \(f(x)=\sqrt{x+2}\) on \([-2,2]\). If \(-2\leq x<y\leq2\), then \(x+2<y+2\), and the square-root function preserves this order on nonnegative numbers. Thus \(f\) is nondecreasing. Its endpoint values are \(f(-2)=0\) and \(f(2)=2\), so it is bounded between \(0\) and \(2\). For the uniform partition with \(n\) subintervals, the exact gap is

$$ U(f,P_n)-L(f,P_n) =\frac{2-(-2)}{n}|2-0| =\frac{8}{n}. $$

For instance, \(n=100\) gives a Darboux gap of \(8/100=0.08\). Given any positive tolerance \(\varepsilon\), we can choose \(n>8/\varepsilon\), making the gap \(8/n<\varepsilon\). The Darboux Criterion therefore confirms integrability without requiring a formula for the integral.

Worked Example: A Decreasing Affine Function

Let \(g(x)=7-3x\) on \([0,2]\). If \(x<y\), then \(g(x)-g(y)=3(y-x)>0\), so \(g\) is nonincreasing. Its endpoint values are \(g(0)=7\) and \(g(2)=1\); hence \(1\leq g(x)\leq7\) on the interval. For a uniform partition into \(n\) cells, the gap is

$$ U(g,P_n)-L(g,P_n) =\frac{2-0}{n}|1-7| =\frac{12}{n}. $$

The function decreases by \(6\) across the full interval, and the total of the cell oscillations is also \(6\). Multiplication by the cell length \(2/n\) gives \(12/n\). For example, \(n=60\) yields a gap of \(12/60=0.2\), so this partition certifies the Darboux Criterion for every tolerance greater than \(0.2\).

Worked Example: An Endpoint Jump

Define \(q:[0,1]\to\mathbb{R}\) by \(q(x)=1\) for \(0\leq x<1\) and \(q(1)=4\). This function is nondecreasing: its values are \(1\) before the endpoint, and the only larger value occurs at \(1\). In particular, the endpoint values are \(q(0)=1\) and \(q(1)=4\), so the exact gap for a uniform partition with \(n\) cells is

$$ U(q,P_n)-L(q,P_n) =\frac{1-0}{n}|4-1| =\frac{3}{n}. $$

All cells other than the last have zero oscillation. The last cell has oscillation \(4-1=3\) and length \(1/n\), giving the same gap \(3/n\). Although the value at the endpoint differs from all values immediately to its left, the cell containing that change becomes short as the partition is refined. Thus the jump does not prevent integrability.

What the Estimate Does—and Does Not—Say

The total change \(|f(b)-f(a)|\) is not the size of the oscillation on each cell; it is the sum of those oscillations for a monotone function. The mesh length then weights that sum in the Darboux gap. This distinction matters: a monotone function can have jumps, and a jump in one cell can still produce a noticeable oscillation there. The guarantee comes from making the cells short, not from assuming the function is continuous.

For a nonmonotone function, the same endpoint difference cannot generally control the sum of cell oscillations. A function may move up and down many times while returning to its starting value. Monotonicity rules out those reversals, which is why the telescoping calculation works. Nor does this argument claim that every bounded function is integrable; boundedness alone does not ensure that its upper and lower sums can be made arbitrarily close.

The result applies to both directions of monotonicity because upper and lower sums depend on the oscillation on each cell, not on whether the function rises or falls. The absolute value in the formula accounts for the direction. The endpoint values also need not describe the function’s behavior continuously at the endpoints; they need only be defined and consistent with monotonicity.

1
Check the direction.
Verify that the function is nondecreasing throughout the interval or nonincreasing throughout it.
2
Use endpoint bounds.
The endpoint values bound every value between them, so the function is bounded on the closed interval.
3
Choose a uniform partition.
On each cell, the oscillation is the difference between the endpoint values of that cell.
4
Apply the gap estimate.
The oscillations telescope, giving a Darboux gap equal to the cell length times the total endpoint change.
5
Invoke the Darboux Criterion.
Make the number of cells large enough that the gap is below the chosen positive tolerance.

A common pitfall is to infer continuity from monotonicity. Monotone functions can be discontinuous, including at endpoints, but continuity is not required for the Darboux Criterion. The useful property is order: it both bounds the function and prevents the cell oscillations from cancelling or repeatedly reversing direction. This is the reason the exact gap formula remains valid even when jumps occur.

Check Your Understanding

Use the endpoint bounds and the uniform-partition gap formula to answer the following questions.

  1. What inequalities bound a nonincreasing function \(f\) on \([a,b]\) in terms of \(f(a)\) and \(f(b)\)?
  2. Why does a monotone function on a closed interval have to be bounded?
  3. If \(f\) is nondecreasing, what is the oscillation on the cell \([x_{i-1},x_i]\) of a uniform partition?
  4. A monotone function on an interval of length \(3\) has endpoint values differing by \(5\). What is the Darboux gap for a uniform partition into \(n\) cells?
  5. Why does an endpoint jump not prevent a monotone function from being Riemann integrable?

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