Monotone Functions Are Riemann Integrable

Why Monotonicity Helps

A bounded function can vary in complicated ways across an interval. But if it is monotone, its values cannot repeatedly rise and fall. On any subinterval, its smallest and largest values occur at the endpoints. This lets us estimate the difference between upper and lower sums by adding endpoint changes, and those changes telescope.

The result is a useful integrability theorem: every bounded monotone function on a closed interval is Riemann integrable. The function need not be continuous. The proof depends on the upper-lower sum criterion and a direct estimate for equal partitions.

Monotonicity and Darboux Sums

Let \(f\) be bounded on \([a,b]\), and let \(P=\{x_0,x_1,\ldots,x_n\}\) be a partition, where \(a=x_0<x_1<\cdots<x_n=b\). On the \(i\)-th subinterval \([x_{i-1},x_i]\), let \(m_i\) be the infimum of the values of \(f\), and let \(M_i\) be their supremum. The lower and upper Darboux sums are

Boundedness ensures these infima and suprema are finite. The upper-lower sum criterion says that a bounded function \(f\) on \([a,b]\) is Riemann integrable if and only if, for every \(\varepsilon>0\), there is a partition \(P\) such that \(U(f,P)-L(f,P)<\varepsilon\).

We will use equal partitions. For a positive integer \(n\), divide \([a,b]\) into \(n\) subintervals of common width \(\Delta x=(b-a)/n\), with partition points \(x_i=a+i\Delta x\) for \(i=0,\ldots,n\).

The Equal-Partition Estimate

Suppose first that \(f\) is nondecreasing. For every \(x\in[x_{i-1},x_i]\), monotonicity gives \(f(x_{i-1})\le f(x)\le f(x_i)\). Thus the infimum on that subinterval is \(f(x_{i-1})\) and its supremum is \(f(x_i)\). These endpoint values give the exact upper-lower sum difference:

The middle equality follows by cancellation: each interior value \(f(x_i)\), for \(1\le i<n\), appears once with a plus sign and once with a minus sign. Since \(f\) is nondecreasing, \(f(b)-f(a)\ge0\).

If \(f\) is nonincreasing, the endpoint roles reverse. On \([x_{i-1},x_i]\), the supremum is \(f(x_{i-1})\) and the infimum is \(f(x_i)\). Consequently,

Both cases can be expressed in one estimate. If \(f\) is monotone, the endpoint change in the direction of its variation is nonnegative, and its size is \(|f(b)-f(a)|\). Therefore, for the equal partition into \(n\) pieces,

Worked Examples

Why Every Bounded Monotone Function Is Integrable

The estimate proves the theorem directly. Boundedness is part of the hypothesis and ensures that the Darboux sums are finite. Monotonicity supplies the endpoint formulas, so we can choose the number of equal subintervals to make the gap as small as desired.

This proof does not require continuity. Monotonicity controls the total variation between the endpoint values, while finer equal partitions divide that total into a smaller upper-lower sum gap. The argument establishes integrability without needing to identify the value of the integral.

A Common Sign Error

For a decreasing function, it is easy to use the increasing-function expression without changing its sign. For \(f(x)=1/(x+1)\) on \([0,3]\), that mistake would give

But an upper sum cannot be smaller than a lower sum, so a negative difference is impossible. The error is using the change in the wrong direction. For a decreasing function, the exact expression is \(\frac{b-a}{n}(f(a)-f(b))\), which here gives \(1(1-\frac14)=\frac34\). Equivalently, use the absolute-value formula \(\frac{b-a}{n}|f(b)-f(a)|\), which works for either direction of monotonicity.

Why the Result Matters

The theorem gives a practical way to establish integrability for functions whose formulas or behavior may be less convenient to analyze through continuity. A bounded monotone function may have discontinuities, but the upper-lower sum criterion still applies: on a fine equal partition, the gap is controlled by the endpoint change divided by the number of pieces.

The key distinction is between the function's total endpoint change and the width of the subintervals. Each subinterval can contribute some gap, but when the function is monotone, the endpoint changes add by telescoping rather than accumulating independently at the full endpoint range. That cancellation is what makes the bound tend to zero.