How to Find the Infimum and Supremum of a Riemann Sum

A Riemann sum depends on two choices: a partition of the interval and a sample point in each subinterval. So the infimum and supremum depend on which choices are allowed to vary. The most common interpretation is to fix a partition and vary the sample points; in that case, the answers are the lower and upper Darboux sums.

Riemann sums for a fixed partition

Let $f$ be bounded on $[a,b]$, and choose a partition

$$a=x_0<x_1<\cdots<x_n=b.$$

Write $\Delta x_i=x_i-x_{i-1}$. A tagged Riemann sum for this partition is

$$S(P,\xi)=\sum_{i=1}^{n} f(\xi_i)\,\Delta x_i,$$

where each sample point $\xi_i$ lies in $[x_{i-1},x_i]$. Define the infimum and supremum of $f$ on the $i$th subinterval by

$$m_i=\inf_{x\in[x_{i-1},x_i]}f(x),\qquad M_i=\sup_{x\in[x_{i-1},x_i]}f(x).$$

Since $m_i\le f(\xi_i)\le M_i$ and each width $\Delta x_i$ is positive, every tagged sum satisfies

$$\sum_{i=1}^{n}m_i\Delta x_i\le S(P,\xi)\le \sum_{i=1}^{n}M_i\Delta x_i.$$

Moreover, the sample points can be chosen independently. They can therefore make the sum as close as desired to either bound. Thus

$$\inf_{\xi} S(P,\xi)=\sum_{i=1}^{n}m_i\Delta x_i, \qquad \sup_{\xi} S(P,\xi)=\sum_{i=1}^{n}M_i\Delta x_i.$$

These are called the lower Darboux sum $L(f,P)$ and upper Darboux sum $U(f,P)$. The infimum or supremum need not be attained: if $f$ does not reach its infimum or supremum on a subinterval, the corresponding bound is approached rather than achieved.

Worked example: $f(x)=x$ on $[0,1]$

Use the partition $P=\{0,\tfrac12,1\}$. On the first subinterval, $[0,\tfrac12]$, the infimum is $0$ and the supremum is $\tfrac12$. On the second, $[\tfrac12,1]$, they are $\tfrac12$ and $1$. Each subinterval has width $\tfrac12$, so

$$L(f,P)=0\cdot\frac12+\frac12\cdot\frac12=\frac14,$$

and

$$U(f,P)=\frac12\cdot\frac12+1\cdot\frac12=\frac34.$$

Therefore, with this partition fixed and the tags allowed to vary, the infimum of the Riemann sums is $\tfrac14$ and the supremum is $\tfrac34$. For example, choosing tags at the midpoints gives the sum $\tfrac12$, but that is only one of the possible sums.

If the partition can also vary

A different interpretation allows all partitions and all choices of tags on $[a,b]$. Let $m=\inf_{[a,b]}f$ and $M=\sup_{[a,b]}f$. Every Riemann sum lies between $m(b-a)$ and $M(b-a)$, because its widths add to $b-a$. A one-subinterval partition, with tags chosen where $f$ is arbitrarily close to its global infimum or supremum, shows that these bounds are sharp:

$$\inf_{P,\xi}S(P,\xi)=(b-a)\inf_{x\in[a,b]}f(x), \qquad \sup_{P,\xi}S(P,\xi)=(b-a)\sup_{x\in[a,b]}f(x).$$

For $f(x)=x$ on $[0,1]$, these overall bounds are $0$ and $1$, not $\tfrac14$ and $\tfrac34$. The distinction is that the latter values came from one particular partition.

How the integral fits in

For a bounded Riemann-integrable function, lower sums can be made arbitrarily close to the integral from below, and upper sums arbitrarily close from above. Equivalently, tagged sums converge to the integral when the partition mesh—the width of its largest subinterval—tends to zero. In the example, refining the partition makes the lower and upper sums approach $\int_0^1 x\,dx=\tfrac12$.

This does not mean the infimum or supremum over every possible tagged sum equals the integral. Coarse partitions are still included in that larger set. The integral describes the common limit of sums on increasingly fine partitions, not generally the bounds over all partitions.

Common mistakes

• Confusing tags with partitions: For a fixed partition, vary the sample points to get the lower and upper Darboux sums. If partitions vary too, the answer may change.
• Calling an infimum a minimum: A function may approach its infimum without attaining it. Use “minimum” or “maximum” only when the extremum is reached.
• Assuming every Riemann sum equals the integral: A finite sum usually does not. Convergence to the integral requires increasingly fine partitions and Riemann integrability.
• Forgetting boundedness: The finite upper and lower Darboux sums described here require $f$ to be bounded on the interval.