Upper and lower sums are built from suprema and infima, which are convenient for proofs and impossible to compute. A Riemann sum instead evaluates the function at one chosen point of each piece — any point at all — and weights it by the width. For an integrable function these sums converge to the integral as the partition gets fine, whatever points are chosen, which is what makes numerical integration possible and what most calculus courses take as the definition.

Rather than finding the tallest and the shortest blade of grass in each square metre, measure one blade per square metre, chosen however you like. If the field is well behaved the estimate is good, and it improves as the squares shrink, however unlucky the choices.

One value is taken from each piece and multiplied by the width of that piece. The level drawn across is the average the total represents; as the pieces narrow, the choice inside each matters less and less.

A tagged partition is a partition together with a chosen point $t_i\in[x_{i-1},x_i]$ for each piece, and its Riemann sum is $S(f,P,t)=\sum f(t_i)\Delta x_i$. The mesh of $P$ is the largest of the widths. For an integrable $f$, the Riemann sums converge to $\int_a^b f$ as the mesh tends to zero, whatever tags are chosen. Every Riemann sum lies between the lower and upper sums of its partition.

• $L(f,P)\le S(f,P,t)\le U(f,P)$ for every choice of tags, since each chosen value lies between the infimum and the supremum on its piece.
• So a partition satisfying the criterion pins every Riemann sum to within the tolerance of the integral, whatever the tags.

Compare the sums piece by piece. On each piece the chosen value lies between the infimum and the supremum, simply because those are the greatest lower bound and least upper bound of the values there. Multiplying by a positive width preserves both inequalities, and adding over the pieces preserves them again. The conclusion is that the tags cannot escape the bracket the partition already provides.

The chosen point lies in the piece, so its value lies between the two extremes there.

Multiply by the width, which is positive.

Add over the pieces; inequalities survive addition.

Both the sum and the integral lie in the same bracket, so they are within its width of each other.

A tag is any point of its piece. The theory does not care which.

For an increasing function the extremes on each piece sit at its two ends.

• A tagged partition chooses one point per piece, freely.
• Every Riemann sum lies between the lower and upper sums of its partition.
• For an integrable function the limit is the integral, whatever the tags.
• Endpoint rules reproduce the upper and lower sums for a monotone function.

Great work! You now know how to:

• Compute a Riemann sum with given tags
• Say why every Riemann sum is trapped by its partition
• Name the mesh and say what it controls
• Say why the theory is built on upper and lower sums