Why the Size of the Subintervals Matters
A Riemann sum approximates accumulated quantity by adding the contributions from small pieces of an interval. Each contribution uses a function value chosen at a tag and the width of its subinterval. If one subinterval remains wide, its tag can represent a large region where the function takes different values. The resulting sum can depend strongly on that single choice.
The issue is not that every coarse sum is wrong. Some coarse sums are exact, and some happen to match the integral. The issue is that coarse partitions do not guarantee reliable results for arbitrary tag choices. The Riemann definition controls this by requiring the largest subinterval to become arbitrarily small.
Partitions, Mesh, and Tagged Sums
A partition \(P\) of a closed interval \([a,b]\) is a finite increasing list of points
Its subintervals are \([x_{i-1},x_i]\), with widths \(\Delta x_i=x_i-x_{i-1}\). The mesh of the partition is the width of its largest subinterval:
A tag \(\xi_i\) for the \(i\)th subinterval is a chosen point in \([x_{i-1},x_i]\). For a function \(f\), the tagged Riemann sum is
Each term uses one function value to represent the function across an entire subinterval. If the interval is wide and the function changes substantially there, this representation can be poor.
The quantifiers are important. Once the mesh is sufficiently small, all permitted tag choices and all partitions must give sums close to the same number. A single sequence of lucky tags is not enough to establish this condition.
A Coarse Interval Can Hide a Jump
Consider the step function on \([0,1]\) defined by
Its graph is zero on the first half of the interval and one on the second half. Its signed area, and hence its integral, is \(1/2\). A partition that does not separate the two regions can combine both function values in a single subinterval.
Worked Example: Two Tag Choices Give Different Coarse Sums
Use the partition \(P=\{0,\frac25,1\}\). Its subinterval widths are \(\frac25\) and \(\frac35\), so
First choose tags \(\xi_1=\frac15\) and \(\xi_2=\frac9{20}\). Both tags are at most \(\frac12\), so \(f(\xi_1)=0\) and \(f(\xi_2)=0\). Thus
Now keep the same partition but choose \(\xi_1=\frac15\) and \(\xi_2=\frac34\). The function values are \(0\) and \(1\), respectively, so
The integral is \(\frac12\). The first sum differs from it by \(\left|0-\frac12\right|=\frac12\); the second differs by \(\left|\frac35-\frac12\right|=\frac1{10}\). Both calculations use valid tags, but the coarse partition gives noticeably different answers.
Check: The second subinterval has width \(1-\frac25=\frac35\), and its tag \(\frac34\) lies in \([\frac25,1]\), where \(f(\frac34)=1\). Therefore its contribution really is \(\frac35\).
The large subinterval \([\frac25,1]\) crosses the jump at \(\frac12\). A tag on its left part assigns value zero to the entire subinterval; a tag on its right part assigns value one to the entire subinterval. The mesh condition prevents this kind of uncertainty from remaining attached to a region of fixed positive width.
A Sequence of Partitions Can Still Fail to Refine
Making a partition more complicated does not necessarily make its mesh smaller. For example, one can keep a wide subinterval fixed while dividing other parts into increasingly many pieces. The number of subintervals then grows, but the largest width does not shrink.
Worked Example: More Subintervals Without a Smaller Mesh
For each positive integer \(n\), divide \([0,\frac25]\) into \(n\) equal subintervals, and append the subinterval \([\frac25,1]\). The first \(n\) widths are \(\frac{2}{5n}\); the last width is \(\frac35\). Since \(\frac{2}{5n}\le\frac25<\frac35\), the mesh is
for every \(n\). Choose the midpoint of each small subinterval in \([0,\frac25]\), and choose \(\frac9{20}\) as the tag in \([\frac25,1]\). All these tags are at most \(\frac12\), so every sampled function value is zero. There are \(n\) small contributions, each equal to \(0\cdot\frac{2}{5n}=0\), and the last contribution is \(0\cdot\frac35=0\). Consequently,
These sums remain at distance \(\frac12\) from the integral for every \(n\). The number of pieces increases, but the mesh stays equal to \(\frac35\), so this sequence does not meet the condition that the mesh tends to zero.
Check: The sum of the first \(n\) widths is \(n\cdot\frac{2}{5n}=\frac25\); adding the final width gives \(\frac25+\frac35=1\), the length of the full interval.
Why Shrinking the Mesh Works for This Step Function
For the step function above, the effect of a tag choice can be bounded directly. Consider any partition and any tags. Every subinterval that lies entirely on one side of \(\frac12\) has a constant function value, so its tagged contribution is exactly the area of that constant piece.
If \(\frac12\) lies inside a partition subinterval, only that subinterval can have a discrepancy between its tagged contribution and its actual area. Write its width as \(w\). The true area within it is between \(0\) and \(w\). Its tagged contribution is either \(0\) or \(w\), since the tag gives function value either \(0\) or \(1\). Thus the absolute discrepancy on this subinterval is at most \(w\), and \(w\le\|P\|\).
If \(\frac12\) is a partition point, all intervals except possibly the interval immediately to its right give their exact constant contribution. On that right-hand interval, a tag at \(\frac12\) could give zero instead of the full width; the discrepancy is at most that interval's width, again no more than \(\|P\|\). In either case, the total error obeys
This argument shows precisely what shrinking the mesh achieves here: any uncertainty caused by the jump is confined to an interval whose width becomes arbitrarily small. For more general functions, the details of integrability require additional reasoning; a small mesh alone does not make every bounded function integrable.
Coarse Sums Can Be Right by Accident
The mesh requirement is a guarantee about all sufficiently fine partitions and all tags. It does not say that every sum from a coarse partition is wrong. A coarse sum may agree with the integral by coincidence, or because the function and partition fit especially well.
Worked Example: Different Sums for a Linear Function
Take \(f(x)=x\) on \([0,1]\), whose integral is the area of a triangle with base \(1\) and height \(1\), namely \(\frac12\). Use the partition \(P=\{0,\frac12,1\}\), with mesh \(\frac12\).
Choosing left-endpoint tags \(0\) and \(\frac12\) gives
Choosing right-endpoint tags \(\frac12\) and \(1\) gives
Choosing midpoint tags \(\frac14\) and \(\frac34\) gives
The midpoint sum matches the integral, while the left- and right-endpoint sums do not. Each tag lies in its assigned subinterval, and the widths sum to \(\frac12+\frac12=1\). This is a useful illustration of the distinction: one correct coarse sum does not establish the required closeness for every tag choice.
The Definition Gives a Robust Limit
For a sequence of partitions \(P_1,P_2,\ldots\), saying that its meshes tend to zero means that for every \(\delta>0\), there is an index \(N\) such that \(\|P_n\|<\delta\) whenever \(n\ge N\). The Riemann definition is stronger than asking one chosen sequence of sums to converge: it requires all tagged sums on sufficiently fine partitions to be close to the same integral.
This robustness also makes the integral unique. Suppose \(I\) and \(J\) both satisfy the definition for the same function. Given any \(\varepsilon>0\), choose partitions fine enough to meet both definitions with error less than \(\varepsilon\). Such partitions exist, for example by dividing \([a,b]\) into sufficiently many equal pieces. Use the same tags for both estimates. The triangle inequality then gives
Since this holds for every positive \(\varepsilon\), the two numbers must be equal. Thus the definition does not permit different integral values arising from different fine partitions or tag choices.
A Common Mistake
A tempting but invalid argument is: “On the single interval \([0,1]\), the left-endpoint tag for the step function gives \(f(0)\cdot1=0\), so the integral is zero.” The calculation of that sum is correct. The conclusion is not: its mesh is \(1\), and one coarse tagged sum does not satisfy the definition's requirement about every sufficiently fine partition and every tag choice.
The correction is to check what happens as the mesh shrinks and tag choices vary. For the step function, the error bound above proves that every tagged sum differs from \(\frac12\) by at most the mesh. That is the kind of control that justifies the integral.
What the Mesh Condition Does—and Does Not—Say
- It rules out permanently wide subintervals. A wide interval can conceal substantial variation or a jump.
- It controls tag sensitivity when the function is integrable. Fine partitions make all permitted tagged sums approach the same value.
- It is not a claim that every bounded function is integrable. For some bounded functions, tagged sums need not settle to one value even as the mesh tends to zero.
- It is not a claim that every coarse sum is inaccurate. A coarse sum may happen to be exact, but that alone gives no general guarantee.
The central point is therefore about reliability, not merely arithmetic accuracy. A Riemann integral is obtained when the sums become stable under both refinement and changes of tags. Requiring the mesh to tend to zero removes the possibility that a fixed-width region can keep exerting uncontrolled influence.
Check Your Understanding
Use the definitions and examples above to answer these questions.
- For the partition \(P=\{0,\frac25,1\}\), what is its mesh, and which subinterval crosses the jump of the step function?
- Why does adding more subintervals to \([0,\frac25]\) in the second example fail to make the mesh tend to zero?
- For the step function, why is the error from a subinterval containing the jump at most the width of that subinterval?
- Does one coarse tagged sum that matches an integral prove that the function is Riemann integrable? Explain.
- In the definition of the Riemann integral, what must be true of the sums once the mesh is sufficiently small?
