From a Limit to a Precise Guarantee
A tagged Riemann sum approximates accumulated quantity by adding rectangle contributions. Saying that these sums “approach” a definite integral is not yet precise enough: we need to specify which sums must be close, how close they must be, and what makes them sufficiently fine.
The key measurement of fineness is the mesh of a partition: the width of its largest subinterval. The epsilon-delta formulation says that, for any requested error tolerance, there is a mesh threshold that guarantees the error is smaller than that tolerance. Crucially, the guarantee must hold for every choice of tags in every partition below that threshold.
Tagged Partitions, Sums, and Mesh
Fix a closed interval \([a,b]\), where \(a<b\). A partition \(P\) of this interval is a finite ordered list of points \(a=x_0<x_1<\cdots<x_n=b\). Write \(\Delta x_i=x_i-x_{i-1}\) for the width of its \(i\)th subinterval. A tag \(\xi_i\) for that subinterval is any point in \([x_{i-1},x_i]\).
The mesh records the largest width, not the average width and not the sum of the widths. The tags may be endpoints or interior points, and different subintervals may use different kinds of tags.
The Epsilon-Delta Definition
Suppose \(f\) is bounded on \([a,b]\). The definite integral is the common limit of its tagged Riemann sums as the mesh tends to zero. The precise condition for convergence to a number \(I\) is:
The order of the quantifiers matters. First, someone specifies any positive error tolerance \(\varepsilon\). Then we must find a positive mesh threshold \(\delta\). After that, every tagged partition with mesh below \(\delta\) must meet the error tolerance. We cannot choose the tags after seeing \(\varepsilon\), or check only one convenient family of partitions.
The threshold \(\delta\) may depend on \(\varepsilon\), on \(f\), and on the interval. It does not depend on a particular partition or its tags. The condition uses \(\|P\|<\delta\), a strict inequality; a proof should establish the stated strict error bound under that hypothesis.
Why the Condition Uses Absolute Error
A tagged sum can lie either above or below the integral. The expression \(S-I\) records the signed error, but convergence requires control in both directions. The absolute value gives exactly that: \(|S-I|<\varepsilon\) is equivalent to \(-\varepsilon<S-I<\varepsilon\).
Nor is it enough to show that one sequence of sums converges to \(I\). For example, uniform partitions with left-endpoint tags are only one possible choice. The definition requires a guarantee for all sufficiently fine partitions and all allowed tags. This universal requirement is what makes the definition a test of convergence rather than a description of one approximation procedure.
Worked Example: A Constant Function
Worked Example: A Constant Function
Let \(f(x)=3\) on \([1,4]\). Show directly that its tagged sums converge to \(9\), and state a valid \(\delta\) for any \(\varepsilon>0\).
For any tagged partition, every tag satisfies \(f(\xi_i)=3\). Therefore $$ S(f,P,\xi) =\sum_{i=1}^{n}3\Delta x_i =3\sum_{i=1}^{n}(x_i-x_{i-1}). $$ The sum of the widths telescopes: $$ \sum_{i=1}^{n}(x_i-x_{i-1})=x_n-x_0=4-1=3. $$ Hence \(S(f,P,\xi)=3\cdot3=9\), regardless of the partition or tags. In particular, $$ |S(f,P,\xi)-9|=|9-9|=0<\varepsilon. $$
For every \(\varepsilon>0\), choose \(\delta=1\). Every tagged partition with \(\|P\|<1\) satisfies the required inequality; in fact, the inequality holds for partitions of any mesh. Checking the exact sum confirms that \(9\) is the integral.
Worked Example: A Linear Function on the Unit Interval
Worked Example: A Linear Function on the Unit Interval
Show that all tagged sums for \(f(x)=x\) on \([0,1]\) converge to \(1/2\), and find a mesh threshold that works for any given \(\varepsilon>0\).
Take an arbitrary partition \(0=x_0<x_1<\cdots<x_n=1\), and put \(\Delta x_i=x_i-x_{i-1}\). Define the left- and right-endpoint sums $$ L=\sum_{i=1}^{n}x_{i-1}\Delta x_i, \qquad U=\sum_{i=1}^{n}x_i\Delta x_i. $$ Because \(x_{i-1}\le \xi_i\le x_i\), multiplication by the positive width \(\Delta x_i\) and summation give \(L\le S(f,P,\xi)\le U\). Moreover, $$ U-L=\sum_{i=1}^{n}(x_i-x_{i-1})\Delta x_i =\sum_{i=1}^{n}(\Delta x_i)^2. $$
Now use the identity \(x_i^2-x_{i-1}^2=2x_{i-1}\Delta x_i+(\Delta x_i)^2\). Summing over all subintervals gives $$ 1=1^2-0^2=2L+\sum_{i=1}^{n}(\Delta x_i)^2. $$ Consequently \(L=\frac12-\frac12\sum_i(\Delta x_i)^2\), and since \(U-L=\sum_i(\Delta x_i)^2\), $$ U=\frac12+\frac12\sum_{i=1}^{n}(\Delta x_i)^2. $$ The tagged sum lies between these two bounds, so $$ \left|S(f,P,\xi)-\frac12\right| \le \frac12\sum_{i=1}^{n}(\Delta x_i)^2. $$
For each \(i\), \(\Delta x_i\le\|P\|\), so \((\Delta x_i)^2\le\|P\|\Delta x_i\). Also \(\sum_i\Delta x_i=1\). Thus $$ \left|S(f,P,\xi)-\frac12\right| \le \frac12\|P\|\sum_{i=1}^{n}\Delta x_i =\frac12\|P\|. $$ Given any \(\varepsilon>0\), choose \(\delta=2\varepsilon\). If \(\|P\|<\delta\), then $$ \left|S(f,P,\xi)-\frac12\right| \le\frac12\|P\|<\frac12(2\varepsilon)=\varepsilon. $$ This proves the required statement for every partition and every tag choice. It also checks the proposed integral value: the bound tends to zero with the mesh, so the sums converge to \(1/2\).
Why the Error Estimate Gives the Epsilon-Delta Choice
The preceding proof illustrates a general step in epsilon-delta reasoning. First prove an estimate that relates the error to the mesh. Then solve the estimate for a sufficient mesh threshold. For the linear example, the estimate was \(|S-\frac12|\le\frac12\|P\|\). Requiring \(\frac12\|P\|<\varepsilon\) suggests choosing \(\|P\|<2\varepsilon\), which is why \(\delta=2\varepsilon\) works.
In general, if an error estimate has the form \(|S-I|\le C\|P\|\) for a fixed constant \(C>0\), then \(\delta=\varepsilon/C\) is sufficient: \(\|P\|<\delta\) implies \(|S-I|\le C\|P\|<C\delta=\varepsilon\). The estimate must apply to all allowed partitions and tags before this choice proves the definition.
Worked Example: Converting an Error Tolerance to a Uniform Partition
Worked Example: Converting an Error Tolerance to a Uniform Partition
For \(f(x)=x\) on \([0,1]\), use the bound from the previous example to ensure that every tagged sum from a uniform partition into \(60\) subintervals is within \(0.01\) of the integral.
Each subinterval has width \(1/60\), so the mesh is $$ \|P\|=\frac{1}{60}<\frac{1}{50}=0.02. $$ The error estimate gives, for every choice of tags, $$ \left|S(f,P,\xi)-\frac12\right| \le\frac12\|P\| =\frac{1}{120} <\frac{1}{100} =0.01. $$ Thus every such tagged sum meets the requested tolerance.
For an explicit check of the bound, the endpoint sums are $$ L=\frac{1}{60^2}\sum_{i=0}^{59}i =\frac{1}{3600}\cdot\frac{59\cdot60}{2} =\frac{59}{120}, \qquad U=\frac{1}{60^2}\sum_{i=1}^{60}i =\frac{1}{3600}\cdot\frac{60\cdot61}{2} =\frac{61}{120}. $$ Every tagged sum lies between \(59/120\) and \(61/120\). Each endpoint differs from \(1/2=60/120\) by \(1/120\), confirming the stated error bound.
A Common Mistake: Choosing Delta Without the Error Constant
A tempting shortcut is to set \(\delta=\varepsilon\) whenever the mesh appears in an error estimate. That is not generally valid: the estimate may multiply the mesh by a constant larger than \(1\).
Worked Example: Correcting an Undersized Error Estimate
Let \(f(x)=10x\) on \([0,1]\), whose integral is \(5\). Scaling the linear estimate by \(10\) gives $$ |S(f,P,\xi)-5|\le 5\|P\|. $$ For \(\varepsilon=0.02\), the mistaken choice \(\delta=0.02\) does not make this bound smaller than \(0.02\): it only gives an error bound below \(5(0.02)=0.1\).
There is a concrete failure. Take \(53\) equal subintervals, so \(\|P\|=1/53<0.02\), and use right-endpoint tags \(\xi_i=i/53\). Then $$ S(f,P,\xi) =\sum_{i=1}^{53}10\frac{i}{53}\frac{1}{53} =\frac{10}{2809}\cdot\frac{53\cdot54}{2} =\frac{270}{53}. $$ Its error is $$ \left|\frac{270}{53}-5\right| =\left|\frac{270-265}{53}\right| =\frac{5}{53} >0.02. $$ So a partition with mesh below the proposed threshold fails the tolerance.
The estimate requires \(5\|P\|<0.02\), which is guaranteed by \(\|P\|<0.02/5=0.004\). Thus the corrected choice is \(\delta=0.004\). The example shows why one must use the actual error estimate rather than match \(\delta\) to \(\varepsilon\) automatically.
Uniqueness of the Limit
The quantifier definition also ensures that there cannot be two different integral values for the same function on the same interval. Suppose the tagged sums converge both to \(I\) and to \(J\). If \(I\ne J\), set \(\varepsilon=|I-J|/3>0\). Each convergence condition supplies a positive mesh threshold; take the smaller of the two thresholds. A uniform partition with sufficiently many subintervals has mesh below that threshold, so choose any tags for it. The two convergence conditions imply $$ |I-J|\le |I-S(f,P,\xi)|+|S(f,P,\xi)-J| <\varepsilon+\varepsilon =\frac{2}{3}|I-J|. $$ This is impossible because \(|I-J|>0\). Therefore \(I=J\).
Keep the Quantifiers in the Right Order
When reading or constructing a proof, check the quantifiers explicitly: Is the threshold chosen after the tolerance? Does it work for all sufficiently fine partitions? Are all allowed tags covered? Is the final inequality strict as required? These checks help distinguish a genuine convergence proof from evidence based on a few approximations.
Check Your Understanding
Use the definition and the estimates in this tutorial to answer the following questions.
- In the tagged-sum convergence condition, what may \(\delta\) depend on, and what may it not depend on?
- Why must the condition apply to every tag choice rather than just to left-endpoint tags?
- For \(f(x)=x\) on \([0,1]\), what error bound follows from \(\|P\|<\delta\), and what choice of \(\delta\) guarantees error below \(\varepsilon\)?
- If an error estimate is \(|S-I|\le 7\|P\|\), give a sufficient positive \(\delta\) for a prescribed \(\varepsilon>0\).
- Why does the triangle inequality rule out two different limits for the same family of tagged sums?
