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Riemann Integration · Tutorial 461 of 1000

Why Integration Requires a Definition

Learn why integration needs a definition based on approximating sums, and how that definition distinguishes reliable limits from misleading samples.

Advanced 9 min read

What You'll Learn

  • Explain why area and antiderivatives alone do not define integration.
  • Define tagged partitions, mesh, and Riemann sums.
  • State Riemann integrability through the behavior of all sufficiently fine tagged sums.
  • Prove uniqueness and basic bounds for the value assigned by the definition.
  • Verify integrability of a linear function and diagnose a misleading sampling method.
  • Use Darboux’s Theorem for Derivatives to see why reverse differentiation has limits.

Why “Add Up the Values” Is Not Yet a Definition

Differentiation gives a precise way to describe local change. Integration asks a different question: how should values of a function across an interval be accumulated into one number? Geometric area offers an initial picture, and antiderivatives provide an effective calculation method in many familiar cases. Neither, by itself, defines integration for the functions we may want to study.

The area picture becomes ambiguous when a function takes negative values, and it does not specify how to handle a graph that oscillates too much for a simple geometric calculation. Reversing differentiation also has limits: not every function is a derivative. For example, the function that equals \(1\) on the rationals and \(0\) on the irrationals takes both values in every interval. It cannot be a derivative, by Darboux’s Theorem for Derivatives from “Darboux Property of Derivatives,” because it does not take every intermediate value between \(0\) and \(1\). Thus an integration theory based only on finding antiderivatives would leave many functions without a definition.

A more flexible idea is to approximate accumulation by finite sums. Divide an interval into short pieces, choose a sample point in each piece, multiply the function value there by the piece’s width, and add. But this plan still needs a rule: how fine must the pieces be, and should the answer be allowed to depend on the sample points? The definition below makes the required independence precise. Partitions will be studied in detail in the next tutorial; here, only the features needed to state the definition are introduced.

Tagged Sums and the Riemann Definition

Definition: Let \(a<b\). A partition of \([a,b]\) is a finite sequence \(a=x_0<x_1<\cdots<x_n=b\). A tag for the subinterval \([x_{i-1},x_i]\) is a point \(\xi_i\in[x_{i-1},x_i]\). A partition together with one tag for each subinterval is a tagged partition. Its mesh is the largest subinterval length, \(\max_i(x_i-x_{i-1})\). For a function \(f:[a,b]\to\mathbb R\), the associated Riemann sum is \(\sum_{i=1}^n f(\xi_i)(x_i-x_{i-1})\).

The mesh measures the largest width in the subdivision. Requiring it to be small prevents a single wide piece from hiding behavior on a substantial part of the interval. Tags may be chosen anywhere in their respective subintervals, including at endpoints. Requiring the conclusion to hold for every choice of tags prevents the result from relying on a lucky sample point.

Definition: A bounded function \(f:[a,b]\to\mathbb R\) is Riemann integrable on \([a,b]\) if there is a real number \(I\) such that, for every \(\varepsilon>0\), there is a \(\delta>0\) for which every tagged partition of \([a,b]\) with mesh less than \(\delta\) has Riemann sum \(S\) satisfying \(|S-I|<\varepsilon\). In that case, \(I\) is the Riemann integral of \(f\) and is denoted by \(\int_a^b f(x)\,dx\).

The quantifiers are essential. A particular sequence of increasingly fine sums might converge, but the definition demands more: once the mesh is small enough, every tagged sum must be close to the same number. That requirement makes the proposed value independent of how the interval is subdivided and where the samples are taken.

Worked Example: A Constant Function

Let \(f(x)=c\) on \([a,b]\), where \(c\) is any real number. For every tagged partition, the sum is exactly

$$ \sum_{i=1}^n f(\xi_i)(x_i-x_{i-1}) = \sum_{i=1}^n c(x_i-x_{i-1}) = c\bigl((x_1-x_0)+\cdots+(x_n-x_{n-1})\bigr) = c(b-a). $$

The sum is the same for every mesh and every tag, so the defining condition holds with \(I=c(b-a)\). This calculation verifies the expected accumulated value without appealing to a geometric picture or an antiderivative.

Worked Example: Arbitrary Samples for \(f(x)=x\)

Consider \(f(x)=x\) on \([0,1]\). Write \(\Delta_i=x_i-x_{i-1}\). Since \(x_{i-1}\leq \xi_i\leq x_i\), multiplying by the positive number \(\Delta_i\) and summing gives

$$ \sum_{i=1}^n x_{i-1}\Delta_i \leq \sum_{i=1}^n \xi_i\Delta_i \leq \sum_{i=1}^n x_i\Delta_i. $$

The endpoint sums can be evaluated exactly. Since \(x_i=x_{i-1}+\Delta_i\),

$$ \sum_{i=1}^n x_i\Delta_i - \sum_{i=1}^n x_{i-1}\Delta_i = \sum_{i=1}^n \Delta_i^2. $$

Also, \((x_i^2-x_{i-1}^2)/2=x_{i-1}\Delta_i+\Delta_i^2/2\). Summing this identity and using \(x_0=0\) and \(x_n=1\) yields

$$ \sum_{i=1}^n x_{i-1}\Delta_i = \frac12-\frac12\sum_{i=1}^n\Delta_i^2, \qquad \sum_{i=1}^n x_i\Delta_i = \frac12+\frac12\sum_{i=1}^n\Delta_i^2. $$

If the mesh is less than \(\delta\), then \(\Delta_i^2<\delta\Delta_i\), so \(\sum_i\Delta_i^2<\delta\sum_i\Delta_i=\delta\). The tagged sum is therefore between \(1/2-\delta/2\) and \(1/2+\delta/2\). Given \(\varepsilon>0\), choosing \(\delta=2\varepsilon\) makes every such sum differ from \(1/2\) by less than \(\varepsilon\) when the mesh is less than \(\delta\). Hence \(x\) is Riemann integrable and its integral is \(1/2\).

The Definition Forces a Unique Value

The definition would not be useful if one function could satisfy it with two different numbers. The next theorem confirms that the common limiting value, when it exists, is determined uniquely.

Theorem (Uniqueness of the Riemann Integral): If a bounded function \(f:[a,b]\to\mathbb R\) satisfies the Riemann integrability definition with both \(I\) and \(J\), then \(I=J\).

Proof. Suppose the definition holds with \(I\) and with \(J\). Let \(\varepsilon>0\). By the definition applied to \(I\), there is a positive mesh bound such that every tagged sum with smaller mesh differs from \(I\) by less than \(\varepsilon\). Applied to \(J\), the definition gives another positive mesh bound. Choose a positive integer \(n\) large enough that the uniform partition into \(n\) equal subintervals has mesh \((b-a)/n\) smaller than both bounds. Choose any tags for this partition and call its sum \(S\). Then

$$ |I-J| \leq |I-S|+|S-J| <\varepsilon+\varepsilon =2\varepsilon. $$

This holds for every \(\varepsilon>0\). If \(|I-J|\) were positive, taking \(\varepsilon=|I-J|/3\) would contradict \(|I-J|<2\varepsilon\). Therefore \(|I-J|=0\), so \(I=J\). \(\square\)

The proof uses a single sufficiently fine tagged partition to compare the two proposed values. Its choice is possible because the definition controls all sufficiently fine tagged sums, not just sums from one favored construction.

Theorem (Bounds for the Integral): Suppose \(f\) is Riemann integrable on \([a,b]\), and \(m\leq f(x)\leq M\) for every \(x\in[a,b]\). Then $$ m(b-a)\leq\int_a^b f(x)\,dx\leq M(b-a). $$

Proof. For every tagged partition, each term satisfies \(m(x_i-x_{i-1})\leq f(\xi_i)(x_i-x_{i-1})\leq M(x_i-x_{i-1})\), because each subinterval has positive length. Adding the inequalities gives

$$ m(b-a)\leq S\leq M(b-a), $$

since the subinterval lengths sum to \(b-a\). Let \(I\) be the integral. Given \(\varepsilon>0\), choose a tagged partition fine enough that \(|S-I|<\varepsilon\). From \(I<S+\varepsilon\leq M(b-a)+\varepsilon\) and \(I>S-\varepsilon\geq m(b-a)-\varepsilon\), we obtain

$$ m(b-a)-\varepsilon<I<M(b-a)+\varepsilon. $$

Because this is true for every \(\varepsilon>0\), \(I\) cannot be below \(m(b-a)\) or above \(M(b-a)\). The stated bounds follow. \(\square\)

Why the Choice of Samples Matters

A tempting shortcut is to select a convenient subdivision and a convenient tag in each piece, then treat the resulting limit as the integral. The next example shows why that is not a definition: different sample choices can give incompatible answers, even when the mesh is arbitrarily small.

Worked Example: Rational and Irrational Tags Give Different Sums

Define \(f:[0,1]\to\mathbb R\) by \(f(x)=1\) when \(x\) is rational and \(f(x)=0\) when \(x\) is irrational. Every interval of positive length contains both rational and irrational numbers. For any partition of \([0,1]\), choose a rational tag in every subinterval. The resulting sum is

$$ \sum_{i=1}^n 1\cdot(x_i-x_{i-1})=1. $$

Instead, choose an irrational tag in every subinterval. The resulting sum is

$$ \sum_{i=1}^n 0\cdot(x_i-x_{i-1})=0. $$

Both choices are possible for partitions of any mesh. If the function had an integral \(I\), choose \(\varepsilon=1/3\) and take a partition whose mesh is smaller than the corresponding \(\delta\) in the definition. Its rational-tag sum would force \(|1-I|<1/3\), while its irrational-tag sum would force \(|I|<1/3\). The triangle inequality would then give \(1\leq|1-I|+|I|<2/3\), a contradiction. Thus this bounded function is not Riemann integrable.

The example explains why a limit along one sequence of sums is not enough. For instance, equal subdivisions with carefully selected rational tags always give the value \(1\), while irrational tags always give \(0\). Neither sequence represents a value independent of the sampling procedure.

Integration Is Not Just Reverse Differentiation

The rational–irrational example also highlights a distinction between two questions. Differentiation asks whether a function can arise as the local rate of change of another function. Integration, as defined here, asks whether all sufficiently fine tagged sums settle near one number. Darboux’s Theorem for Derivatives rules out the rational–irrational function as a derivative, but that fact alone does not define an integral or decide integrability. The tagged-sum definition supplies an independent test.

For functions that are Riemann integrable, antiderivatives can later provide a powerful way to compute the integral when the required hypotheses hold. That calculation method depends on a previously defined integral and a theorem connecting it to differentiation; it is not a replacement for the definition. The definition also gives immediate safeguards: uniqueness prevents competing values, and the bounds theorem ensures that the integral respects pointwise bounds.

The central idea is therefore not simply to add sampled values, but to prove that every sufficiently fine sampling scheme gives nearly the same result. The next step is to examine partitions more closely: how they are organized, how their meshes behave, and how refinements help control sums.

Check Your Understanding

Use the tagged-sum definition and the examples in this tutorial to answer these questions.

  1. Why does the definition require control of every tagged partition with sufficiently small mesh, rather than just one sequence of sums?
  2. For a constant function \(f(x)=c\) on \([a,b]\), why does every tagged sum equal \(c(b-a)\)?
  3. In the proof of uniqueness, why can one choose a single tagged partition that works for both proposed values?
  4. What bounds does the integral satisfy if \(m\leq f(x)\leq M\) throughout \([a,b]\)?
  5. Why do rational tags and irrational tags show that the indicator of the rationals on \([0,1]\) is not Riemann integrable?

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