Riemann's Criterion

When Do All Fine Tagged Sums Agree?

A tagged partition records both the subintervals and the points at which a function is sampled. The previous tutorial showed that for an integrable function, tagged sums on sufficiently fine partitions are close to the integral, regardless of the choice of tags. Riemann’s criterion reverses this perspective: if every tagged sum on every sufficiently fine partition is close to one fixed number, then the function must be integrable.

The quantifiers matter. We require a single mesh threshold to work for every partition and every choice of tags. It is not enough that one carefully chosen sequence of tagged sums converges. A sequence may select tags that avoid the places where a function behaves badly.

Definition (Riemann’s Criterion): Let $f:[a,b]\to\mathbb{R}$ be bounded. We say its tagged sums satisfy the fixed-value criterion with value $A\in\mathbb{R}$ if, for every $\varepsilon>0$, there exists $\delta>0$ such that whenever $(P,\xi)$ is a tagged partition of $[a,b]$ with $\|P\|<\delta$, $$ |S(f;P,\xi)-A|<\varepsilon. $$

The number $A$ must be the same for every tolerance, partition, and tag choice. The definition does not assume in advance that $A$ is an integral. The criterion says that fine tagged sums identify that value on their own.

Riemann’s Criterion: The Fixed-Value Form

Theorem (Riemann’s Criterion): A bounded function $f:[a,b]\to\mathbb{R}$, where $a<b$, is Riemann integrable if and only if there exists $A\in\mathbb{R}$ such that its tagged sums satisfy the fixed-value criterion with value $A$. When this holds, $A=\int_a^b f(x)\,dx$.

Proof. First suppose $f$ is Riemann integrable, with integral $I$. By the Convergence of Riemann Sums from the previous tutorial, for every $\varepsilon>0$ there exists $\delta>0$ such that every tagged partition with mesh less than $\delta$ satisfies

|S(f;P,\xi)-I|<\varepsilon.

Thus the fixed-value criterion holds with $A=I$.

Conversely, suppose the criterion holds with a value $A$. Fix $\varepsilon>0$, and apply the criterion with tolerance $\varepsilon/4$. It gives $\delta>0$ such that every tagged sum on a partition with mesh less than $\delta$ lies within $\varepsilon/4$ of $A$.

Choose a partition $P=\{x_0,\ldots,x_n\}$ with $\|P\|<\delta$. Such a partition exists by the result on uniform partitions having arbitrarily small mesh. For each interval $[x_{i-1},x_i]$, let $M_i$ and $m_i$ be the supremum and infimum of $f$ there. For any $\eta>0$, boundedness and the definitions of supremum and infimum let us choose tags $\xi_i,\zeta_i\in[x_{i-1},x_i]$ such that

f(\xi_i)>M_i-\eta \qquad\text{and}\qquad f(\zeta_i)<m_i+\eta.

The tagged sum using the $\xi_i$ is at least $U(f,P)-\eta(b-a)$, and the sum using the $\zeta_i$ is at most $L(f,P)+\eta(b-a)$. Both sums are within $\varepsilon/4$ of $A$, so their difference in absolute value is less than $\varepsilon/2$. Consequently,

U(f,P)-L(f,P) < \frac{\varepsilon}{2}+2\eta(b-a).

This holds for every $\eta>0$. Letting $\eta$ decrease to zero gives $U(f,P)-L(f,P)\leq\varepsilon/2<\varepsilon$. The Darboux Criterion now implies that $f$ is Riemann integrable.

It remains to identify its integral. Write $I=\int_a^b f(x)\,dx$. By the Convergence of Riemann Sums, sufficiently fine tagged sums are arbitrarily close to $I$; by the assumed criterion, they are also arbitrarily close to $A$. For any $\rho>0$, choose a partition fine enough that both distances are less than $\rho/2$. The triangle inequality gives $|A-I|<\rho$. Since this holds for every $\rho>0$, $A=I$. $\square$

The converse proof uses tags chosen near the supremum and infimum on each subinterval. These tags need not attain either value; the $\eta$-approximation accounts for that possibility. In particular, the argument does not silently assume that a bounded function attains its extrema on a subinterval.

The Cauchy Form: Comparing Sums Directly

Sometimes no candidate value $A$ is known. In that situation, we can ask whether any two tagged sums on sufficiently fine partitions are close to each other. This is the Cauchy form of Riemann’s criterion.

Theorem (Cauchy Criterion for Tagged Sums): A bounded function $f:[a,b]\to\mathbb{R}$, where $a<b$, is Riemann integrable if and only if, for every $\varepsilon>0$, there exists $\delta>0$ such that any two tagged partitions $(P,\xi)$ and $(Q,\eta)$ satisfying $\|P\|<\delta$ and $\|Q\|<\delta$ also satisfy $$ |S(f;P,\xi)-S(f;Q,\eta)|<\varepsilon. $$

Proof. Suppose first that $f$ is Riemann integrable, with integral $I$. Apply the Convergence of Riemann Sums with tolerance $\varepsilon/3$. For sufficiently small mesh, both sums lie within $\varepsilon/3$ of $I$, and hence

|S(f;P,\xi)-S(f;Q,\eta)| \leq |S(f;P,\xi)-I|+|S(f;Q,\eta)-I| <\frac{2\varepsilon}{3}<\varepsilon.

Now suppose the stated Cauchy condition holds. Fix $\varepsilon>0$, and choose $\delta>0$ for the tolerance $\varepsilon/2$. Take any partition $P$ with $\|P\|<\delta$. As in the proof of Riemann’s Criterion, choose tags near the suprema and infima on each subinterval, with an arbitrary $\eta>0$ as the approximation error. The difference between these two tagged sums is less than $\varepsilon/2$. Since that difference is at least

U(f,P)-L(f,P)-2\eta(b-a),

we obtain $U(f,P)-L(f,P)<\varepsilon/2+2\eta(b-a)$ for every $\eta>0$, and therefore $U(f,P)-L(f,P)\leq\varepsilon/2<\varepsilon$. The Darboux Criterion implies that $f$ is Riemann integrable. $\square$

The fixed-value and Cauchy forms express the same phenomenon. In the first, sums approach a specified number. In the second, sums approach one another, and integrability supplies the number they approach. The Cauchy form is useful when comparing sums is easier than guessing their limit.

Worked Examples: Applying the Criterion

Worked Example: A Lipschitz Function on a Closed Interval

Let $f(x)=\sin x$ on $[0,1]$. The Mean Value Theorem gives $|\sin x-\sin y|\leq|x-y|$ for $x,y\in[0,1]$, since the absolute value of the derivative of sine is at most $1$. On a partition $P$, this implies that the oscillation of $f$ on $[x_{i-1},x_i]$ is at most $\Delta x_i$. The oscillation formula for the Darboux gap therefore gives

U(f,P)-L(f,P) \leq\sum_{i=1}^n(\Delta x_i)^2 \leq\|P\|\sum_{i=1}^n\Delta x_i =\|P\|.

For $\|P\|<\varepsilon$, the gap is less than $\varepsilon$. The Darboux Criterion shows that $f$ is integrable. Moreover, the integral lies between $L(f,P)$ and $U(f,P)$, and every tagged sum lies in that same interval. Thus

\left|S(f;P,\xi)-\int_0^1\sin x\,dx\right| \leq U(f,P)-L(f,P) \leq\|P\|.

Taking $\delta=\varepsilon$ verifies the fixed-value criterion directly, with $A=\int_0^1\sin x\,dx$. The argument controls every tag choice, not just a selected sequence of sample points.

Worked Example: A Function Nonzero at One Point

On $[0,2]$, define $g(\sqrt{2})=1$ and $g(x)=0$ for $x\ne\sqrt{2}$. In any tagged sum, only subintervals whose tag equals $\sqrt{2}$ can contribute. At most two subintervals can have that tag: if $\sqrt{2}$ is a partition point, they are the intervals immediately to its left and right; otherwise it lies in just one interval. Each such interval has length at most $\|P\|$. Therefore

0\leq S(g;P,\xi)\leq 2\|P\|.

Given $\varepsilon>0$, choose $\delta=\varepsilon/2$. If $\|P\|<\delta$, then every tagged sum satisfies $0\leq S(g;P,\xi)<\varepsilon$. Thus the fixed-value criterion holds with $A=0$, and Riemann’s Criterion proves integrability with integral zero. The bound also handles the edge case where the exceptional point is a shared endpoint and can be chosen as the tag on both adjacent intervals.

Worked Example: The Dirichlet Function Fails the Criterion

Define $d:[0,1]\to\mathbb{R}$ by $d(x)=1$ when $x$ is rational and $d(x)=0$ when $x$ is irrational. For any partition $P$, every subinterval of positive length contains both a rational and an irrational number. Choose rational tags on all subintervals to obtain

S(d;P,\xi)=\sum_{i=1}^n 1\cdot\Delta x_i=1.

Choosing irrational tags on all subintervals instead gives $S(d;P,\eta)=0$. These two sums differ by $1$, no matter how small the mesh is. In particular, the Cauchy criterion fails: for any proposed $\delta>0$, choose a partition with mesh less than $\delta$ and make the two tag choices just described. Riemann’s Criterion shows that $d$ is not Riemann integrable.

What the Criterion Does—and Does Not—Require

The criterion is uniform over all sufficiently fine partitions and all their tags. This is stronger than convergence along one chosen sequence. For example, selecting only rational tags for the Dirichlet function would always produce the sum $1$, but irrational tags on the very same partition produce $0$. A conclusion based on only one tag-selection rule would miss the failure of integrability.

The proof also explains why the Darboux Criterion is the right bridge between tagged sums and integrability. On each fixed partition, tags can approximate the supremum and infimum, so control over all tagged sums forces the upper and lower sums to be close. Conversely, once the upper and lower sums are close, every tagged sum and the integral lie between them.

A common pitfall is to confuse “there exists a fine partition with a small Darboux gap” with “every sufficiently fine partition has a small gap.” Riemann’s criterion concerns every tagged partition below the mesh threshold. The proof uses this uniform control to establish a small gap on any chosen fine partition; it does not rely on selecting a favorable one.

Check Your Understanding

Use the fixed-value and Cauchy forms of Riemann’s criterion to answer the following questions.

In the fixed-value criterion, which choices must be controlled by the same mesh threshold?
Why do the tags near the suprema and infima prove a bound on the Darboux gap even if those extrema are not attained?
How does integrability imply that two sufficiently fine tagged sums are close to each other?
Why does the Dirichlet function fail the Cauchy criterion on every partition, regardless of its mesh?
For the function nonzero only at $\sqrt{2}$, why can two subintervals contribute to one tagged sum?

What You'll Learn