How to Prove That a Function Is Riemann Integrable

To prove that a function is integrable, first specify the kind of integral and the interval. In introductory calculus, “integrable” usually means Riemann integrable on a closed, bounded interval $[a,b]$. The most general elementary test is to compare upper and lower sums. In routine problems, a known theorem—such as continuity or monotonicity—often gives a shorter proof.

The upper–lower sum criterion

A Riemann-integrable function on $[a,b]$ must be bounded there. Take a partition

$$P: a=x_0<x_1<\cdots<x_n=b.$$

On each subinterval $[x_{i-1},x_i]$, let $M_i$ be the supremum of $f$ and $m_i$ its infimum. The upper sum $U(f,P)$ and lower sum $L(f,P)$ are

$$U(f,P)=\sum_{i=1}^n M_i(x_i-x_{i-1}), \qquad L(f,P)=\sum_{i=1}^n m_i(x_i-x_{i-1}).$$

The Darboux criterion says that a bounded function is Riemann integrable if and only if, for every $\varepsilon>0$, there is a partition $P$ such that

$$U(f,P)-L(f,P)<\varepsilon.$$

This is a useful proof target: the upper and lower sums can be made arbitrarily close. Along partitions whose upper–lower gaps tend to zero, both sums converge to the integral.

A direct proof for $f(x)=x^2$ on $[0,1]$

Divide $[0,1]$ into $n$ equal subintervals, each of width $1/n$. Since $x^2$ is increasing on this interval, its minimum and maximum on the $i$th subinterval are attained at the left and right endpoints. Thus

$$\begin{aligned} U(f,P)-L(f,P) &=\sum_{i=1}^n \left[\left(\frac{i}{n}\right)^2-\left(\frac{i-1}{n}\right)^2\right]\frac{1}{n}\\ &=\frac{1}{n^3}\sum_{i=1}^n(2i-1)\\ &=\frac{1}{n}. \end{aligned}$$

Given any $\varepsilon>0$, choose an integer $n>1/\varepsilon$. Then $U(f,P)-L(f,P)=1/n<\varepsilon$. The criterion proves that $x^2$ is Riemann integrable on $[0,1]$. This argument proves integrability; evaluating the integral is a separate step.

Shortcuts from standard theorems

Use a theorem when its hypotheses are easy to verify:

• Continuous functions: Every function continuous on $[a,b]$ is Riemann integrable there. Continuity on the whole interval is sufficient, not necessary.
• Monotone functions: Every bounded monotone function on $[a,b]$ is Riemann integrable. A monotone function may have jump discontinuities and still pass this test.
• Discontinuities on a negligible set: The Lebesgue criterion for Riemann integrability states that a bounded function on $[a,b]$ is Riemann integrable exactly when its set of discontinuities has measure zero. This is a powerful theorem, but it usually requires more background than the Darboux criterion.

For example, define $g(0)=1$ and $g(x)=0$ for $0<x\leq1$. This function is discontinuous at $0$, so the continuity theorem does not apply. Yet it is integrable: use a partition whose first interval is $[0,1/n]$ and whose other intervals lie to its right. The upper–lower sum gap is at most $1/n$, which can be made smaller than any positive tolerance. The value at a single point does not affect the integral.

A practical proof template

When a problem asks you to prove integrability directly, follow these steps:

1. State the interval and verify that $f$ is bounded on it.
2. Choose a partition, often an equal-width partition or one adapted to the function’s discontinuities.
3. Bound $M_i-m_i$ on each subinterval.
4. Add the bounds to estimate $U(f,P)-L(f,P)$.
5. Given $\varepsilon>0$, choose the partition fine enough to make that estimate less than $\varepsilon$.
6. Invoke the Darboux criterion.

The key is not merely to say that the intervals are small. You must show quantitatively that the total upper–lower gap becomes arbitrarily small.

Common mistakes

• Checking continuity when it is not given: Continuity is a convenient sufficient condition, not the definition. Discontinuous functions can be integrable.
• Forgetting boundedness: A function unbounded on $[a,b]$ is not Riemann integrable there in the ordinary proper sense. An improper integral may still converge, but that is a different definition and requires a separate convergence argument.
• Showing only that the mesh shrinks: The mesh is the width of the largest partition interval. A fine partition alone is not a proof unless you connect it to a bound on the upper–lower sum gap.
• Confusing integrability with finding an antiderivative: Riemann integrability concerns the existence of the definite integral. It does not require an elementary antiderivative.

If the function is continuous or monotone, cite the relevant theorem and verify its conditions. Otherwise, the upper–lower sum criterion gives a systematic route: construct partitions that squeeze the possible integral values together.