Why Boundedness Matters
A definite integral on a finite interval is built from sums of rectangle areas. Each rectangle has a width and a height given by a sampled value of the function. If the function’s values can grow without limit on the interval, then even a single rectangle of positive width can have an arbitrarily large signed contribution, depending on the sample point.
This observation is more than a practical warning about choosing samples. It leads to a necessary condition for the proper Riemann integral: on a nondegenerate closed interval, an integrable function must be bounded. We will state the condition precisely and prove it using tagged Riemann sums.
Boundedness on a Closed Interval
The same condition can be written as \(-M\le f(x)\le M\) for every \(x\in[a,b]\). The bound need not be the smallest possible one. To establish boundedness, it is enough to find any finite \(M\) that works uniformly across the whole interval. Checking only a few values does not establish such a bound.
For example, saying that \(f(x)\) is finite at every point is not the same as saying that \(f\) is bounded. A real-valued function assigns a finite real number to each individual input, but those values can still grow without limit as the input varies through the interval.
Tagged Sums and the Proper Integral
A partition of \([a,b]\) is a finite list \(a=x_0<x_1<\cdots<x_n=b\). Its subinterval widths are \(\Delta x_i=x_i-x_{i-1}\), and its mesh is the largest of those widths. A tagged partition also chooses a sample point \(t_i\in[x_{i-1},x_i]\) for each subinterval. Its tagged Riemann sum is
For this tutorial, a function \(f:[a,b]\to\mathbb{R}\) is properly Riemann integrable if there is a finite real number \(I\) such that tagged sums approach \(I\) as the mesh approaches zero, uniformly over the choices of partition and tags. Precisely, for every \(\varepsilon>0\), there must be a \(\delta>0\) such that every tagged partition \(P\) with mesh less than \(\delta\) satisfies
When this condition holds, \(I\) is the definite integral \(\int_a^b f(x)\,dx\). The requirement applies to every allowed choice of tags, not just to sample points that appear convenient.
The Necessary Condition
The key idea is that integrability forces all sufficiently fine tagged sums to remain close to one finite number. If \(f\) were unbounded, we could choose a sufficiently fine partition and then vary one tag in a subinterval where \(f\) is unbounded. That one term would make the sums unbounded, contradicting the integral condition.
Proof of the Theorem
Suppose \(f\) is properly Riemann integrable with integral \(I\), but is unbounded on \([a,b]\).
Apply the definition with \(\varepsilon=1\). There is a \(\delta>0\) such that every tagged sum from a partition of mesh less than \(\delta\) satisfies \(|S(f,P)-I|<1\). In particular, each such sum lies between \(I-1\) and \(I+1\).
Since \(a<b\), we can partition \([a,b]\) into finitely many subintervals, each of positive width and with mesh less than \(\delta\). The subintervals cover \([a,b]\). If \(f\) were bounded on each one, the largest of the finitely many bounds would bound \(f\) on the entire interval. Therefore, at least one subinterval, say \([x_{j-1},x_j]\), must contain values of \(f\) that are unbounded above or unbounded below.
Fix any allowed tag in every other subinterval. Their contributions have a fixed finite sum \(C\). The remaining contribution is \(\Delta x_j f(t_j)\), where \(\Delta x_j=x_j-x_{j-1}>0\). If \(f\) is unbounded above on this subinterval, choosing \(t_j\) with increasingly large values makes \(C+\Delta x_j f(t_j)\) unbounded above. If \(f\) is unbounded below, the same expression is unbounded below. In either case, tagged sums from this fine partition cannot all lie between \(I-1\) and \(I+1\).
The partition has mesh less than \(\delta\), so the integral condition requires every tagged sum for it to satisfy \(|S(f,P)-I|<1\). The unbounded sums just constructed violate that requirement. Thus \(f\) cannot be unbounded, and it must be bounded on \([a,b]\).
The argument depends on the subinterval having positive width: multiplying unbounded function values by \(\Delta x_j>0\) does not make them bounded. The theorem concerns a nondegenerate interval \(a<b\), as stated.
Worked Examples
Worked Example: Finding a Bound on a Closed Interval
Show that \(f(x)=x^2-2x\) is bounded on \([0,3]\), and find a bound that cannot be reduced.
Rewrite the function by completing the square:
For \(0\le x\le 3\), we have \(-1\le x-1\le 2\), so \(0\le (x-1)^2\le 4\). Subtracting \(1\) gives \(-1\le f(x)\le 3\). Hence \(|f(x)|\le 3\), and \(M=3\) is a bound.
To check that this bound cannot be reduced, evaluate the function at \(x=3\): \(f(3)=3^2-2(3)=9-6=3\), so \(|f(3)|=3\). The bound \(M=3\) is attained and is therefore the smallest possible bound.
Worked Example: An Unbounded Function Near an Endpoint
Define \(f:[0,1]\to\mathbb{R}\) by \(f(0)=0\) and \(f(x)=1/\sqrt{x}\) when \(x>0\). See directly why its tagged sums cannot obey the proper-integral condition.
Use the partition \(0<\frac12<1\). Choose the tag \(t\) in the first subinterval \([0,\frac12]\), and fix the tag \(\frac34\) in the second. The second contribution is
For each positive integer \(k\), choose \(t_k=\frac{1}{2k^2}\), which lies in \((0,\frac12]\). Then
Thus the tagged sum for these tags is \(S_k=\frac{k}{\sqrt{2}}+\frac{1}{\sqrt{3}}\), which grows without bound as \(k\) increases. The individual function values are finite for every chosen \(t_k>0\), but they have no common finite bound. This calculation illustrates the obstruction; the theorem’s proof ensures the same contradiction for any required mesh size by choosing a fine partition first.
Check the tag calculation: \(t_k\le\frac12\), and substituting \(t_k\) into the first rectangle’s width-times-height contribution gives exactly \(k/\sqrt{2}\), as claimed.
Worked Example: An Unbounded Function Near an Interior Point
Define \(g:[0,2]\to\mathbb{R}\) by \(g(1)=0\) and \(g(x)=1/|x-1|\) for \(x\ne1\). For the partition \(0<1<2\), fix the first tag at \(1/2\) and choose the second tag \(t_k=1+1/k\), where \(k\) is a positive integer.
The first rectangle contributes \(1\cdot g(1/2)=1\cdot 2=2\). The second has width \(1\), and its contribution is
Therefore the tagged sum is \(S_k=2+k\), which is unbounded as \(k\) increases. Each tag belongs to \([1,2]\), including \(t_1=2\), and the displayed substitution verifies the contribution. The unbounded behavior is not limited to endpoints: it can occur near an interior point as well.
A Necessary Condition Is Not a Guarantee
The theorem gives a one-way implication: proper Riemann integrability on a nondegenerate closed interval implies boundedness there. It does not say that every bounded function is integrable. Boundedness ensures that the function’s values do not grow without limit, but integrability also requires all sufficiently fine tagged sums to converge to the same finite number, regardless of the tags. Therefore, finding a bound is an important check, not a complete test for integrability.
This distinction prevents a common logical error. If a function is unbounded, the theorem lets us conclude that it is not properly Riemann integrable on that interval. If a function is bounded, we have established only the necessary condition; a separate argument is needed to establish convergence of its tagged sums.
Common Pitfall: Checking Only Endpoint Values
Consider again the function \(f(0)=0\), \(f(x)=1/\sqrt{x}\) for \(x>0\), on \([0,1]\). A mistaken argument might check the two endpoints, observe \(f(0)=0\) and \(f(1)=1\), and conclude that \(|f(x)|\le1\) throughout the interval.
That conclusion is false. For example, \(f(1/100)=10\), and \(f(1/10000)=100\). More generally, for every positive integer \(m\),
These values exceed any proposed fixed bound when \(m\) is chosen large enough. Endpoint checks alone do not control the values between the endpoints. Boundedness requires one finite bound that works for every input in the interval.
Why This Condition Matters
Boundedness is an early structural check for a proposed proper integral. It explains why a finite interval alone does not guarantee that the tagged rectangle sums have a finite limit: the sampled heights must also be controlled. In practice, the theorem gives a quick way to rule out proper Riemann integrability. If the function is unbounded on the interval, there is no need to search for a limit of its tagged sums under the proper definition.
Keep the direction of the result clear. Boundedness is necessary for proper Riemann integrability on \([a,b]\) when \(a<b\), but it is not sufficient. The integral still has to satisfy the tagged-sum convergence condition.
Check Your Understanding
Answer these questions using the definition and theorem from this tutorial.
- What does it mean for \(f:[a,b]\to\mathbb{R}\) to be bounded?
- Why does proper Riemann integrability require all sufficiently fine tagged sums to remain close to one finite number?
- In the proof, why must at least one subinterval of a finite partition contain unbounded values if \(f\) is unbounded on the full interval?
- Why does varying one tag in a positive-width subinterval make the tagged sums unbounded when \(f\) is unbounded there?
- If a function is bounded on a closed interval, what additional fact must be established before concluding that it is properly Riemann integrable?
