Why Boundedness Matters
A definite integral can represent accumulated quantity, signed area, or another net total. For the proper Riemann integral, however, the function must meet an essential requirement: it must be bounded throughout the interval of integration.
This condition can seem restrictive when a function becomes arbitrarily large near one point. The key is that a proper Riemann integral is built from finite upper and lower sums over partitions. If a function has a vertical blow-up, at least one interval in every finite partition still contains values with no finite bound. Its upper or lower sum therefore cannot be finite.
Partitions and Proper Riemann Integrability
A partition of a closed interval \([a,b]\) is a finite list of points \(a=x_0<x_1<\cdots<x_m=b\). Its subintervals are \([x_{i-1},x_i]\), with widths \(\Delta x_i=x_i-x_{i-1}\).
For a bounded function \(f:[a,b]\to\mathbb{R}\), let \(M_i\) be the supremum of \(f\) on the \(i\)-th subinterval and \(m_i\) its infimum there. The upper sum and lower sum for the partition are
The proper Riemann integral is defined for bounded functions. A bounded function is Riemann integrable if, for every \(\varepsilon>0\), there is a partition \(P\) such that \(U(f,P)-L(f,P)<\varepsilon\). Its integral is the common value approached by the upper and lower sums.
The upper and lower sums track the largest and smallest function values on each subinterval. If the function is unbounded above near a point, the supremum on any subinterval containing arbitrarily large values is \(+\infty\). The corresponding upper sum is not finite. If the function is unbounded below, the corresponding lower sum is not finite.
Why a Finite Partition Cannot Avoid an Unbounded Blow-Up
There is a simple reason that subdividing the interval cannot solve the problem. A finite partition divides \([a,b]\) into only finitely many subintervals. If the function were bounded on each one, then it would be bounded on the whole interval: take the largest of the finitely many bounds. Therefore, if the function is unbounded on \([a,b]\), at least one partition subinterval must still contain unbounded function values.
Suppose \(f\) has no finite upper bound on \([a,b]\).
If \(f\) had a finite upper bound on every subinterval, the largest of those finitely many bounds would bound \(f\) on the whole interval.
Since \(f\) is unbounded on the whole interval, at least one subinterval has supremum \(+\infty\).
The subinterval has positive width, so its upper-sum contribution is infinite. Thus the upper sum cannot be a finite quantity.
This also explains why placing a partition point at the location of the blow-up does not help. For a vertical blow-up, values become unbounded on approach to the point from at least one side. One of the neighboring subintervals therefore still has no finite upper bound.
Worked Example: A Blow-Up at the Left Endpoint
Worked Example: The Function \(1/x\) on \([0,1]\)
Define \(f(0)=0\), and set \(f(x)=1/x\) for \(0<x\leq 1\). The function is defined at every point of \([0,1]\), but it is unbounded above near \(0\). Indeed, for every positive integer \(n\), \(f(1/n)=n\).
We can also see the problem in tagged Riemann sums. For an integer \(n\geq 2\), use the partition \(P_n=\{0,1/n,1\}\). Choose the tag \(1/n^2\) in the first subinterval \([0,1/n]\), and choose the tag \(1\) in \([1/n,1]\).
The first subinterval has width \(1/n\), and its tag value is
Its contribution to the tagged sum is \((1/n)n^2=n\). The second subinterval has width \(1-1/n\), and \(f(1)=1\), so its contribution is \(1-1/n\). The full sum is therefore
As \(n\) increases, this sum is unbounded. For example, at \(n=2\) it is \(5/2\), and at \(n=4\) it is \(19/4\). This sequence of sums is a concrete warning that the proper-integral boundedness condition fails. The value \(f(0)=0\) does not change the values \(f(x)\) at positive \(x\) arbitrarily close to zero.
Check: the chosen tag satisfies \(0<1/n^2\leq1/n\) for \(n\geq2\), and the two subinterval widths add to \(1/n+(1-1/n)=1\), the length of \([0,1]\).
Worked Example: A Different Rate of Blow-Up
Worked Example: The Function \(1/\sqrt{x}\) on \([0,1]\)
Define \(g(0)=0\), and let \(g(x)=1/\sqrt{x}\) when \(0<x\leq1\). At \(x=1/n^2\), the function value is \(n\), so the values are not bounded above as \(n\) increases. The function is not properly Riemann integrable on \([0,1]\).
To display large tagged sums, take the same partition \(P_n=\{0,1/n,1\}\), for \(n\geq2\). Choose the tag \(1/n^3\) on \([0,1/n]\), and choose \(1\) on \([1/n,1]\). The first tag is in its subinterval because \(0<1/n^3\leq1/n\). Its function value is
The first subinterval contributes \((1/n)n^{3/2}=\sqrt{n}\). The second contributes \((1-1/n)g(1)=1-1/n\). Thus the total tagged sum is
For \(n=4\), this is \(2+1-1/4=11/4\); for \(n=16\), it is \(4+1-1/16=79/16\). The sum grows without bound along these choices. Most importantly, \(g\) is unbounded on the interval, so it fails the defining boundedness requirement regardless of the particular sequence of sums.
Check: the first contribution is \((1/4)(4^{3/2})=2\) when \(n=4\), and the second is \(3/4\), giving \(11/4\) as stated.
Worked Example: A Blow-Up Inside the Interval
Worked Example: The Function \(1/|x-1/2|\) on \([0,1]\)
Define \(h(1/2)=0\), and for \(x\ne1/2\) let \(h(x)=1/|x-1/2|\). The function is unbounded on both sides of \(1/2\): as \(x\) approaches \(1/2\), the denominator can be made arbitrarily small while remaining positive.
For an integer \(n\geq3\), put \(\delta=1/n\) and use the partition with endpoints \(0\), \(1/2-\delta\), \(1/2+\delta\), and \(1\). Choose the midpoint as the tag on each outer subinterval. On the central subinterval, choose the tag \(1/2+\delta^2\), which lies inside it since \(0<\delta^2<\delta\).
The central interval has width \(2\delta=2/n\). At its tag, the distance from \(1/2\) is \(\delta^2=1/n^2\), so the function value is \(n^2\). Its contribution is therefore
Each outer interval has width \(1/2-1/n=(n-2)/(2n)\). The midpoint of the left interval is \(1/4-1/(2n)\), whose distance from \(1/2\) is \(1/4+1/(2n)=(n+2)/(4n)\). Its function value is \(4n/(n+2)\), so its contribution is
The right midpoint is \(3/4+1/(2n)\), also at distance \((n+2)/(4n)\) from \(1/2\). Its contribution is the same. The complete tagged sum is
For \(n=4\), this is \(8+4/3=28/3\). For \(n=8\), it is \(16+24/10=88/5\). These sums grow without bound. The calculation also shows why choosing \(1/2\) itself as a partition point cannot remove the blow-up: values still become arbitrarily large on the neighboring intervals.
Check: when \(n=4\), the three interval widths are \(1/4\), \(1/2\), and \(1/4\), which add to \(1\). The central contribution is \(2n=8\), and the two outer contributions are each \(2(4-2)/(4+2)=2/3\), giving \(8+2/3+2/3=28/3\).
A Common Mistake: Changing the Value at the Blow-Up Point
A tempting but incorrect argument is that a function becomes bounded once a finite value is assigned at the point where its formula is undefined. For example, one might write, “Set \(f(0)=0\); now \(f(x)=1/x\) is bounded on \([0,1]\).” The calculation \(f(1/n)=n\) for every positive integer \(n\) shows the error: those values remain unbounded, even though \(f(0)\) has been specified.
Another important distinction is that “no proper Riemann integral” is a statement about the proper integral on the specified closed interval. It does not, by itself, settle whether another notion of integral can be used or whether some limiting procedure has a finite value. Those questions require separate definitions and analysis.
Why This Requirement Is Useful
Boundedness makes the upper and lower sums meaningful as finite quantities. It ensures that each subinterval contributes a finite upper and lower estimate, so the question of bringing those estimates close together can be asked. With a vertical blow-up, no finite partition can provide finite upper and lower bounds throughout. The obstacle arises before any attempt to make the partition finer.
When examining a function on a closed interval, first check whether its values have a finite bound. If it has a vertical blow-up anywhere in the interval, it is not properly Riemann integrable there. Do not infer otherwise just because the function is defined at the point, or because calculations on some selected partitions appear finite.
Check Your Understanding
Use boundedness and upper or lower sums to answer these questions.
- State the boundedness requirement in the definition of a proper Riemann integral.
- Why must every finite partition of an unbounded function have a subinterval on which the function is unbounded?
- For \(f(0)=0\) and \(f(x)=1/x\) when \(x>0\), what is \(f(1/n)\), and what does this tell you?
- Why does assigning a finite value at \(x=1/2\) fail to bound \(1/|x-1/2|\) near that point?
- Does failure to have a proper Riemann integral automatically determine whether a different integral notion applies?
