Why Bound Sums Instead of Choosing Sample Points?
A tagged Riemann sum uses one selected function value from each subinterval. Its value can therefore depend on the chosen tags. Upper and lower Darboux sums take a different approach: on each subinterval, they use the greatest and least values the function can approach there. These bounds enclose every tagged sum for that partition, whatever tags are chosen.
This construction applies to any bounded function on a closed interval. The function need not be continuous, and it need not attain its largest or smallest value on each subinterval. The key quantities are the supremum and infimum, which remain defined even when a bound is not attained.
Partitions and Intervalwise Bounds
Let \(f:[a,b]\to\mathbb{R}\) be bounded, where \(a<b\). A partition \(P\) of \([a,b]\) is a finite list of points
The \(i\)th subinterval is \(I_i=[x_{i-1},x_i]\), and its width is \(\Delta x_i=x_i-x_{i-1}\). On that interval, define
The infimum is the greatest lower bound of the function values on the interval; the supremum is their least upper bound. Because \(f\) is bounded, both numbers are finite. They need not be values of \(f\): a supremum need not be a maximum, and an infimum need not be a minimum.
The subintervals are written as closed intervals, so neighboring intervals share an endpoint. This causes no difficulty: the definitions are made separately on each interval, and a shared endpoint simply belongs to both of them. Since each width is positive and \(m_i\leq M_i\), we immediately have \(L(f,P)\leq U(f,P)\).
Worked Example: A Square Function on Two Subintervals
Worked Example: A Square Function on Two Subintervals
Let \(f(x)=x^2\) on \([0,2]\), and take \(P=\{0,1,2\}\). The two subintervals are \([0,1]\) and \([1,2]\), each of width \(1\).
On \([0,1]\), the function values range from \(0\) to \(1\), so \(m_1=0\) and \(M_1=1\). On \([1,2]\), they range from \(1\) to \(4\), so \(m_2=1\) and \(M_2=4\). Thus
Check the calculation interval by interval: the first interval contributes \(0\) to the lower sum and \(1\) to the upper sum; the second contributes \(1\) and \(4\), respectively. Adding gives \(1\) and \(5\), as stated. Every function value on each interval is bounded by that interval’s listed infimum and supremum.
Worked Example: A Supremum That Is Not a Maximum
Worked Example: A Supremum That Is Not a Maximum
Define \(f:[0,1]\to\mathbb{R}\) by \(f(x)=x\) for \(0\leq x<1\), and \(f(1)=0\). Use the partition \(P=\{0,\frac12,1\}\), whose two subintervals each have width \(\frac12\).
On \([0,\frac12]\), the values range from \(0\) to \(\frac12\), so \(m_1=0\) and \(M_1=\frac12\). On \([\frac12,1]\), the values include \(0\) at \(x=1\), and approach \(1\) as \(x\) approaches \(1\) from below. No point in this interval has value \(1\), so its infimum is \(m_2=0\) and its supremum is \(M_2=1\).
Check the second interval carefully: \(1\) is an upper bound, and values of \(f\) on the interval get arbitrarily close to \(1\), so no smaller number is an upper bound. Therefore its supremum is \(1\), even though the function never takes that value there. The lower and upper sums are \(0\) and \(\frac34\).
Why Every Tagged Sum Lies Between the Bounds
A tagged partition chooses a point \(t_i\in[x_{i-1},x_i]\) for each subinterval. Its tagged Riemann sum is \(\sum_{i=1}^n f(t_i)\Delta x_i\). The pointwise definition of infimum and supremum gives \(m_i\leq f(t_i)\leq M_i\). Multiplying by the positive width \(\Delta x_i\) preserves the inequalities. Adding them gives the following result.
This also proves \(L(f,P)\leq U(f,P)\) directly: each interval’s lower bound is no greater than its upper bound, and all widths are positive. The argument does not require the function to attain its bounds, because an infimum is a lower bound and a supremum is an upper bound whether or not they are attained.
For every tag \(t_i\), the definition gives \(m_i\leq f(t_i)\leq M_i\).
Since \(\Delta x_i>0\), we have \(m_i\Delta x_i\leq f(t_i)\Delta x_i\leq M_i\Delta x_i\).
Summing over all subintervals gives the lower and upper Darboux bounds for the tagged sum.
Worked Example: Unequal Widths and Negative Values
Worked Example: Unequal Widths and Negative Values
Let \(f(x)=2-x\) on \([0,3]\), and use the nonuniform partition \(P=\{0,1,3\}\). The first subinterval has width \(1\), and the second has width \(2\).
On \([0,1]\), the values run from \(2\) down to \(1\), so \(m_1=1\) and \(M_1=2\). On \([1,3]\), they run from \(1\) down to \(-1\), so \(m_2=-1\) and \(M_2=1\). Therefore
Check each contribution: the first interval contributes \(1\) to the lower sum and \(2\) to the upper sum; the second contributes \(-2\) and \(2\). The totals are \(-1\) and \(4\). In particular, negative values are allowed: multiplying the negative infimum \(-1\) by the positive width \(2\) gives the lower-sum contribution \(-2\).
What Happens When a Partition Is Refined?
A partition \(Q\) refines a partition \(P\) if it contains every point of \(P\), possibly with additional points inserted. Refinement divides some intervals into smaller intervals. On a smaller interval contained in a larger one, the infimum cannot decrease below the larger interval’s infimum, and the supremum cannot increase above the larger interval’s supremum.
To see why, consider one interval \(I=[u,v]\) of \(P\), split by \(Q\) into intervals \(J_1,\ldots,J_k\). Let \(m_I\) and \(M_I\) be the infimum and supremum on \(I\), and let \(m_{J_j}\) and \(M_{J_j}\) be those on \(J_j\). Since \(J_j\subseteq I\), every value on \(J_j\) is also a value on \(I\), so \(m_I\leq m_{J_j}\) and \(M_{J_j}\leq M_I\). Multiplication by the positive width of \(J_j\) gives
where \(|J_j|\) denotes the width of \(J_j\). Add these inequalities over the smaller intervals. Their widths sum to the width of \(I\), so
Applying this argument to every interval of \(P\) proves the theorem. Thus adding partition points can only raise the lower sum and lower the upper sum. The two sums move inward, although this fact alone does not say that they eventually meet or determine an integral.
A Common Pitfall: Replacing Supremum with Maximum
A common error is to use a maximum where the function does not attain one. In the second example, on \([\frac12,1]\) the function values approach \(1\), but none equals \(1\). A mistaken calculation might say “the maximum is the value at \(x=1\), so \(M_2=f(1)=0\),” and then assign an upper-sum contribution of \(0\cdot\frac12=0\). That is incorrect: \(f(1)=0\) is not the largest value, and values such as \(f(\frac{99}{100})=\frac{99}{100}\) exceed it.
The correction is to use the least upper bound of all values on the interval. Since every value is at most \(1\), and values get arbitrarily close to \(1\), the supremum is \(1\). Its contribution is therefore \(1\cdot\frac12=\frac12\), as used in the correct upper sum. When a function does attain its bounds, maximum and supremum agree, as do minimum and infimum; the definitions of Darboux sums do not assume this happens.
Why Darboux Sums Matter
Darboux sums turn intervalwise information about a bounded function into numerical lower and upper bounds. For a fixed partition, the lower sum is no greater than any tagged Riemann sum, and the upper sum is no smaller. Refining the partition improves these bounds in a definite direction. These properties make Darboux sums useful for studying how sums behave as partitions are made finer and for formulating precise conditions connected with Riemann integrability.
For now, keep the construction distinct from the question of whether a particular function is integrable. The sums are defined for every bounded function on a closed interval. Their definitions require only the partition, the intervalwise infima and suprema, and the widths.
Check Your Understanding
Use the definitions and the bounds proved above to answer these questions.
- For a bounded function on a partition interval, what do the infimum and supremum describe?
- Why is the width of each subinterval multiplied by its local infimum or supremum?
- Can a supremum be used in an upper Darboux sum even when the function never attains it? Explain.
- Why must every tagged sum for a partition lie between its lower and upper Darboux sums?
- When a partition is refined, what happens to its lower and upper Darboux sums?
