9. Integration and Darboux Sums
≪ 8. Derivatives and Local-to-Global Correspondences | Table of ContentsBefore we get started with our discussion, let us refresh our memories on some preliminary definitions and notation. For the duration of today, the domain of integration is always the closed interval \([0,1]\).
Definitions 1.
- A partition of \([0, 1]\) is a finite ordered list of points \(0 = t_0 < t_1 < \cdots < t_n = 1\). We often write \(\mathcal{P} = \left( t_0, \ldots, t_n \right)\).
- If \(f : [0, 1]\to \mathbb{R}\) is a function and \(\mathcal{P}\) is a partition, the upper Darboux sum of \(f\) with respect to \(\mathcal{P}\) is given by \[U(f; \mathcal{P}) = \sum _{i=1}^{n} \left( t_i - t _{i-1} \right) \cdot \sup _{x\in \left[ t _{i-1}, t_i \right]} f(x).\] The lower Darboux sum \(L(f; \mathcal{P})\) is defined analogously with \(\inf\) in place of \(\sup\).
- The upper Darboux integral of \(f\) is defined as \[U(f) = \inf _{\mathcal{P}} U(f; \mathcal{P}),\] where the \(\inf\) is taken over all possible partitions. Likewise, the lower Darboux integral of \(f\) is defined as \[L(f) = \sup _{\mathcal{P}} L(f; \mathcal{P}).\]
- The Darboux integral of \(f\) is defined as \(U(f)\) or \(L(f)\), whenever they agree.
To ease notation, I will often denote by \(\Delta \mathcal{P}_i = t _{i} - t _{i-1}\) and \(\mathcal{P}_i = \left[ t _{i-1}, t _{i} \right]\) for \(i = 1,\ldots, n\) and a given partition.
Proving that something is Darboux integrable (or not) may seem like it involves some horrible mess of computation, what with the nested \(\sup\)’s and \(\inf\)’s and all. But there is an extremely good theorem that greatly, greatly simplifies our life:
Theorem 2.
A function \(f : [0, 1]\to \mathbb{R}\) is Darboux integrable if and only if for every \(\epsilon > 0\), there exist partitions \(\underline{\mathcal{P}}\) and \(\overline{\mathcal{P}}\) such that \[U \left( f; \overline{\mathcal{P}} \right) - L \left( f; \underline{\mathcal{P}} \right) < \epsilon.\]
I’ll mention here that in practise, we often take \(\overline{\mathcal{P}} = \underline{\mathcal{P}}\) and things will work out nicely.
Let’s consider our intuition of integrals as “areas under a curve”. A partition \(\mathcal{P}\) is a government mandate to approximate the area using a fixed number of rectangles with a prescribed set of bases, e.g. the first rectangle must lie on \(\mathcal{P}_1\). Given this partition, \(U(f; \mathcal{P})\) is the absolute largest possible overestimate of the area beneath the curve, while \(L(f; \mathcal{P})\) is the absolute smallest possible underestimate possible. The above theorem (AKA Cauchy’s criterion, of which there are many) states that if the two “worst-case-scenarios” ever get close, even on different partitions, then you are Darboux integrable.
You should think about a lot of problems in this guise: can I cook up a partition \(\mathcal{P}\) that makes it impossible to overestimate or underestimate the area by a lot? Here’s a first example:
Example 3.
Show that the function \[f(x) = \begin{cases} 0 & x \neq \frac{1}{2}, \\ 1 & x = \frac{1}{2} \end{cases}\] is Darboux integrable. What is its Darboux integral?
Solution
If you draw the graph of this guy, you should intuit that the area must be zero. Moreover, we observe that it’s impossible to underestimate the area. Precisely state, we have for all partitions \(\mathcal{P}\) that \(L(f;\mathcal{P}) = 0\). This is because every subinterval of \(\mathcal{P}\) will contain more than one point and in particular will contain a point that’s not \(\frac{1}{2}\). Thus \(\inf _{x\in \mathcal{P}_i}f(x) = 0\) for all \(i = 1,\ldots,n\), regardless of what \(\mathcal{P}\) is, and it follows from the definition that \(L(f, \mathcal{P})=0\).
Now our goal is to find a partition that makes it very hard to overestimate the nonexistent area. The only obstacle is the point \(f\left( \frac{1}{2} \right)\). If \(\mathcal{P}_i\) doesn’t contain this point, then there’s no chance we’ll overestimate the area in that particular subinterval. So we’ll make the one interval that does contain \(\frac{1}{2}\) very, very thin, producing a miniscule contribution to the Darboux sum.
Let’s be a bit more precise now and apply the Cauchy criterion. Given any \(\epsilon > 0\), we seek partitions \(\overline{\mathcal{P}}\) and \(\underline{\mathcal{P}}\) so that \[U\left( f, \overline{\mathcal{P}} \right) - L \left( f, \underline{\mathcal{P}} \right) < \epsilon .\] We’ve already established that the lower sum is always zero, so we only need to find the partition \(\overline{\mathcal{P}}\) that makes \(U\left(f, overline{\mathcal{P}}\right)<\epsilon .\)
Given this \(\epsilon > 0\), we take the partition \(\overline{\mathcal{P}} = \left( 0, \frac{1- \epsilon }{2}, \frac{1+\epsilon }{2}, 1 \right).\) Then \(\sup _{x\in \mathcal{P}_i}f(x) = 0\) when \(i = 1\) and \(i = 3\) (these are the outer two subintervals), and it’s equal to \(1\) when \(i = 2\). Thus \[U \left( f, \overline{\mathcal{P}} \right) = \Delta \mathcal{P}_2 \sup _{x\in \mathcal{P}_2} f(x) = \Delta \mathcal{P}_2 = \epsilon. \] Augh, replace \(\epsilon \) with \(\frac{\epsilon }{2}\) to get the strict inequality and we’re done!
Some functions may seem difficult to work with, but certain properties like monotonicity can make them a lot more approachable.
Example 4.
Show that \(f(x) = \sin(x)\) is Darboux integrable on \([0, 1]\). You may use the fact that \(f\) is increasing on this domain.
Solution
If you draw any picture with both Darboux sums present, you should notice a pretty peculiar pattern that can guide you through the proof presented.
Let \(\mathcal{P} = \left( t_0, \ldots, t_n \right)\) be any partition for now. Since \(f\) is increasing, \(\sup _{x\in \mathcal{P}_i} f(x) = f \left( t_i \right)\): the maximal value is attained at the right endpoint. Likewise, we have that \(\inf _{x\in \mathcal{P}_i}f(x) = f \left( t _{i-1} \right).\) Now our formula for the two Darboux sums simplify: \[\begin{align*} L(f; \mathcal{P}) = \sum _{i=1}^{n} \Delta \mathcal{P}_i f \left( t _{i-1} \right) && \textrm{and} && U(f; \mathcal{P}) = \sum _{i=1}^{n} \Delta \mathcal{P}_i f \left( t _{i} \right). \end{align*} \]
If we shoot for the moon and try to apply Cauchy’s criterion with the same partition on both Darboux sums, we can then collect some like terms when we subtract. Namely, we get \[\begin{align*} U(f; \mathcal{P}) - L(f; \mathcal{P}) = -f(0) \Delta \mathcal{P}_1 + f\left( t_1 \right) \left( \Delta \mathcal{P}_2 - \Delta \mathcal{P}_1 \right) + f \left( t_2 \right) \left( \Delta \mathcal{P}_3-\Delta \mathcal{P}_2 \right) +\\ \cdots + f\left( t _{n-1} \right) \left( \Delta \mathcal{P}_n -\Delta \mathcal{P} _{n-1}\right) + f\left( 1 \right) \Delta \mathcal{P}_n. \end{align*}\]
In particular, if we just choose \(\mathcal{P}\) to be the partition with \(n+1\) evenly spaced points, we get \(\Delta P_i = \frac{1}{n}\) for all \(i\). All of the junk in the middle cancels and we’re left with \[U(f; \mathcal{P}) - L(f; \mathcal{P}) = \frac{f(1)-f(0)}{n} \] Given any \(\epsilon > 0\), we can clearly choose \(n\) very large so that \(\frac{f(1)-f(0)}{n} < \epsilon \), and we conclude that \(f\) is Darboux integrable. \(\square\)
I promise pictures help a lot with these problems.
One more, but this time a negative example.
Example 5.
Show that the function \[f(x) = \begin{cases} 1 & x\in \mathbb{Q}, \\ 0 & x\notin \mathbb{Q} \end{cases}\] is not Darboux integrable on \([0, 1]\).
Solution
The graph of \(f\) looks like two parallel horizontal lines. Though it seems like the area could be \(1\), it’s very easy to underestimate this area due to the bottom line, and we can exploit that during our proof.
Specifically, we have for any partition \(\mathcal{P} = \left( t_0,\ldots, t_n \right)\) that \(\mathcal{P}_i\) always contains at least one rational and irrational number. Thus \(\sup _{x\in \mathcal{P}_i} f(x) = 1\) and \(\inf _{x\in \mathcal{P}_i} f(x) = 0\), regardless of what \(\mathcal{P}\) and \(i\) are! We therefore have that \[L(f; \mathcal{P}) = 0\] for all \(\mathcal{P}\) (since every summand is just \(0\)) and that \[U(f; \mathcal{P}) = \sum _{i=1}^{n} \Delta \mathcal{P}_i = \sum _{i=1}^{n} \left( t_i - t _{i-1} \right) = t_n - t_0 = 1. \] (Note that the sum telescopes a lot like the preceeding example.)
In particular, if \(\epsilon = 1\), there exist no partitions \(\overline{\mathcal{P}}\) and \(\underline{\mathcal{P}}\) such that \[U\left( f; \overline{\mathcal{P}} \right) -L\left( f; \underline{\mathcal{P}} \right) < \epsilon .\] It follows that \(f\) is not Darboux integrable. \(\square\)
If we have time, please chew on the following problem:
Problem 6.
Let \(f: [0, 1]\to [0, 1]\) be the function \(f(x) = \frac{1}{q}\) if \(x = \frac{p}{q}\) is a rational number and \(p, q\) are coprime (i.e. \(x\) is written in reduced form) and \(f(x) = 0\) whenever \(x\) is irrational.
Is \(f\) Darboux integrable? You don’t need to prove or disprove it, but try to give a high-level explanation involving Darboux sums about how you arrived at your answer.