10. A Little Bit More on Integrals
≪ 9. Integration and Darboux Sums | Table of ContentsWow, I can’t believe it’s already week 10. Congrats on making it this far!
Let’s tie off the course by getting some more practise in with proving things about integrals. Look back to last week’s notes if you’ve forgotten what partitions, lower Darboux sums, etc. are. But one thing I will repeat is the main theorem that drives all of these proofs forward:
Theorem 1.
A function \(f : [0, 1]\to \mathbb{R}\) is Darboux integrable if and only if for every \(\epsilon > 0\), there exists a partition \(\mathcal{P}\) such that \[U \left( f; \mathcal{P} \right) - L \left( f; \mathcal{P} \right) < \epsilon.\]
For today, I think we’ll be applying this theorem with the same partition in both the upper and lower Darboux sums. One common type of problem that we encounter are those of the form, “given one or more Darboux integrable functions, modify or combine them somehow and show that the resulting function is also Darboux integrable.” For instance,
Example 2.
Let \(f, g : [0, 1]\to \mathbb{R}\) be two Darboux integrable functions. Show that \(f + g\) is also Darboux integrable.
I think you guys have seen and proven this theorem in class, so let’s take the opportunity to remember the proof and refresh ourselves on some tools and terminology while we’re at it.
As usual, we’ll try to apply Cauchy’s criterion. Let’s first fix \(\epsilon > 0\), then use the integrability of \(f\) and \(g\) to find two partitions \(\mathcal{P}_f\) and \(\mathcal{P}_g\) such that \[\begin{align*} U \left( f; \mathcal{P}_f \right) - L \left( f; \mathcal{P}_f \right) < \epsilon && \textrm{and} && U \left( g; \mathcal{P}_g \right) - L \left( g; \mathcal{P}_g \right) < \epsilon. \end{align*}\]
The idea is to now “merge” the two partitions together: let \(\mathcal{P}\) be the partition formed out of all the endpoints in \(\mathcal{P}_f\) and \(\mathcal{P}_g\). We say \(\mathcal{P}\) is a refinement of the latter two! Now let’s compute \[\begin{align*} U(f+g; \mathcal{P}) &= \sum \Delta \mathcal{P}_i \cdot \sup _{x\in \mathcal{P}_i} \left( f(x)+g(x) \right) \\ & \leq \sum \Delta \mathcal{P}_i \sup _{x\in \mathcal{P}_i} f(x) + \sum \Delta \mathcal{P}_i \sup _{x\in \mathcal{P}_i} g(x) \\ &= U(f; \mathcal{P}) + U(g; \mathcal{P}) \\ & \leq U \left( f; \mathcal{P}_f \right) + U \left( g; \mathcal{P}_g \right). \end{align*}\] The first inequality is because in general \(\sup _{x\in S} (f(x) + g(x) \leq \sup _{x\in S}f(x) + \sup _{x\in S} g(x)\) for any set \(S\) contained in the common domains of \(f\) and \(g\). (Those of you with good memory will remember this from week 3 or so.) The second inequality is because \(\mathcal{P}\) is a refinement of both \(\mathcal{P}_f\) and \(\mathcal{P}_g\) — the upper Darboux sums cannot become worse overestimates of the integral by using finer rectangles.
Using an identical argument, we can show that \[L(f+g; \mathcal{P}) \geq L\left( f;\mathcal{P}_f \right) + L \left( g; \mathcal{P}_g \right). \] Thus, putting the two together and using the construction of \(\mathcal{P}_f\) and \(\mathcal{P}_g\) yields \[\begin{align*} U(f+g; \mathcal{P}) - L(f+g; \mathcal{P}) & \leq U\left( f; \mathcal{P}_f \right) - L \left( f; \mathcal{P}_f \right) + U \left( g; \mathcal{P}_g \right) - L \left( g; \mathcal{P}_g \right) \\ & < 2 \epsilon. \end{align*} \]
Replacing \(\epsilon \) with \(\frac{\epsilon }{2}\) yields the Cauchy criterion for \(f + g\).
Think about the structure of this argument. We first applied the Cauchy criterion for \(f\) and \(g\) separately to obtain two distinct partitions. We then put those two partitions together into a common “refinement” and used the properties of \(\sup\) and \(\inf\) to relate the Darboux sums for the sum and the individual functions.
Problem 3.
Show that if \(f, g : [0, 1]\to \mathbb{R}\) are both Darboux integrable, then the function \[h(x) = \begin{cases} f(x) & x \leq \frac{1}{2}, \\ g(x) & x > \frac{1}{2} \end{cases} \] is also Darboux integrable.
There is a significant subtlety to this problem: if you follow the path of the preceeding example and take the common refinement of the partitions for \(f\) and \(g\), you will have to deal with whichever subinterval contains the point \(\frac{1}{2}\). There, \(\sup h\) and \(\inf h\) may have a significant disparity that’s not addressed by the close-ness of the Darboux sums for \(f\) and \(g\) themselves. This manifests visually when there is a jump discontinuity of \(h\) at the point \(x=\frac{1}{2}\).
There are several ways around this. You can either take the route of a direct proof and just handle this potential discrepancy by excising a very, very thin rectangle around \(\frac{1}{2}\), or you can take an indirect proof by expressing \(h\) as a sum, product, sum of products, product of sums, etc. of functions that we already know are Darboux integrable. The latter is a much cleaner proof, but you may still be asked to prove things “from the definitions” on the final exam.
Solution 1 (Indirect Method)
If, after reading the above proof, you are extremely dissatisfied, do not despair. While it’s very clean, I find it quite unnatural…
Solution 2 (Direct Method)
Let \(\epsilon > 0\) be fixed. We will build our partition around \(\epsilon \). By Cauchy’s criterion, there exist partition \(\mathcal{P}_f\) and \(\mathcal{P}_g\) such that \[\begin{align*} U \left( f; \mathcal{P}_f \right) - L \left( f; \mathcal{P}_f \right) < \epsilon && \textrm{and} && U \left( g; \mathcal{P}_g \right) - L \left( g; \mathcal{P}_g \right) < \epsilon. \end{align*}\]
Since \(f\) and \(g\) are Darboux integrable, they are bounded, hence there exists \(M \in \mathbb{R}\) such that \(\left\lvert f(x) \right\rvert \leq M\) and \(\left\lvert g(x) \right\rvert \leq M\) for all \(x\in [0, 1]\). Let \(\delta > 0\) be a very small parameter, to be chosen later. Let \(\mathcal{P}\) be the partition formed by:
- The endpoints in \(\mathcal{P}_f\) to the left of \(\frac{1}{2} - \delta \);
- the points \(\frac{1}{2}- \delta \) and \(\frac{1}{2} + \delta \); and
- the endpoints in \(\mathcal{P}_g\) to the right of \(\frac{1}{2} + \delta \).
Then, one can show that \[U(h; \mathcal{P}) - L(h; \mathcal{P}) < 2 \epsilon + 4M \delta .\] The \(2 \epsilon \) comes from the discrepancy between the two Darboux sums outside of the interval \(\left[ \frac{1}{2}-\delta ,\frac{1}{2}+\delta \right]\), whereas the \(4M \delta \) is the largest discrepancy from the one subinterval containing \(\frac{1}{2}\)! I’ll leave the fine details as an exercise. Picking \(\delta = \frac{\epsilon }{M}\) (or \(\delta = \frac{1}{2}\) if this is bigger than \(\frac{1}{2}\)) gives an upper bound of \(6 \epsilon \). Replacing \(\epsilon \) with \(\frac{\epsilon }{6}\) finishes the proof.
Problem 4.
Let \(f, g : [0, 1] \to \mathbb{R}\) be two Darboux integrable functions. Show that \(\max(f, g)\) is Darboux integrable.
Hint
Proof
Let’s follow the idea presented in the hint. Suppose \(h\) is Darboux integrable on \([0, 1]\), and let \(\epsilon > 0\). Let \(\hat h(x) = \max(h, 0)\). Let \(\mathcal{P}\) be a partition so that \(U(h; \mathcal{P}) - L(h; \mathcal{P}) < \epsilon \). Then, for any subinterval \(\mathcal{P}_i\), we have \[\sup _{x\in \mathcal{P}_i}h(x) - \inf _{x\in \mathcal{P}_i}h(x) \leq \sup _{x\in \mathcal{P}_i} \hat h(x) - \inf _{x\in \mathcal{P}_i} \hat h(x).\] Unravelling the definitions yields \[U \left( \hat h; \mathcal{P} \right) - L \left( \hat h; \mathcal{P} \right) \leq U(h; \mathcal{P}) - L(h; \mathcal{P}) < \epsilon.\]
For the remainder of the argument, we have \(f - g\) is integrable, hence \(\max(f - g, 0)\) is integrable, hence \(\max(f - g, 0) + g = \max(f, g)\) is integrable! Checking this last equality is left as a cheeky exercise.