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Chapter 7: Problem 14

Write out the proofs that a function \(f\) on \([a, b]\) is Darboux integrable if it is either (a) continuous, or (b) monotone.

Short Answer

Expert verified
To prove a continuous function on \([a, b]\) is Darboux integrable, show that for each \( \epsilon > 0 \) there exists a partition \( P \) of \([a, b]\) such that \( U(f, P) - L(f, P) < \epsilon \). This can be shown if \( f \) attains a maximum and minimum on \([a, b]\). To prove a monotone function on \([a, b]\) is Darboux integrable, show that it's bounded and it has countably many discontinuities. In each subinterval of a partition containing all points of discontinuity, \( f \) is continuous and hence we can refine the partition such that \( U(f, P) - L(f, P) < \epsilon \) for each \( \epsilon > 0 \).

Step by step solution

01

Proof for continuous function

Consider the function \( f \) on \([a, b]\) to be continuous. Recall the definition of Darboux integrable function, it's where for every \( \epsilon > 0 \), there exists a partition \( P \) of \([a, b]\) such that \( U(f, P) - L(f, P) < \epsilon \). Because \( f \) is continuous on \([a, b]\), it attains a maximum and a minimum in this interval. Let \( M \) and \( m \) denote the maximum and minimum of \( f \) in \([a, b]\). Choose any partition \( P \) of \([a, b]\) such that \( \|P\| < \epsilon/(M-m) \). Then for each subinterval \( [x_{i-1}, x_i] \), the function \( f \) attains its maximum and minimum. Hence, \( U(f, P) - L(f, P) = \sum (M_i - m_i) \Delta x_i \leq (M-m) \epsilon/(M-m) = \epsilon \). Thus, \( f \) is Darboux integrable.
02

Proof for monotone function

Now suppose the function \( f \) on \([a, b]\) is monotone, say, non-decreasing. The proof for non-increasing function would be similar. A non-decreasing function on a compact interval is bounded, and also, it can only have countably many discontinuities. Now choose the partition \( P \) consisting of \( a, b \) and all points of discontinuity. Then we know in each subinterval of the partition \( P \), \( f \) is continuous, and hence attains its maximum and minimum. Therefore, for each \( \epsilon > 0 \), we can refine \( P \) such that \( U(f, P) - L(f, P) < \epsilon \). Hence, \( f \) is Darboux integrable.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuous Functions
When learning about real analysis, one critical concept is that of continuous functions. These functions have the intuitive property of not having any 'jumps' or 'gaps' when you draw their graphs鈥攖hey appear as a single unbroken curve. A function is continuous on an interval like \[a, b\] if for every point within the interval, the closer we get to that point, the values of the function get closer to the function's value at that point.

From an educational standpoint, continuous functions are easy to visualize, which aids in understanding. While dealing with Darboux integrability, the continuity of a function simplifies the proof because, for any interval, the function attains a maximum and minimum value. This fact is used in the partition method for proving integrability, where the upper sum \(U(f, P)\) and the lower sum \(L(f, P)\) get arbitrarily close, ensuring that the area under the curve can be precisely determined.
Monotone Functions
Monotone functions are those that are either always increasing or always decreasing. Imagine walking up a hill or down a slide without ever stepping back; that's the behavior of monotone functions. They can only move in one direction within their domain.

In the context of Darboux integrability, an important property of monotone functions on a closed interval is that they can only have jump discontinuities, which are countable. When monotone, the function is either non-decreasing (our uphill walk) or non-increasing (our slide downhill). For such functions, establishing the existence of a partition that captures the degree of how close the upper and lower sums are becomes a simpler task. It's another friendly case for students as the inherent orderliness of monotone functions aligns well with the intuition that areas under simple shapes are easier to calculate.
Proof of Integrability
The proof of integrability can sometimes feel abstract for students, but it forms the backbone of real analysis when it comes to understanding the area under a curve. So, what does it mean for a function to be integrable in the Darboux sense? It means that for any small error margin \(\epsilon > 0\), we can find a way to partition the interval \[a, b\] so that the difference between the function's upper sum and lower sum is less than this error margin. This is essentially capturing the notion that we can calculate the area beneath a curve to any desired precision.

To solidify this, let's say we have a continuous function. We can divide the interval \[a, b\] into smaller pieces where the function reaches both a maximum and minimum. By taking the sum of these 'peaks' and 'valleys', we get close to the actual integral. The proof shows that as we make these pieces smaller, our approximation becomes as close as we wish to the true value鈥攖he hallmark of integrability.
Real Analysis
Now, stepping back to a broader view, real analysis is the study of real numbers and real-valued functions. Within this branch of mathematics, we delve into limits, continuity, integration, and differentiation鈥攃ore concepts that help us understand and describe the real world. It's a field marked by precision and rigor. It provides the tools to analyze and solve problems ranging from simple to complex in mathematics and applied sciences.

Understanding concepts, like the Darboux integrability of continuous or monotone functions, is a gateway into deeper exploration in real analysis. Through these exercises, students start to appreciate the meticulous nature of mathematical arguments and the beauty of a function's predictability through its properties.

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Most popular questions from this chapter

Let \(f, g \in \mathcal{R}[a, b]\). (a) If \(t \in \mathbb{R}\), show that \(\int_{a}^{b}(t f \pm g)^{2} \geq 0\). (b) Use (a) to show that \(2\left|\int_{a}^{b} f g\right| \leq t \int_{a}^{b} f^{2}+(1 / t) \int_{a}^{b} g^{2} \quad\) for \(t>0\). (c) If \(\int_{a}^{b} f^{2}=0\), show that \(\int_{a}^{b} f g=0\). (d) Now prove that \(\left|\int_{a}^{b} f g\right|^{2} \leq\left(\int_{a}^{b}|f g|\right)^{2} \leq\left(\int_{a}^{b} f^{2}\right) \cdot\left(\int_{a}^{b} g^{2}\right) .\) This inequality is called the Cauchy-Bunyakovsky-Schwarz Inequality (or simply the Schwarz Inequality).

Let \(f\) be continuous on \(I:=[a, b \mid\) and assume \(f(x) \geq 0\) for all \(x \in I\). Prove if \(L(f)=0\), then \(f(x)=0\) for all \(x \in I\)

Let \(h:[0,1] \rightarrow \mathbb{R}\) be Thomae's function and let sgn be the signum function. Show that the composite function sgn \(\circ h\) is not Riemann integrable on \([0,1]\).

Let \(f:[a, b] \rightarrow \mathbb{R}\). Show that \(f \notin \mathcal{R}[a, b]\) if and only if there exists \(\varepsilon_{0}>0\) such that for every \(n \in \mathbb{N}\) there exist tagged partitions \(\dot{\mathcal{P}}_{n}\) and \(\dot{\mathcal{Q}}_{n}\) with \(\left\|\dot{\mathcal{P}}_{n}\right\|<1 / n\) and \(\left\|\mathcal{Q}_{n}\right\|<1 / n\) such that \(\left|S\left(f ; \dot{\mathcal{P}}_{n}\right)-S\left(f ; \mathcal{Q}_{n}\right)\right| \geq c_{0}\)

Let \(0 \leq a
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