Problem 38 - A Course in Calculus and Real Analysis

Question

Let \(f:[a, b] \rightarrow \mathbb{R}\) be integrable and let \(g:[m(f), M(f)] \rightarrow \mathbb{R}\) be continuous. Show that \(g \circ f:[a, b] \rightarrow \mathbb{R}\) is integrable. (Hint: Given \(\epsilon>0\), find \(\delta>0\) using the uniform continuity of \(g .\) There is a partition \(P\) of \([a, b]\) such that \(U(P, f)-L(P, f)<\delta^{2}\). Divide the summands in \(U(P, f)-L(P, f)\) into two parts depending on whether or not \(M_{i}(f)-m_{i}(f)<\delta .\) Use the Riemann condition for \(g \circ f .\) )

Step-by-Step Solution

Answer

To show \(g \circ f\) is integrable, we need to find a partition \(P\) such that the upper and lower sums satisfy \(U(P, g \circ f)-L(P, g \circ f)<\epsilon\). Using the uniform continuity of \(g\), we find a partition \(P\) so that \(U(P, f)-L(P, f)<\delta^{2}\). We divide the summands in \(U(P, f) - L(P, f)\) into two parts based on whether \(M_i(f) - m_i(f) < \delta\) or not. Applying the uniform continuity of \(g\) and estimating the difference \(U(P, g \circ f) - L(P, g \circ f)\) for both parts, we choose \(\epsilon_1 = \frac{\epsilon}{2(b-a)}\) to ensure the total difference is smaller than \(\epsilon\), concluding that \(g \circ f\) is integrable on the interval \([a, b]\).

Step 1: Definition of integrability and uniform continuity

To show that \(g \circ f\) is integrable, we need to show that for any given \(\epsilon>0\), we can find a partition \(P\) such that the upper and lower sums satisfy \(U(P, g \circ f)-L(P, g \circ f)<\epsilon\). We are also given that \(g\) is continuous and hence uniformly continuous on the closed interval \([m(f), M(f)]\). By the definition of uniform continuity, for any \(\epsilon_1>0\), there exists \(\delta>0\) such that for all \(x, y \in [m(f), M(f)]\), if \(|x-y|<\delta\), then \(|g(x)-g(y)|<\epsilon_1\).

Step 2: Finding a partition P satisfying the given condition

As given in the hint, we need to find a partition \(P\) of interval \([a, b]\) such that \(U(P, f)-L(P, f)<\delta^{2}\). Since \(f\) is integrable, we can find such a partition by choosing an appropriate partition size that sub-divides the interval into smaller subintervals.

Step 3: Dividing summands in U(P, f) - L(P, f)

Now, we need to divide the summands in the expression \(U(P, f) - L(P, f)\) into two parts. The first part consists of the summands for which \(M_{i}(f) - m_{i}(f) < \delta\), and the second part consists of the summands for which \(M_{i}(f) - m_{i}(f) \geq \delta\). Since the summands for the first part are already smaller than \(\delta\), we will mainly need to focus on the second part in the next steps.

Step 4: Applying Riemann condition for g ∘ f

For summands with \(M_{i}(f) - m_{i}(f) < \delta\), we can use the uniform continuity of \(g\) to ensure that the oscillation of \(g \circ f\) is also small. To do this, we apply the uniform continuity of \(g\) to the interval \([m_{i}(f), M_{i}(f)]\). Since \(|m_{i}(f) - M_{i}(f)| < \delta\), we have \(|g(m_{i}(f)) - g(M_{i}(f))| < \epsilon_1\). In other words, the oscillation of \(g\) is small when the difference between \(m_{i}(f)\) and \(M_{i}(f)\) is small.

Step 5: Estimating U(P, g ∘ f) - L(P, g ∘ f)

Now, we need to estimate the difference \(U(P, g \circ f) - L(P, g \circ f)\) by combining the information we have obtained so far. Since the oscillation of \(g \circ f\) is small when the oscillation of \(f\) is small, the summands in the first part of the division will also contribute small amounts to the difference \(U(P, g \circ f) - L(P, g \circ f)\). For the second part, we notice that since \(M_{i}(f) - m_{i}(f) \geq \delta\), the oscillation of \(g \circ f\) is bounded by \(M(g \circ f) - m(g \circ f)\), which is a finite number.

Step 6: Choosing ε₁ and showing integrability

To complete the proof, we need to choose an appropriate value for \(\epsilon_1\) such that the sum of the differences \(U(P, g \circ f) - L(P, g \circ f)\) for both parts is smaller than \(\epsilon\). We can do this by choosing \(\epsilon_1 = \frac{\epsilon}{2(b-a)}\), and noting that the difference \(U(P, g \circ f) - L(P, g \circ f)\) for the second part is also smaller than \(\frac{\epsilon}{2}\) due to the chosen partition size. Hence, the total difference is smaller than \(\epsilon\), and we can conclude that \(g \circ f\) is integrable on the interval \([a, b]\).

Key Concepts

Uniform Continuity

Uniform continuity is a strengthening of the continuity concept that is especially important for ensuring the integrability of composite functions like g ∘ f. As we encounter in calculus, a function g is continuous on an interval if for every point within this interval and for every ε>0, there is a δ>0 such that small changes in input (less than δ) produce changes in output less than ε. However, uniform continuity takes this one step further by requiring that the chosen δ works uniformly for every point in the interval.

What does this mean for our problem? If a function g is uniformly continuous on the range of f, we can guarantee that small oscillations in f will not cause large jumps in the values of g(f(x)). As a result, the function g ∘ f will not oscillate too wildly, which is a key ingredient in showing its integrability.

Riemann Integral

The Riemann integral is a method of integrating functions in a very tangible manner—by approximating the area under the curve as a sum of the areas of many thin rectangles. It rests on the idea that if we can make the heights of these rectangles approach the function's true value closely enough, and the total width of all rectangles small enough, we have a good approximation of the integral.

A function is Riemann integrable if we can make the difference between the upper and lower sums (sums of the areas of rectangles either just above or just below the curve) arbitrarily small. To demonstrate that the composite function g ∘ f is integrable in our scenario, we want to show that we can find a partition of the domain such that the difference between these sums is less than any predefined ε. This involves ensuring that our function does not oscillate too much, as large oscillations can make it difficult to pin down a good approximation—hence the connection to uniform continuity.

Upper and Lower Sums

When working with the concept of the Riemann integral, upper and lower sums are pivotal. To clarify, the upper sum is calculated by taking the supremum (or the least upper bound) of the function values on each subinterval of a partition, then multiplying each by the subinterval's width and summing these products. The lower sum uses the infimum (or the greatest lower bound) instead.

What we're doing is building a 'staircase' that hugs the function from above and below. For g ∘ f to be integrable, as the exercise demands, these two 'staircases' need to come arbitrarily close together as we refine our partition. The smaller the difference between these sums, the more accurately they trap the true area under the curve, which is what we mean by integrability of the function.

Oscillation of a Function

Oscillation of a function at a point describes how much a function's value varies near that point. In a precise sense, for any interval centered at a point, the oscillation is the difference between the function's largest and smallest value over that interval. A function with small oscillation doesn't vary much—it's relatively stable or flat.

Why does this matter? When proving a composite function like g ∘ f is integrable, we rely on the idea that if f has small oscillation, then g will follow suit because of its uniform continuity. When we partition the domain and look at the function's behavior on these small subintervals, if the oscillation of f is within our chosen δ then, due to the uniform continuity of g, g(f(x)) will have small oscillation as well. This low oscillation across the subintervals helps us ensure the upper and lower sums of g ∘ f are close to each other, satisfying the Riemann integrability criteria.