Question

(a) Give an example where \(\int_{a}^{b} f(x) d g(x)\) exists even though \(f\) is unbounded on \([a, b]\). (Thus, the analog of Theorem 3.1 .2 does not hold for the RiemannStieltjes integral.) (b) State and prove an analog of Theorem 3.1 .2 for the case where \(g\) is increasing.

Short answer

#tag_title#Short Answer #tag_content# In this exercise, we found an example where an unbounded function, \(f(x) = \begin{cases} n^2 & \text{if } x = \frac{1}{n} \text{ with } n\in \mathbb{N} \\ 0 & \text{otherwise} \end{cases}\), has a finite Riemann-Stieltjes integral with respect to the function \(g(x) = x\) on the interval \([0,1]\). We have shown that the integral exists and is equal to 1. We also stated and proved an analog of Theorem 3.1.2 for the case where g(x) is increasing. The theorem states that if a function \(f\) is Riemann integrable on \([a, b]\), and \(g\) is increasing on \([a, b]\), then the Riemann-Stieltjes integral of \(f\) with respect to \(g\) exists on \([a, b]\).

Step-by-step solution

Example of an unbounded function with finite Riemann-Stieltjes integral

Consider the function \(f(x) = \begin{cases} n^2 & \text{if } x = \frac{1}{n} \text{ with } n\in \mathbb{N} \\ 0 & \text{otherwise} \end{cases}\) and the function \(g(x) = x\). Integrate \(f(x)\) with respect to \(g(x)\) on the interval \([0,1]\). Notice that \(f(x)\) is unbounded on the interval \([0, 1]\), as the value of \(f(x)\) increases without bound when \(x\) approaches 0. Now, we will compute the Riemann-Stieltjes integral: \(\int_{0}^{1} f(x) dg(x) = \lim_{n\to\infty} \sum_{k=1}^{n} f\left(\frac{1}{k}\right) \left(g\left(\frac{1}{k}\right) - g\left(\frac{1}{k+1}\right)\right)\). Since \(g(x) = x\), we can substitute that into the expression: \(\int_{0}^{1} f(x) dg(x) = \lim_{n\to\infty} \sum_{k=1}^{n} f\left(\frac{1}{k}\right) \left(\frac{1}{k} - \frac{1}{k+1}\right) = \lim_{n\to\infty} \sum_{k=1}^{n} k^2 \left(\frac{1}{k^2} - \frac{1}{k^2+k}\right)\). Now, we can simplify the expression: \(\int_{0}^{1} f(x) dg(x) = \lim_{n\to\infty} \sum_{k=1}^{n} \left(1 - \frac{1}{k+1}\right) = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{k}{k+1}\). This is a convergent series, and the Riemann-Stieltjes integral exists, equal to \(\lim_{n\to\infty} \sum_{k=1}^{n} \frac{k}{k+1} = 1\). For part (b):

Analog of Theorem 3.1.2 for increasing g(x)

We will state and prove the following theorem, which is an analog of Theorem 3.1.2, considering g(x) is increasing on the interval [a, b]: Theorem: Let \(f\) and \(g\) be functions defined on the closed interval \([a, b]\). If \(f\) is Riemann integrable on \([a, b]\), and \(g\) is increasing on \([a, b]\), then \(f\) is Riemann-Stieltjes integrable with respect to \(g\) on \([a, b]\) Proof: Since \(f\) is Riemann integrable on \([a, b]\) and g(x) is increasing on \([a, b]\), for any partition \(P = \{x_0, x_1, \ldots, x_n\}\) of \([a, b]\) and any set of points \(T=\{t_1,t_2,\ldots,t_n\}\) in \([x_{k-1},x_k]\), with \(k =1,2,\ldots,n\), we can define \(\Delta g_k = g(x_k) - g(x_{k-1})\) as the difference in \(g\) value in each subinterval. We want to show that the Riemann-Stieltjes integral exists. In other words, we want to show that for every \(\epsilon > 0\), there exists a partition \(P\) such that \(U(P, f, g) - L(P, f, g) < \epsilon\), where \(U\) and \(L\) are the upper and lower sums, respectively. Since \(f\) is Riemann integrable, we know that for every \(\epsilon > 0\), there exists a partition \(P\) such that \(U(P, f) - L(P, f) < \frac{\epsilon}{g(b) - g(a)}\). Now, observe that: \(U(P, f, g) - L(P, f, g) = \sum_{k=1}^n(M_k - m_k)\Delta g_k \leq \sum_{k=1}^n(M_k - m_k)(g(b) - g(a))\) \(\Rightarrow U(P, f, g) - L(P, f, g) \leq (U(P, f) - L(P, f))(g(b) - g(a)) < \epsilon\). So, we have shown that \(\forall \epsilon>0\), \(\exists P\) such that \( U(P, f, g) - L(P, f, g) < \epsilon \) Hence, the Riemann-Stieltjes integral of \(f\) with respect to the increasing function \(g\) exists on \([a, b]\).

Key concepts

Riemann Integrable

The concept of a function being Riemann integrable is central to understanding the Riemann-Stieltjes integral. A function is considered Riemann integrable over an interval if the area under the curve of the function within that interval can be accurately approximated by sums of rectangles. More formally, a function is Riemann integrable if the upper and lower sums of the function approach the same limit as the partition of the interval becomes finer (the width of the rectangles decreases).

This limit is what we call the Riemann integral of the function over that interval. Importantly, for a function to be Riemann integrable, it must be bounded and have no discontinuities that result in an infinite number of rectangles of non-zero width. However, it's possible for a bounded function to have point discontinuities and still be Riemann integrable.

When discussing the Riemann-Stieltjes integral, if the function to be integrated, denoted as 'f', is Riemann integrable and the second function 'g' acts as an integrator and is increasing on the given interval, then 'f' is also Riemann-Stieltjes integrable with respect to 'g'. This relationship extends the applicability of Riemann integrals and meshes well with cumulative distribution functions in probability, as will be seen in examples and proofs related to increasing functions.

Unbounded Function

An unbounded function is one that does not have an upper or lower bound on its values. In other words, the function can take on infinitely large positive or negative values within its domain. For the Riemann-Stieltjes integral, the traditional requirement for the function to be bounded, as it is for Riemann integration, is relaxed in certain cases.

One might think that if a function, such as 'f', becomes unbounded on an interval, it automatically becomes non-integrable. However, this isn't necessarily the case for the Riemann-Stieltjes integral. An illuminating instance is when 'f' is unbounded on an interval '[a, b]', yet the integral of 'f' with respect to a well-behaved function 'g' is finite. The behavior of 'f' at a countable number of points doesn't spoil the integral provided 'g' varies little at those points (

Example in Practice

If 'g' is the identity function, 'f(x)=n^2' for 'x=1/n' and '0' otherwise, the integral can still exist despite 'f' being unbounded near zero).

In conclusion, the existence of a Riemann-Stieltjes integral does not automatically disqualify an unbounded function. The critical factor lies in how the increasing function 'g' behaves and how significantly that behavior affects the unboundedness of 'f'.

Increasing Function

An increasing function is one that does not decrease as its input increases; in precise terms, if 'x1' < 'x2', then 'g(x1)' ≤ 'g(x2)' for all 'x1' and 'x2' in the domain. In the context of Riemann-Stieltjes integrals, when 'g', the integrator function, is increasing, the integration process mirrors the Riemann integral more closely.

This property of being increasing is critical when considering the Riemann-Stieltjes integral. If 'g' is increasing and 'f' is Riemann integrable, 'f' will also be Riemann-Stieltjes integrable with respect to 'g'. The theorem for increasing 'g' mentioned in the exercise solution formalizes this relationship and provides a way to assert the integrability of 'f' within the framework of the Riemann-Stieltjes integral (

Theorem in Action

This theorem extends our understanding of integrals by relating the Riemann integral of 'f' and the Riemann-Stieltjes integral where 'g' is an increasing function, ensuring a more generalized approach to integration).

Understanding the importance of 'g' being an increasing function opens up versatile applications for the Riemann-Stieltjes integral, notably in fields such as economics and probability theory, where integrals with respect to non-constant functions are quite common.