Question

Show that \(\int_{a}^{b} f(x) d g(x)\) exists if \(f\) is of bounded variation and \(g\) is continuous on \([a, b] .\)

Short answer

Based on the given problem, provide a brief explanation of why the integral \(\int_a^b f(x) dg(x)\) exists as long as \(f\) is of bounded variation and \(g\) is continuous on the interval \([a, b]\). In order to show that the integral exists, we first consider the properties of functions of bounded variation and continuous functions. A function of bounded variation guarantees that there exists a constant \(M\) such that the sum of the absolute differences between consecutive function values is bounded. This in turn implies that the function is bounded on the given interval. Since \(g\) is continuous on \([a, b]\), there exists a \(\delta\) such that when the partition size \(\Delta x\) is smaller than \(\delta\), the differences between consecutive values of \(g\) are also small. By considering the sum of the product of these differences and the function values of \(f\), we can show that this sum is bounded and approaches a finite limit as \(\Delta x\) tends to 0. Consequently, the integral \(\int_a^b f(x) dg(x)\) exists as long as \(f\) is of bounded variation and \(g\) is continuous on the interval \([a, b]\).

Step-by-step solution

Recall Definitions

We need to recall the definitions of a function of bounded variation and an existing integral. A function \(f\) is of bounded variation on \([a, b]\) if there exists a constant \(M\) such that for any finite sequence \(a = x_0 < x_1 < ... < x_n = b\), we have $$\sum_{i=1}^n \left| f(x_i) - f(x_{i-1})\right| \le M.$$ For the integral \(\int_a^b f(x) dg(x)\) to exist, it means that the limit $$\lim_{\Delta x\to 0} \sum_{i=1}^n f(x_i) \Delta g(x_i)$$ exists and is finite, where \(x_i = a + i\Delta x\), \(\Delta g(x_i) = g(x_i) - g(x_{i-1})\), and \(\Delta x = \frac{b-a}{n}\).

Use the Definition of Bounded Variation

Since \(f\) is a function of bounded variation, we know that there exists a constant \(M\) such that for any finite sequence \(a = x_0 < x_1 < ... < x_n = b\), we have $$\sum_{i=1}^n \left| f(x_i) - f(x_{i-1})\right| \le M.$$ This implies that \(f\) is also bounded on \([a, b]\), as for any \(x\in [a, b]\), there exists \(a\le x_{i-1}\le x\le x_i\le b\) such that $$\left| f(x) - f(x_{i-1})\right| \le |f(x)|+|f(x_{i-1})|\le M.$$

Use the Property of Continuous Functions

The given function \(g\) is continuous on \([a, b]\). This means that for any \(\epsilon > 0\), there exists a \(\delta > 0\) such that if \(|x-y| < \delta\), then \(|g(x)-g(y)| < \epsilon\). Now, set the partition size \(\Delta x\) to satisfy \(\Delta x < \delta\).

Show that the Integral Exists

Now, we want to show that the limit $$\lim_{\Delta x\to 0} \sum_{i=1}^n f(x_i) \Delta g(x_i)$$ exists and is finite. Using the previously mentioned properties of \(f\) and \(g\), we write: $$\left|\sum_{i=1}^n f(x_i) \Delta g(x_i)\right| \le \sum_{i=1}^n |f(x_i)| |\Delta g(x_i)| \le \sum_{i=1}^n M|\Delta g(x_i)|,$$ where the last inequality follows from \(|f(x_i)|\le M\) as shown in Step 2. Now, $$\sum_{i=1}^n |\Delta g(x_i)| \le \epsilon \sum_{i=1}^n 1 = n\epsilon \le \frac{(b-a)\epsilon}{\delta} \Longrightarrow \sum_{i=1}^n M|\Delta g(x_i)|\le \frac{M(b-a)\epsilon}{\delta}.$$ As \(\epsilon\) can be made arbitrarily small by choosing smaller \(\delta\) and \(\Delta x<\delta\), we have: $$\lim_{\Delta x\to 0} \sum_{i=1}^n f(x_i) \Delta g(x_i) \le \frac{M(b-a)\epsilon}{\delta}.$$ Since \(\epsilon\) can be any positive number, this limit is finite, and thus the integral \(\int_a^b f(x) dg(x)\) exists.

Key concepts

Real Analysis Fundamentals

Real Analysis is a branch of mathematics that deals with real numbers and real-valued functions. In this context, an important concept is that of bounded variation. A function f is said to be of bounded variation on a closed interval [a, b] if there is a ceiling to the total 'variability' of f within those bounds - meaning, no matter how you choose points between a and b, the sum of the absolute differences of f at these points is always less than some constant M. This property of bounded variation plays a pivotal role in the study of the Riemann-Stieltjes integral. It ensures that the function doesn't oscillate wildly, which in turn guarantees that certain types of integrals will have well-defined, finite values. Consequently, in problems requiring the evaluation of an integral, establishing that a function is of bounded variation can be a critical step.

Continuous functions, another cornerstone of real analysis, simply mean that the function's graph has no breaks, jumps, or points of discontinuity. The value of the function at a point smoothly transitions to its value at nearby points. This behavior has profound implications for integration, as continuity often simplifies the calculation and interpretation of an integral's value.

Understanding the Riemann-Stieltjes Integral

The Riemann-Stieltjes integral is a generalization of the Riemann integral, and it's a tool that's particularly useful in situations where a function is being integrated with respect to another function, symbolized as t(x) dg(x). For such an integral to exist, the limit of a particular kind of sum must converge to a finite value as the intervals in the sum become infinitely small - this is the heart of Riemann-Stieltjes integration.

One might visualize this by imagining a series of rectangles, where their widths are determined by the changes in g(x) (g(x_i) - g(x_{i-1})) and their heights by the function f(x). The integral calculates the 'combined effect' of f over the changes in g. It's essential that f is of bounded variation and g is continuous. If f has bounded variation, we avoid unmanageable oscillations, and if g is continuous, we can ensure that the widths of our rectangles (the changes in g) behave nicely as they become smaller. The Riemann-Stieltjes integral thus crafted is a powerful construction that captures intricate interplays between functions and their magnitudes of change, allowing for the evaluation of more complex integrals.

The Role of Continuous Functions in Analysis

Continuous functions are the bread and butter of real analysis. Their properties facilitate not only the understanding of functions themselves but also the study of integrals, limits, and other analytic concepts. In the context of the Riemann-Stieltjes integral, the continuity of g(x), the function with respect to which we integrate, cannot be understated. The definition of a continuous function - one that has no sudden jumps or interruptions over its domain - means that small changes in the input (x) result in small and predictable changes in (g(x)).

For an integral like (t(x) dg(x)), the interplay between the bounded variation of f and the continuity of g becomes critical. It helps us guarantee that the sums involved in assessing the Riemann-Stieltjes integral will not spiral out of control as the partitioning becomes finer. Therefore, the notion that g is continuous over [a, b] is not just an abstract nicety - it's a fundamental condition that ensures the integral behaves well, converging to a real, finite number, a conclusion vital to many areas of analysis, from calculus to more advanced topics.