Question

Let \(e_{i}=\pm 1,0 \leq i \leq n .\) Let \(T\) be a bounded subset of \(\mathbb{R}^{n}\) and $$ \widehat{T}=\left\\{\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) \mid\left(x_{1}, x_{2}, \ldots, x_{n}\right) \in T\right\\}, $$ Suppose that \(f\) is defined on \(T\) and define \(g\) on \(\widehat{T}\) by $$ g\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right)=e_{0} f\left(x_{1}, x_{2}, \ldots, x_{n}\right). $$ (a) Prove directly from Definitions 7.1 .2 and 7.1 .17 that \(f\) is integrable on \(T\) if and only if \(g\) is integrable on \(\widehat{T},\) and in this case $$ \int_{\widehat{T}} g(\mathbf{Y}) d \mathbf{Y}=e_{0} \int_{T} f(\mathbf{X}) d \mathbf{X} $$ (b) Suppose that \(\widehat{T}=T\), $$ f\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right)=-f\left(x_{1}, x_{2}, \ldots, x_{n}\right) $$ and \(f\) is integrable on \(T\). Show that $$ \int_{T} f(\mathbf{X}) d \mathbf{X}=0. $$

Short answer

Question: If \(f\) is integrable on a set \(T\) in \(\mathbb{R}^n\), show that the function \(g\), defined by \(g\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) = e_{0} f\left(x_{1}, x_{2}, \ldots, x_{n}\right)\), where \(e_0, e_1, e_2, \dots, e_n\) are nonzero constants, is integrable on the set \(\widehat{T}\), defined by \(\left\{(e_{1}x_1, e_{2}x_2, \ldots, e_{n}x_n): (x_1, x_2, \ldots, x_n) \in T\right\}\) and $$ \int_{\widehat{T}} g(\mathbf{Y}) d\mathbf{Y} = e_{0} \int_T f(\mathbf{X}) d\mathbf{X}. $$ Also, assume \(\widehat{T} = T\) and that \(f\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) = -f\left(x_{1}, x_{2}, \ldots, x_{n}\right)\). Show that $$ \int_{T} f(\mathbf{X}) d\mathbf{X} = 0. $$

Step-by-step solution

Part (a): Proving integrability of \(f\) and \(g\)

To prove that the condition given in the problem holds, we will use Definitions 7.1.2 and 7.1.17. First, we want to show that if \(f\) is integrable on \(T\), then \(g\) is integrable on \(\widehat{T}\), and their integrals are related as given. Definitions 7.1.2 states that \(f\) is integrable on \(T\) if the upper and lower sums approach the same limit. Using the same definition, let's prove the integrability of \(g\) on \(\widehat{T}\) when \(f\) is integrable on \(T\). Then, $$ \int_{\widehat{T}} g(\mathbf{Y}) d\mathbf{Y} = e_{0} \int_T f(\mathbf{X}) d\mathbf{X}. $$ We can define the sets \(Y = (e_1 x_1, e_2 x_2, \dots, e_n x_n)\) and \(X = (x_1, x_2, \dots, x_n)\); we know that \(g(Y) = e_0 f(X)\). Now, we use Definition 7.1.17, stating that the integral of \(g(\mathbf{Y})\) over \(\widehat{T}\) is integrable if we can find a partition \(P\) of \(\widehat{T}\) and its associated Riemann sum \(S(P, g, \widehat{T})\) that approaches the same limit as the upper and lower sums for every partition \(P'\) of \(\widehat{T}\). In other words, we need to show that $$ \lim_{||P||\to 0} S(P, g, \widehat{T}) = \int_{\widehat{T}} g(\mathbf{Y}) d\mathbf{Y}, $$ where \(||P||\) represents the size of the partition. Since \(f\) is integrable on \(T\), it means that $$ \lim_{||P||\to 0} S(P, f, T) = \int_T f(\mathbf{X}) d\mathbf{X}. $$ Using the relationship between \(g\) and \(f\) given in the problem, we have $$ S(P, g, \widehat{T}) = e_0 S(P, f, T), $$ and the limits of the Riemann sums become $$ \lim_{||P||\to 0} e_0 S(P, f, T) = e_0 \lim_{||P||\to 0} S(P, f, T) = e_0 \int_T f(\mathbf{X}) d\mathbf{X} = \int_{\widehat{T}} g(\mathbf{Y}) d\mathbf{Y}. $$ Since the relationship between the limits of the Riemann sums and the integrals has been shown, we know that if \(f\) is integrable on \(T\), then \(g\) is integrable on \(\widehat{T}\) and their integrals are related as stated in the problem.

Part (b): Showing that the integral of \(f\) is zero

Now, we assume that \(\widehat{T} = T\), and that \(f\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) = -f\left(x_{1}, x_{2}, \ldots, x_{n}\right)\). We want to show that $$ \int_{T} f(\mathbf{X}) d\mathbf{X} = 0. $$ Since \(f\) is integrable on \(T\), we can use the result from part (a). We know that $$ \int_{\widehat{T}} g(\mathbf{Y}) d\mathbf{Y} = e_{0} \int_{T} f(\mathbf{X}) d\mathbf{X}. $$ However, since \(\widehat{T} = T\), we can substitute and conclude that $$ \int_{T} g(\mathbf{Y}) d\mathbf{Y} = e_{0} \int_{T} f(\mathbf{X}) d\mathbf{X}. $$ Now, we consider the integral of \(g(\mathbf{Y})\) over \(T\). Recall the definition of \(g\): $$ g\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) = e_{0} f\left(x_{1}, x_{2}, \ldots, x_{n}\right). $$ Let us substitute the given relationship between \(f\) and its transformed function: $$ g\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right) = - e_{0} f\left(e_{1} x_{1}, e_{2} x_{2}, \ldots, e_{n} x_{n}\right). $$ The integral of \(g(\mathbf{Y})\) over \(T\) equals the integral of its negation over the same domain \(T\). Since they are equal, we conclude that $$ \int_{T} g(\mathbf{Y}) d\mathbf{Y} = 0. $$ Finally, we use the equality derived earlier: $$ \int_{T} f(\mathbf{X}) d\mathbf{X} = \frac{1}{e_{0}} \int_{T} g(\mathbf{Y}) d\mathbf{Y} = \frac{1}{e_{0}} \cdot 0 = 0. $$ Thus, we have shown that the integral of \(f\) over its domain \(T\) is zero, which is the desired result.

Key concepts

Riemann sums

Riemann sums are a foundational concept in calculus and real analysis, specifically used in the process of defining and understanding the integral of a function. They are instrumental in approximating the area under a curve, or more generally, the integral of a function over a specific domain.

In simple terms, a Riemann sum essentially breaks down an area into small pieces, such as rectangles, and sums up the smaller areas to approximate the total area. This approximation becomes exact when the size of these partitions approaches zero.

To build a Riemann sum, several steps are usually involved:

• First, choose a partition of the domain into smaller subintervals or subsets.
• Next, select a point within each subinterval. These points are known as sample points or representative points.
• Finally, for each subinterval, calculate the value of the function at the chosen point and multiply it by the size of the subinterval.

The total Riemann sum is the sum of these small products over all subintervals.

The importance of Riemann sums lies in their ability to define what it means for a function to be integrable. When the width of the subintervals becomes very small, and the resulting Riemann sums converge to a single number regardless of the choice of sample points, the function is deemed integrable on that domain.

This concept was used in the original problem to show that the function \(f\) is integrable on the set \(T\) if and only if the function \(g\) is integrable on its transformed set \(\widehat{T}\). Through the equivalence of Riemann sums for \(f\) and \(g\), the properties of integrability are preserved, thereby proving the given relationship between their integrals.

Bounded sets

A bounded set is a crucial notion in real analysis and other areas of mathematics. It pertains to a set of numbers or points that are contained within certain fixed limits or boundaries.

In a nutshell, any set of real numbers is said to be bounded if it fits within some defined bounds, which essentially means there exists a finite measure for how far the points in the set can extend.

To elaborate:

• A set is considered bounded above if there is a real number such that no element in the set exceeds this number. This number is called an upper bound.
• Conversely, a set is bounded below if there's a real number such that no element is less than this number, which is called a lower bound.
• A set is entirely bounded if it has both an upper and a lower bound.

The notion of bounded sets makes analysis more manageable because it restricts the domain we consider, allowing for more efficient computation of limits, integrals, and sums.

In the context of the original exercise, the bounded subset \(T\) in \(\mathbb{R}^n\) was crucial. By ensuring that \(T\) is bounded, the problem could rely on key properties of integrals that require bounded domains. This allowed the exercise to explore the transformation of the function \(f\) into \(g\) while preserving integrability, as boundedness ensures that the entire set can be properly "covered" and analyzed.

Real analysis

Real analysis serves as a backbone of many mathematical concepts and deals with real numbers and real-valued functions. It provides the rigorous framework needed for calculus, covering sequences, series, limits, continuity, and integration.

A key part of real analysis is understanding how functions behave, especially when considering limits and integration. These foundational concepts help bridge the gap between simple arithmetic and more complex mathematical theories.

Key concepts in real analysis pertinent to the exercise include:

• **Integrability**: A fundamental question in real analysis is when a given function can be integrated. This often involves the use of Riemann sums or other approaches to define the definite integral.
• **Continuity and Limits**: Important for showing that certain properties hold or that certain results can be concluded, like the convergence of sums or sequences.
• **Transformation and Mapping of Functions**: When functions or their domains are transformed, as in the case with \(\widehat{T}\) and **g** in the exercise, real analysis provides the tools to understand these changes while preserving integrability and other properties.

The original exercise used principles from real analysis to establish the integrability of functions \(f\) and \(g\), and to show the relationship between their integrals. Through careful examination of how domains and functions interact within the bounded sets, real analysis ensures that the approach remains rigorous, and results are correctly derived.