Question

Let \(A\) be the set of points of the form \(\left(2^{-m} p, 2^{-m} q\right),\) where \(p\) and \(q\) are odd integers and \(m\) is a nonnegative integer. Let $$ f(x, y)=\left\\{\begin{array}{ll} 1, & (x, y) \notin A \\ 0, & (x, y) \in A \end{array}\right. $$ Show that \(f\) is not integrable on any rectangle \(R=[a, b] \times[c, d],\) but $$ \int_{a}^{b} d x \int_{c}^{d} f(x, y) d y=\int_{c}^{d} d y \int_{a}^{b} f(x, y) d x=(b-a)(d-c) . $$

Short answer

Based on the given solution, here is a short answer question you can use for your students: Question: Consider the function f(x, y), defined as follows: f(x, y) = 0 if $(x,y) \in A$, and f(x, y) = 1 if $(x,y) \notin A$, where A is the set of all points with coordinates $(2^{-m}p, 2^{-m}q)$, m $\geq$ 0, p and q are odd integers. Explain why f(x, y) is not integrable on any rectangle R = [a, b] × [c, d], and show that the iterated integrals evaluate to (b-a)(d-c). Answer: The function f(x, y) is not integrable on any rectangle R = [a, b] × [c, d], because the points in A are dense on the xy-plane, causing the value of f to oscillate between 0 and 1 without any continuous partitioning. By evaluating the iterated integrals, we find that they are equal to the intervals' lengths, (b-a) and (d-c), which results in the product (b-a)(d-c) when considering both nested integrals.

Step-by-step solution

Rewrite f(x, y) using indicator functions

To make it easier to work with the function, we can rewrite the definition of f(x, y) using an indicator function. The indicator function is defined as: $$ I_A (x) = \left\{ \begin{array}{ll} 1, & x \notin A \\ 0, & x \in A \end{array} \right. $$ So, we can rewrite f(x, y) as: $$ f(x, y) = I_A (x, y) $$ This will make it easier for us to manipulate the function in the next steps.

Determine the density of points in A

The set A is formed by points (2^{-m}p, 2^{-m}q), where p and q are odd integers and m is a nonnegative integer. We see that, for any two distinct points P and Q in the xy-plane, there will always be a point R in A that lies between P and Q. This is true because we can always find an integer m large enough such that decreasing the coordinates of P or Q by a power of 2 smaller than 2^{-m} would still result in a point between P and Q, and we can choose odd integers p and q to match those coordinates. This means the points in A are dense on the xy-plane.

Show that f is not integrable on any rectangle R

Because the points of A are dense on the xy-plane and f(x, y) takes the value 0 on points from A and 1 on all other points, it's clear that the value of f will oscillate between 0 and 1 and there will be no continuous partitioning which can properly approximate the value of an integral. Therefore, f(x, y) is not integrable on any rectangle R = [a, b] × [c, d].

Evaluate the iterated integrals

Now we need to show that the iterated integrals are equal to (b-a)(d-c). We can use the indicator function and focus on the intervals [a, b] and [c, d]. Let's first evaluate the nested integral with respect to x: $$ \int_{a}^{b} f(x, y) dx = \int_{a}^{b} I_A(x, y) dx $$ As f(x, y) = 1 for all (x, y) not in A, the value of this integral is equal to (b-a). Now, we can find the second nested integral with respect to y: $$ \int_{c}^{d} f(x, y) dy = \int_{c}^{d} I_A(x, y) dy $$ Similarly, as f(x, y) = 1 for all (x, y) not in A, the value of this integral is equal to (d-c).

Show the iterated integrals equality

Since both nested integrals evaluate to their corresponding intervals' lengths, we can now assert that the iterated integrals are equal: $$ \int_{a}^{b} dx \int_{c}^{d} f(x, y) dy = \int_{c}^{d} dy \int_{a}^{b} f(x, y) dx = (b-a)(d-c) $$ We have successfully demonstrated that the function f(x, y) is not integrable on any rectangle R = [a, b] × [c, d] and that the iterated integrals are equal to (b-a)(d-c).

Key concepts

Oscillating Functions

Oscillating functions are unique in the way they behave. These functions fluctuate between values, like 0 and 1, as seen in the function \( f(x, y) \) from the exercise. This function is defined as 0 for points in set \( A \) and 1 elsewhere. An oscillating function rapidly switches between its potential values, which makes calculating a definite integral challenging on any set domain.
Regarding \( f(x, y) \), since the points of \( A \) are dense (meaning they fill the entire space densely), the function cannot maintain a consistent value over any rectangle. Such oscillations make it impossible to find a uniform partition to approximate an integral effectively. The function simply cannot settle on a single value over any tiny patch of the rectangle. Thus, \( f(x, y) \) is oscillatory and non-integrable over rectangles.

Dense Sets

Dense sets are an essential concept to understand in this context. A set is considered dense if every point in the space can be approximated arbitrarily closely by elements of the set. In our case with \( A \), any point \((x, y)\) on the plane finds a member \((2^{-m}p, 2^{-m}q)\) from \( A \) no matter how close or small a distance between any two points. The terms \( p \) and \( q \) being odd integers, combined with varying \( m \), ensure this notion.
The denseness of \( A \) means there is minimum space between any point in \( A \) and any given coordinate. This capability of \( A \) to fill the plane so densely indicates why \( f(x, y) \), as a function relying on set \( A \), results in its non-integrability. The dense nature causes \( f(x, y) \) to fluctuate endlessly between 0 and 1.

Iterated Integrals

Iterated integrals involve integrating a function in multiple steps, usually one variable at a time. In this exercise, although \( f(x, y) \) is not integrable as one might hope, its iterated integrals still produce a particular result. The integral first considers one variable (say \( x \)), treating the other \( (y) \) as constant, and then the reverse. Here, we see the elegance of iterated integration: despite the oscillations and dense sets causing trouble initially, you end up with simple products of lengths of intervals. For instance:

• First consider \( \int_{a}^{b} d x \int_{c}^{d} f(x, y) d y \), focus on \( x \) integration first, then \( y \).
• Then \( \int_{c}^{d} d y \int_{a}^{b} f(x, y) d x \), where \( y \) goes first.

Each approach yields \((b-a)(d-c)\), reiterating the intended effect of these integrals: covering the entire area of the rectangle. Iterated integrals overcome the limits of the function's irregular behavior, underlining a powerful application of calculus.