Problem 39 - Calculus Early Transcendentals: Pearson New International Edition

Question

Prove that the function \(f\) defined by $$f(x)=\left\\{\begin{array}{ll} 1 & \text { if } x \text { is rational } \\ 0 & \text { if } x \text { is irrational } \end{array}\right.$$ is not integrable on \([0,1] .\) Hint \(:\) Show that no matter how small the norm of the partition, \(\|P\|\), the Riemann sum can be made to have value either 0 or 1

Step-by-Step Solution

Answer

The function is not Riemann integrable on \([0,1]\) because the upper and lower sums are always 1 and 0, respectively, for any partition.

Step 1: Understand the Function Definition

The function \( f(x) \) is defined on the interval \([0, 1]\). It assigns a value of 1 if \( x \) is rational and a value of 0 if \( x \) is irrational. Rational numbers are plentiful, as are irrational ones, in any interval, including \([0, 1]\).

Step 2: Define a Partition of the Interval

Consider a partition \( P = \{ x_0, x_1, x_2, \, \ldots, x_n \} \) of the interval \([0,1]\), where \( x_0 = 0 \) and \( x_n = 1 \). The norm of the partition \( \|P\| \) is the length of the largest subinterval \( [x_{i-1}, x_i] \).

Step 3: Consider Upper and Lower Sums

The upper sum \( U(f, P) \) is obtained by using the supremum of \( f(x) \) on each subinterval, while the lower sum \( L(f, P) \) uses the infimum of \( f(x) \). Since \( f(x) \) equals 1 on rational points and 0 on irrational points, on any subinterval, the supremum is 1 and the infimum is 0.

Step 4: Calculate the Upper and Lower Sums

For any partition \( P \), \( U(f, P) = \sum_{i=1}^{n} M_i \Delta x_i = \sum_{i=1}^{n} 1 \cdot (x_i - x_{i-1}) = 1 \). Similarly, \( L(f, P) = \sum_{i=1}^{n} m_i \Delta x_i = \sum_{i=1}^{n} 0 \cdot (x_i - x_{i-1}) = 0 \).

Step 5: Conclude Non-Integrability

Since \( U(f, P) = 1 \) and \( L(f, P) = 0 \) for any partition, their difference is 1, which does not tend to zero as \( \|P\| \to 0 \). Therefore, \( f \) is not Riemann integrable on \([0,1]\).

Key Concepts

Partition of Interval

A partition of an interval refers to dividing a given interval into smaller subintervals. Consider the interval [0, 1]. A partition, denoted as \( P \), can be represented as a set of points \( \{ x_0, x_1, x_2, \ldots, x_n \} \), where \( x_0 = 0 \) and \( x_n = 1 \). Each pair of consecutive points \( x_{i-1} \) and \( x_{i} \) create a subinterval.

The size or the norm of a partition, \( \|P\| \), is determined by the largest length among these subintervals. Mathematically, \( \|P\| = \max(x_i - x_{i-1}) \). Understanding partitions is crucial as they form the foundation for calculating sums that lead to the understanding of integrals.

• Key Elements: Set of division points
• Norm: Length of the largest subinterval

These concepts are used to explore properties such as the behavior of sums over an interval.

Upper Sum

In the context of Riemann sums and integrability, the upper sum represents an approach using the supremum (or least upper bound) of the function in each subinterval. For a function \( f(x) \), the upper sum with respect to a partition \( P \) is calculated by taking the highest value \( M_i \) that the function attains in the subinterval \([x_{i-1}, x_i]\).

Thus, the upper sum \( U(f, P) \) is given by:\[ U(f, P) = \sum_{i=1}^{n} M_i (x_i - x_{i-1}) \]
In our exercise, the function \(f(x)\) takes the value 1 on rational points. Therefore, the supremum \( M_i \) is always 1 within each subinterval. This results in the upper sum being equal to 1, regardless of the partition or its norm.

• Usage: Calculating the supremum of \( f(x) \) in each subinterval
• Outcome: Reflects upper bound contributions

Lower Sum

The lower sum acts as a counterpart to the upper sum by utilizing the infimum (greatest lower bound) of the function within each subinterval of a partition. For function \( f(x) \), the lower sum \( L(f, P) \) is similarly computed by considering the smallest value \( m_i \) that \( f(x) \) achieves on the subinterval \([x_{i-1}, x_i]\).

The expression for the lower sum is given as:\[ L(f, P) = \sum_{i=1}^{n} m_i (x_i - x_{i-1})\]
In the case of the exercise's function, \( f(x) \) always takes a minimum value of 0 on irrational numbers. Thus, the infimum \( m_i \) is 0 for each subinterval, leading to a total lower sum of 0.

• Purpose: Evaluating the minimum contribution of the function
• Result: Serves as the lower bound

Supremum and Infimum

Supremum and infimum are fundamental concepts in the analysis of real numbers, particularly in determining upper and lower bounds within intervals.

Supremum is the least upper bound of a set. It’s the smallest number that is greater than or equal to every number in the set. In terms of functions, on any given subinterval, the supremum is the largest possible value that a function can take.
An example from the exercise: for the function \( f(x) \) on any subinterval, since it reaches 1 (at least on the rational numbers), the supremum is 1.

Infimum, conversely, is the greatest lower bound, which is the largest number that is less than or equal to every number in the set. For functions, it represents the smallest value obtainable within the subinterval.
In the same exercise, since the function \( f(x) \) reaches 0 (at least on the irrational numbers), the infimum is 0.

• Supremum: Determines the largest possible function values
• Infimum: Assesses the smallest possible values