Question

If \(f(x)=0\) for all irrational \(x, f(x)=1\) for all rational \(x\), prove that \(f \notin \mathscr{R}\) on \([a, b]\) for any \(a

Short answer

Question: Prove that the function defined by f(x) = 0 for all irrational x and f(x) = 1 for all rational x is not Riemann integrable on any interval [a, b] where a < b.

Step-by-step solution

Define the function and intervals

Let the function \(f(x)\) be defined by \(f(x) = 0\) for all irrational \(x\), and \(f(x) = 1\) for all rational \(x\). Given an interval \([a, b]\) where \(a < b\), let us partition it into n equal subintervals.

Analyze the upper and lower sums

Since there are rational numbers and irrational numbers in every non-empty open interval of the real numbers, we can say that each subinterval of our partition contains both rational and irrational points. This means that in each subinterval, the maximum value of the function is 1 (when x is rational) and the minimum value of the function is 0 (when x is irrational).

Calculate the limit of the difference between upper and lower sums

Next, we want to calculate the upper Riemann sum (U) and lower Riemann sum (L) of the partition. The upper and lower sums for the function f are defined by the following: \begin{align*} U = \sum_{i=1}^{n}(M_i \Delta x_i) \\ L = \sum_{i=1}^{n}(m_i \Delta x_i) \end{align*} where \(M_i\) is the maximum value of f in the ith subinterval, and \(m_i\) is the minimum value of f in the ith subinterval. In this case, every \(M_i = 1\) and every \(m_i = 0\). This simplifies the upper and lower sums to: \begin{align*} U = \sum_{i=1}^{n}(1 \Delta x_i) \\ L = \sum_{i=1}^{n}(0 \Delta x_i) \end{align*} As we let the partition size go to infinity, the differences between the subintervals, \(\Delta x_i\), approach 0. Therefore, the lower sum will approach 0 (\(L \rightarrow 0\)). Since the upper sum consists of the sum of n equal partitions of width \(\Delta x_i\), it approaches the total width of the interval, which is \((b-a)\) (\(U \rightarrow b-a\)).

Determine Riemann integrability

A function is Riemann integrable if the following condition holds: $$\lim_{n\to\infty}(U-L) = 0$$ In this case, for any given interval, we have: $$ \lim_{n\to\infty}(U-L) = \lim_{n\to\infty}( (b-a) - 0) = b-a $$ And since \(a 0\). Therefore, the function \(f\) is not Riemann integrable, which we can represent as \(f\notin \mathscr{R}\) on any interval \([a, b]\) where \(a < b\).

Key concepts

Upper Riemann Sum

The upper Riemann sum is a method used to approximate the total area under a curve, or more formally, the integral of a function over a specific interval. In this context, it involves choosing subintervals within a partitioned interval. For each subinterval, we identify the maximum value of the function, denoted as \(M_i\), and that maximum is then multiplied by the width of the subinterval \(\Delta x_i\). These products are summed for all subintervals to get the upper Riemann sum, represented as

• \(U = \sum_{i=1}^{n}(M_i \Delta x_i)\).

In our example, since \(f(x) = 1\) for rational points and \(0\) for irrational points, each subinterval's maximum (\(M_i\)) over the interval is \(1\). As a result, the sum simply equals the total length of the interval \((b-a)\) as \(n\) approaches infinity.

Lower Riemann Sum

The lower Riemann sum offers another approximation by taking the lowest possible values of a function in each partitioned subinterval. The lower sum focuses on the minimum value of the function in each subinterval \(m_i\), and multiplies it by the subinterval width \(\Delta x_i\), resulting in the lower Riemann sum. This sum can be expressed as:

• \(L = \sum_{i=1}^{n}(m_i \Delta x_i)\).

In scenarios with a function like \(f(x)\), where \(f\) is zero for irrational values and one for rational values, the minimum value in any subinterval is \(0\), making the lower sum effectively zero across all subintervals as the partition becomes finer, hence \(L \to 0\). This stark contrast with the upper Sum indicates lack of integrability.

Partition of an Interval

The concept of a partition involves dividing a continuous interval into a series of smaller, non-overlapping subintervals. This is crucial for calculating Riemann sums, as it facilitates the approximation process. A partition usually comprises a finite sequence of subintervals which altogether span the original interval \([a, b]\). Over each of these partitions:

• Each subinterval width \(\Delta x_i\) gets smaller as the partition is refined.
• Both the upper and lower sums are calculated based on these subintervals.

By making these subintervals smaller and more plentiful (as \(n\) goes to infinity), the idea is to achieve a more accurate approximation of the area under the curve, or the integral. In the given function example, it is clear that this partition refinement doesn't bring the upper and lower sums to converge, highlighting the challenge in defining a single integral value.

Rational and Irrational Numbers

Understanding the difference between rational and irrational numbers is key in this exercise. Rational numbers are numbers which can be expressed as the quotient of two integers (e.g., fractions like \(\frac{1}{2}\), \(-4\)). Irrational numbers, by contrast, cannot be written as a simple fraction—examples include \(\sqrt{2}\), \(\pi\), and \(e\) (Euler's number).

• Rational numbers have terminating or repeating decimal expansions.
• Irrational numbers have non-terminating and non-repeating decimal expansions.

In the context of the function \(f(x)\) from the exercise, the distinguishing factor is important: the function is defined differently for rational values \(f(x) = 1\) and irrational values \(f(x) = 0\). Because every subinterval of \([a, b]\) contains both types of numbers, the integrability through Riemann sums is obstructed, as each subinterval never allows for approaching a single value.