Question

A nonintegrable function Consider the function defined on [0,1] such that \(f(x)=1\) if \(x\) is a rational number and \(f(x)=0\) if \(x\) is irrational. This function has an infinite number of discontinuities, and the integral \(\int_{0}^{1} f(x) d x\) does not exist. Show that the right, left, and midpoint Riemann sums on regular partitions with \(n\) sub-intervals equal 1 for all \(n .\) (Hint: Between any two real numbers lie a rational and an irrational number.)

Short answer

Question: Prove that the right, left, and midpoint Riemann sums of the function \(f(x) = \left\{ \begin{array}{ll} 1, & \mbox{if }x \mbox{ is rational} \\ 0, & \mbox{if } x \mbox{ is irrational} \end{array} \right. \) are equal to 1 for all \(n\), on regular partitions with \(n\) sub-intervals of the interval [0,1].

Step-by-step solution

Calculate the right Riemann sum

To calculate the right Riemann sum, we need to evaluate the function at the right endpoints of each subinterval, and then sum them up, multiplying by the sub-interval width \(\Delta x\). First, let's recall that between any two real numbers there are a rational number and an irrational number. Since the width of each partition is strictly greater than zero, there is always a rational and irrational point within each interval where the function evaluates to either 1(rational) or 0(irrational). For the right Riemann sum, the function f(x) is evaluated at the right endpoints of the sub-intervals. By the given hint, we can always choose a rational point for each right endpoint, so f evaluates to 1 on these points. Right Riemann sum: $$S_R = \Delta x \sum_{k=1}^{n} f(k\Delta x) = \Delta x \sum_{k=1}^{n} 1 = \Delta x \cdot n = 1 \cdot 1 = 1$$

Calculate the left Riemann sum

Similarly, to calculate the left Riemann sum, we need to evaluate the function at the left endpoints of each subinterval, and then sum them up, multiplying by the sub-interval width \(\Delta x\). We can again choose rational points for each left endpoint and f evaluates to 1 on these points. Left Riemann sum: $$S_L = \Delta x \sum_{k=0}^{n-1} f(k\Delta x) = \Delta x \sum_{k=0}^{n-1} 1 = \Delta x \cdot n = 1 \cdot 1 = 1$$

Calculate the midpoint Riemann sum

Lastly, we need to calculate the midpoint Riemann sum. For this, we evaluate the function at the midpoint of each sub-interval and sum them up, multiplying by the sub-interval width \(\Delta x\). Recalling the hint from before, we can choose all midpoints to be rational points so f evaluates to 1 at these points. Midpoint Riemann sum: $$S_M = \Delta x \sum_{k=1}^{n} f(\Delta x \cdot(k - 0.5)) = \Delta x \sum_{k=1}^{n} 1 = \Delta x \cdot n = 1 \cdot 1 = 1$$ In conclusion, we have shown that the right, left, and midpoint Riemann sums of the considered function are equal to 1 for all \(n\), on regular partitions with \(n\) sub-intervals of the interval [0,1].

Key concepts

Nonintegrable Function

A nonintegrable function is one for which the integral over a certain interval does not exist. This might seem strange, especially if you're used to simple functions like polynomials that integrate nicely. However, in certain cases, functions can behave too wildly or erratically for standard integration techniques, like the Riemann integral, to apply.
\[\int_{0}^{1} f(x) \, dx = \text{does not exist}\]
The exercise shows a classic example by defining a function with a significant number of discontinuities. This can be quite puzzling, but let's break it down. The function is defined on the interval \(0,1\) such that it assigns the value 1 to any rational input, and 0 to any irrational input. Since rational and irrational numbers are densely packed between any two real numbers, the function jumps between 1 and 0 frequently, making it discontinuous at every point.
In this context, the function behaves in a way that its integral cannot converge to a single value. Despite the fact that the Riemann sums give us estimates, they cannot calculate an integral where such wild behavior occurs consistently over \(0,1\).

Discontinuous Functions

Discontinuous functions are those that have breaks, jumps, or holes in their domain. These are functions that do not smoothly transition from one value to the next as their input changes slightly.
A great example of a discontinuous function is the one described in the exercise. It alternates between 1 and 0 depending on whether the input is rational or irrational. Since every rational number is near an irrational number (and vice versa), this causes infinite discontinuities in our function.
The implications are important:

• They might not be integrable using conventional methods like the Riemann integral.
• They are often considered pathological or extreme cases in mathematical analysis.

Such functions challenge basic assumptions of calculus, particularly those about limit behavior and the existence of integrals. These functions require more advanced methods, such as Lebesgue integration, to understand and evaluate.

Rational and Irrational Numbers

The distinction between rational and irrational numbers is crucial in understanding the function in the exercise. Rational numbers are numbers that can be expressed as a fraction of two integers (like \(\frac{3}{4}\) or \(5\)). Irrational numbers can't be expressed this way — numbers like \(\sqrt{2}\) or \(\pi\) are examples.
Here's why this matters: in the exercise's function, every rational input returns a 1, while every irrational input returns a 0. Given that between any two real numbers there exists both a rational and an irrational number, this function is littered with jumps and lacks continuity everywhere.
This juxtaposition of rational and irrational numbers means that:

• For any given interval, there is no segment of real numbers over which the function does not jump between 1 and 0.
• It highlights how densely packed numbers can be on the number line, contributing to the function’s non-integrability.

In essence, understanding the distributions of these numbers helps explain why the function doesn't stabilize at any numeric value across the interval, leading to the Riemann sums existing but the integral not. Thus, the interplay of rational and irrational numbers is integral to this function's chaotic behavior.