Problem 82 - Calculus Early Transcendentals: Pearson New International Edition

Question

Give an example to show that the accumulation function \(G(x)=\int_{a}^{x} f(x) d x\) can be continuous even if \(f\) is not continuous.

Step-by-Step Solution

Answer

Choose \( f(x) \) as the Dirichlet function; its integral \( G(x) = 0 \) is continuous.

Step 1: Choose a Discontinuous Function

Consider the function \( f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \ 0 & \text{if } x \text{ is irrational} \end{cases} \). This function, known as the Dirichlet function, is discontinuous everywhere on the real line.

Step 2: Define the Accumulation Function

Define the accumulation function as \( G(x) = \int_{a}^{x} f(t) \, dt \). For simplicity, let's choose \( a = 0 \) so that \( G(x) = \int_{0}^{x} f(t) \, dt \).

Step 3: Analyze the Integral of \( f(x) \)

The Dirichlet function \( f(x) \) has the property that at any interval, the measure of the rational numbers is 0. Therefore, no matter where \( x \) is, the integral \( \int_{a}^{x} f(t) \, dt \) is 0 because the integral over any interval of a set of measure 0 is 0.

Step 4: Verify Continuity of \( G(x) \)

Since the integral of \( f(x) \) is 0 for any \( x \), the function \( G(x) = 0 \) for all \( x \). A constant function like \( G(x) \) is continuous because it does not vary with \( x \). Thus, \( G(x) \) remains continuous even though \( f \) is discontinuous everywhere.

Key Concepts

Dirichlet Function

The Dirichlet function is an interesting example of a function that is discontinuous at every point on the real line. It is defined as follows:

• For any rational number \( x \), \( f(x) = 1 \).
• For any irrational number \( x \), \( f(x) = 0 \).

This means that no matter how you zoom in on the real number line, the function jumps between 1 and 0 depending on whether the point is rational or irrational. Because rationals and irrationals are densely interwoven in the real numbers, the Dirichlet function never "settles down" to a value as you approach any given point. It is a classic example used to illustrate a function that is discontinuous everywhere.

Discontinuous Function

A function is termed discontinuous at a point if there is a sudden jump in its values near that point. To determine discontinuity, consider the behavior of the function as it approaches a point from different directions. A function can be discontinuous in different ways:

• Jump Discontinuity: When \( f(x) \) jumps from one value to another as \( x \) approaches a point.
• Infinite Discontinuity: When \( f(x) \) becomes infinite at a certain point.
• Point Discontinuity: When \( f(x) \) has a gap.

The Dirichlet function is discontinuous at every point because it cannot align with a single value due to the alternating nature of rational and irrational numbers densely packed on the real line.

Continuity

Continuity in mathematical terms refers to the smooth, unbroken nature of a function. A function \( f(x) \) is continuous at a point \( c \) if the following conditions are satisfied:

• The function \( f(x) \) is defined at \( c \).
• The limit of \( f(x) \) as \( x \) approaches \( c \) from both sides exists.
• \( \ \lim_{x \to c} f(x) = f(c) \ \).

In the example provided with the Dirichlet function, the accumulation function \( G(x) = \int_{0}^{x} f(t) \, dt \) is continuous even though \( f \) is not. Remarkably, \( G(x) \) manages to be continuous because it results in a constant zero value for all \( x \), showing that integration can smooth out the discontinuities of \( f \).

Integral of Measure Zero

The concept of measure zero is crucial when dealing with integrals of functions like the Dirichlet function. In simple terms, a set is of measure zero if it can be "covered" by a collection of intervals whose total length is arbitrarily small. For example,

• The set of rational numbers has measure zero because, although there are infinitely many rationals, they can be covered by a sequence of intervals with diminishing lengths.

In the context of the Dirichlet function, this means that when taking the integral over any interval, the contribution from the rational numbers, which carry the value 1, becomes negligible because they include an infinitesimally small portion of the entire set. The integral over a set of measure zero, therefore, sums to zero, which explains why \( G(x) \), the accumulation function, remains zero and consequently continuous.