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Chapter 23: Problem 25

Find a sequence of integrable functions \(\left\\{f_{n}\right\\}\) which converges to the (nonintegrable) function \(f\) that is 1 on the rationals and 0 on the irrationals. Hint: Each \(f_{n}\) will be 0 except at a few points.

Short Answer

Expert verified
The sequence \( f_{n} (x) \) where \( f_{n}(x) = 1 \) for the first \( n \) rational numbers and \( f_{n}(x) = 0 \) otherwise converges to the target function and is integrable.

Step by step solution

01

- Define the Target Function

Understand the target function: It is given that the function \( f \) is 1 on the rationals and 0 on the irrationals.
02

- Properties of Rational and Irrational Numbers

Recall that the rationals are dense in the real numbers but the set of irrationals is uncountable. This means any interval of real numbers contains both rationals and irrationals.
03

- Constructing the Sequence

Construct the sequence of functions \( \left\{ f_{n} \right\} \) where each \( f_{n} \) has finite support on rational numbers. For each positive integer \( n \), let \( f_{n}(x) = 1 \) if \( x \) is one of the first \( n \) rational numbers (in a some enumeration). Set \( f_{n}(x) = 0 \) otherwise.
04

- Checking Integrability

Verify that each \( f_n \) is integrable. Since \( f_n \) is non-zero at only a finite number of points, their integrals with respect to the Lebesgue measure are zero.
05

- Demonstrate Convergence

Show that \( f_n \) converges pointwise to \( f \). For every real number \( x \), if \( x \) is rational, there will eventually be an \( n \) such that \( f_n(x) = 1 \). For irrationals, \( f_n(x) = 0 \) for all \( n \), thereby matching the target function \( f \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rational Numbers
Rational numbers are numbers that can be expressed as the quotient of two integers, meaning they can be written in the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers, and \( q eq 0 \). Examples of rational numbers include 1/2, -4, and 3. They are interesting because they can have either terminating or repeating decimals.
Rational numbers are dense in the real numbers. This means every interval of real numbers contains an infinite number of rational numbers.
This density property is crucial in constructing sequences of functions that approximate integrable functions.
Irrational Numbers
Irrational numbers cannot be expressed as a simple fraction \( \frac{p}{q} \). They include numbers such as \( \sqrt{2} \), \( \pi \), and \(\text{e}\).
Unlike rational numbers, the decimal expansion of irrational numbers neither terminates nor repeats. This is part of what makes them unique and sometimes challenging to work with in analysis.
The set of irrational numbers is uncountable; that is, there are 'more' irrationals than rationals. This uncountability is essential for understanding how \( f_n \) functions converge to a given function.
Lebesgue Measure
The Lebesgue measure is a way of assigning a length, area, or volume to subsets of a given space, such as the real numbers. Unlike other measures, it can assign a 'measure zero' to sets with many points, like the rational numbers.
In this exercise, the Lebesgue measure helps us understand that although the rational numbers are dense, they have a measure of zero. This is significant for proving that the sequence of functions \( f_n \) is integrable.
Pointwise Convergence
Pointwise convergence concerns the behavior of a sequence of functions at individual points. Here, \( f_n \) converges pointwise to a function \( f \) if, for every \( x \) in the domain, \( \lim_{{n \to \infty}} f_n(x) = f(x) \).
In our example, the sequence of functions \( f_n \) is constructed so that each function is non-zero only on a finite number of rational points. Over time, \( f_n(x) \) gets closer to \( f(x) \), where \( f \) is 1 on the rational points and 0 on the irrationals.
Pointwise convergence also means that while each \( f_n \) might be integrable, the limit function \( f \) itself may not be integrable, illustrating some subtleties of Lebesgue integration.

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Most popular questions from this chapter

(a) Prove that the series $$f(x)=\sum_{n=0}^{\infty} \frac{n x}{1+n^{4} x^{2}}$$ converges uniformly on \([a, \infty)\) for \(a>0 .\) Hint: First find the maximum of \(n x /\left(1+n^{4} x^{2}\right)\) on \([0, \infty)\) (b) Show that $$f\left(\frac{1}{N}\right) \geq \frac{N}{2} \sum_{n \geq \sqrt{N}} \frac{1}{n^{3}}$$ and by using an integral to estimate the sum, show that \(f\left(1 / N^{2}\right) \geq 1 / 4\) Conclude that the series does not converge uniformly on \(\mathbf{R}\). (c) What about the series \(\sum_{n=0}^{\infty} \frac{n x}{1+n^{5} x^{2}} ?\)

(a) Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for all \(x\) in some interval \((-R, R)\) and that \(f(x)=0\) for all \(x\) in \((-R, R) .\) Prove that each \(a_{n}=0\) (If you remember the formula for \(a_{n}\) this is easy.) (b) Suppose we know only that \(f\left(x_{n}\right)=0\) for some sequence \(\left\\{x_{n}\right\\}\) with \(\lim _{n \rightarrow \infty} x_{n}=0 .\) Prove again that each \(a_{n}=0 .\) Hint: First show that \(f(0)=a_{0}=0 ;\) then that \(f^{\prime}(0)=a_{1}=0,\) etc. This result shows that if \(f(x)=e^{-1 / x^{2}} \sin 1 / x\) for \(x \neq 0,\) then \(f\) cannot possibly be written as a power series. It also shows that a function defined by a power series cannot be 0 for \(x \leq 0\) but nonzero for \(x>0\) -thus a power series cannot describe the motion of a particle which has remained at rest until time \(0,\) and then begins to move! (c) Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) and \(g(x)=\sum_{n=0}^{\infty} b_{n} x^{n}\) converge for all \(x\) in some interval containing 0 and that \(f\left(t_{m}\right)=g\left(t_{m}\right)\) for some sequence \(\left\\{t_{m}\right\\}\) converging to \(0 .\) Show that \(a_{n}=b_{n}\) for each \(n\)

A sequence \(\left\\{a_{n}\right\\}\) is called Abel summable if \(\lim _{x \rightarrow 1^{-}} \sum_{n=0}^{\infty} a_{n} x^{n}\) exists; Problem 19 shows that a summable sequence is necessarily Abel summable. Find a sequence which is Abel summable, but which is not summable. Hint: Look over the list of Taylor series until you find one which does not converge at 1 even though the function it represents is continuous at 1

(a) Suppose that \(\left.| f_{n}\right\\}\) is a sequence of bounded (not necessarily continuous) functions on \([a, b]\) which converge unifornily to \(f\) on \([a, b] .\) Prove that \(f\) is bounded on \([a, b]\) (b) Find a sequence of continuous functions on \([a, b]\) which converge pointwise to an unbounded function on \([a, b]\)

Suppose that \(f(x)=\sum_{n=0}^{\infty} a_{n} x^{n}\) converges for some \(x_{0},\) and that \(a_{0} \neq 0\) for simplicity, we'll assume that \(a_{0}=1 .\) Let \(\left\\{b_{n}\right\\}\) be the sequence defined recursively by $$\begin{aligned} b_{0} &=1 \\ b_{n} &=-\sum_{k=0}^{n-1} b_{k} a_{n-k} \end{aligned}$$ The aim of this problem is to show that \(\sum_{n=0}^{\infty} b_{n} x^{n}\) also converges for some \(x \neq 0,\) so that it represents \(1 / f\) for small enough \(|x|\) (a) If all \(\left|a_{n} \cdot x_{0}^{\prime \prime}\right| \leq M,\) show that $$ \left|b_{n} x_{0}^{n}\right| \leq M \sum_{k=0}^{n-1}\left|b_{k} x_{0}^{k}\right| $$ (b) Choose \(M\) so that \(\left|a_{n} x_{0}^{\prime \prime}\right| \leq M,\) and also so that \(M /\left(M^{2}-1\right) \leq 1 .\) Show that $$ \left|b_{n} x_{0}^{n}\right| \leq M^{2 n} $$ (c) Conclude that \(\sum_{n=0}^{\infty} b_{n} x^{n}\) converges for \(|x|\) sufficiently small.
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