Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 5

Show that every countable subset of \(R\) is measurablc and has Lebesgue measure zero.

Short Answer

Expert verified
Every countable subset of real numbers is measurable with Lebesgue measure zero because we can cover the set by a family of open intervals whose total measure is less than any positive number.

Step by step solution

01

Define countable set

First, let's enumerate the countable set \(S\) which can be expressed as \(S = \{x_1, x_2, x_3, \ldots\}\), where \(x_n\) represents the elements in the set.
02

Construct the family of cover

Now, for each \(n\) in natural numbers, construct an open interval centered at \(x_n\) with small length \(l_n < \frac{1}{2^n}\), specifically \( (x_n - l_n, x_n + l_n)\). This becomes a family of open intervals that covers \(S\).
03

Define the measure of the family of open intervals

Utilize the definition of the Lebesgue measure which states that for a countable collection of disjoint sets, the measure is the sum of the individual measures. In our case, the measure of the family of open intervals is \(\sum_{n=1}^{\infty} 2l_{n}\). Since \(2l_{n} < \frac{2}{2^n} = \frac{1}{2^{n-1}} \), it implies that \(\sum_{n=1}^{\infty} 2l_{n} < \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = 2\).
04

Conclude that the set is measurable with Lebesgue measure zero

From the previous step, it's clear that we can cover \(S\) by a family of open intervals whose total measure is less than any positive number, here it is less than 2. Hence, by definition, the Lebesgue measure of \(S\) should be zero. As \(S\) is a subset of \(R\), it is also measurable.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that the union of a sequence of sets of measure zero is a set of Lebesgue measure zero,

The inner measure \(\mu_{*}(E)\) of a set \(E\) is defined as the least upper bound of the measures of all measurable subsets of \(E\). Show that \(\mu_{*}(E) \leq \mu^{*}(E)\). For any open set \(U \supset E\), show that $$ \mu(U)=\mu_{*}(U \cap E)+\mu^{*}(U-E) $$ and that \(E\) is measurable with finite measure if and only if \(\mu_{*}(E)=\mu^{*}(E)<\infty\).

Show that a subset \(E\) of \(\mathbb{R}\) is measurable if for all \(\epsilon>0\) there exists an open set \(U \supset E\) such that \(\mu^{*}(U-E)<\epsilon\)

If \(E\) is bounded and there exists an interval \(I \supset E\) such that $$ \mu^{*}(I)=\mu^{*}(I \cap E)+\mu^{*}(I-E) $$ then this holds for all intervals, possibly even those overlapping \(E\).

A measure is said to be complete if every subset of a sct of measure zero is measurable. Show that if \(A \subset \mathbb{R}\) is a set of outer measure zero, \(\mu^{*}(A)=0\), then \(A\) is Lebesgue measurable and has measur zero. Hence shew that Lebesgue measure is complcte.
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free