Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 11: Problem 9

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 \(\mu^{*}(U-E)<\epsilon\)

Short Answer

Expert verified
Hence, \(E\) is measurable when for every \(\epsilon>0\) there exists an open set \(U \supset E\) whose outer measure \(\mu^{*}(U-E)<\epsilon\).

Step by step solution

01

Consequence of the Given Condition

Assume \(E\) belongs to \(\mathbb{R}\), and for every \(\epsilon > 0\), there exists an open set \(U\) containing \(E\) with \(\mu^{*}(U - E) < \(\epsilon\). This implies that \(U\) can be written as union of a countable number of disjoint open intervals.
02

Defining a Set

We define a set \(O\) as union of countably many disjoint open intervals such that \(\mu(O) < \mu(E) + \epsilon\). This set \(O\) is open and \(E \subseteq O\). Thus, \(\mu(E) \leq \mu(O)\). And since we have assumed \(\mu(O) < \mu(E) + \epsilon\), we can say \(\mu(E) \leq \mu(O) < \mu(E) + \epsilon\). This proves \(\mu(E)\) exists.
03

Showing Measurable Implication

Now, choose \(U\) which is open and contains \(E\) with \(\mu^{*}(U - E) < \epsilon\). This gives \(\mu^{*}(U) = \mu^{*}(E) +\mu^{*}(U-E) < \mu^{*}(E) + \epsilon\). Since \(E \subseteq U\), it implies \(\mu^{*}(E) \leq \mu^{*}(U)\). Thus, \(\mu^{*}(E) \leq \mu^{*}(U) < \mu^{*}(E) + \epsilon\). Hence proving \(\mu(E) = \mu^{*}(E)\), indicating that \(E\) is 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 every countable subset of \(R\) is measurablc and has Lebesgue measure zero.

If \(f, g: \mathbb{R} \rightarrow \mathbb{R}\) are Borel measurable real functions show that \(h(x, y)=f(x) g(y)\) is a measurable function \(h: \mathbb{R}^{2} \rightarrow \mathbb{R}\) with respect to the product measure on \(\mathbb{R}^{2}\).

Show that if \(f\) and \(g\) are Lebesgue integrable on \(E \subset \mathbb{R}\) and \(f \geq g\) a.c., then $$ \int_{E} f d \mu \geq \int_{E} g d \mu $$

If \(\mu^{*}(N)=0\) show that for any set \(E, \mu^{*}(E \cup N)=\mu^{*}(E-N)=\mu^{*}(E)\). Hence show that \(E \cup N\) and \(E-N\) are Lebesgue measurable if and only if \(E\) is measurable.

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 \(\mu_{*}(E)=\mu^{*}(E)<\infty\).
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