Chapter 11: Problem 8

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.

Short Answer

Expert verified

Subset \(A\) of \(\mathbb{R}\) with outer measure zero is proven to be Lebesgue measurable and to have measure zero. This allows us to show that the Lebesgue measure is complete, as per the definition of completeness, i.e., every subset of a set of measure zero is measurable.

Step by step solution

Prove \(A\) is Lebesgue measurable

To show \(A\) is Lebesgue measurable, first consider a set \(E\) which lies in the real numbers. We know that \(\mu^{*}(E)=\mu^{*}(E \cap A) + \mu^{*}(E \cap A^{C})\), where \(A^{C}\) represents the complement of \(A\). Given that \(\mu^{*}(A)=0\), replacing \(E \cap A\) with \(A\) gives \(\mu^{*}(E) \geq \mu^{*}(A) + \mu^{*}(E \cap A^{C})\). This implies \(\mu^{*}(E) \geq \mu^{*}(E \cap A^{C})\). Due to the sub-additivity of the outer measure, \(\mu^{*}(E) \leq \mu^{*}(E \cap A^{C})\). Therefore, \(\mu^{*}(E) = \mu^{*}(E \cap A^{C})\). This demonstrates that \(A\) is Lebesgue measurable.

Prove \(A\) has measure zero

Given that \(A\) is Lebesgue measurable, we must verify it also has measure zero. We know \(\mu^{*}(A) = 0\), and because \(A\) is measurable, its outer and inner (Lebesgue) measures are equal; i.e., \(\mu(A) = \mu^{*}(A) = 0\). Thus, \(A\) has measure zero.

Prove Lebesgue measure is complete

Finally, we need to show the Lebesgue measure is complete. Complete measure guarantees that if a set \(A\) has measure zero, every subset of \(A\) is measurable. We have proven any set \(A\) with outer measure zero is Lebesgue measurable and has measure zero. As any subset of \(A\) will also have outer (and therefore Lebesgue) measure zero, and thus will be measurable. This confirms that the Lebesgue measure is complete.