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Chapter 11: Problem 7

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.

Short Answer

Expert verified
Given a null set \(N\), we have the following: \(\mu^{*}(E \cup N) = \mu^{*}(E)\) and \(\mu^{*}(E-N) = \mu^{*}(E)\). Furthermore, if \(E\) is measurable, the measurability is preserved under union and difference with a null set. Hence, both \(E \cup N\) and \(E-N\) are also Lebesgue measurable.

Step by step solution

01

Show that \(\mu^{*}(E \cup N) = \mu^{*}(E)\)

By sub-additivity of an outer measure \(\mu^{*}\), we have \(\mu^{*}(E \cup N) \leq \mu^{*}(E) + \mu^{*}(N)\). Since \(\mu^{*}(N) = 0\), we get \(\mu^{*}(E \cup N) \leq \mu^{*}(E)\). We also know that \(E \subseteq E \cup N\) so \(\mu^{*}(E) \leq \mu^{*}(E \cup N)\) by monotonicity of the outer measure. Hence, \(\mu^{*}(E \cup N) = \mu^{*}(E)\).
02

Show that \(\mu^{*}(E-N) = \mu^{*}(E)\)

We know that \(E = (E-N) \cup (E \cap N)\). By sub-additivity, \(\mu^{*}(E) \leq \mu^{*}(E-N) + \mu^{*}(E \cap N)\). Since \(E \cap N\) is subset of \(N\) which is a null set, \(\mu^{*}(E \cap N) = 0\). Therefore, \(\mu^{*}(E) \leq \mu^{*}(E-N)\). By monotonicity, since \(E-N \subseteq E\), \(\mu^{*}(E-N) \leq \mu^{*}(E)\). Hence, \(\mu^{*}(E-N) = \mu^{*}(E)\).
03

Prove the final assertion

Now, E is measurable if and only if for all \(A \subset \mathbb{R}\), \(\mu^{*}(A) = \mu^{*}(A \cap E) + \mu^{*}(A \cap E^c)\). Replace \(A\) with \(A \cup N\) and \(A-N\) in this expression and use the results from step 1 and 2 to show that this condition is also fulfilled for \(E \cup N\) and \(E-N\) if \(E\) is measurable.

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Most popular questions from this chapter

Find a step function \(s(x)\) that approximates \(f(x)=x^{2}\) uniformly to within \(\varepsilon>0\) on \([0,1]\), in the sense that \(|f(x)-s(x)|<\varepsilon\) cierywhere in \([0,1]\).

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

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 \(E\).

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