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Chapter 11: Problem 10

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 \(E\).

Short Answer

Expert verified
Yes, if a bounded set \(E\) has the property that the measure of any interval containing \(E\) can be partitioned into the measures of the intersection and difference between \(E\) and the interval, then this property extends to all intervals, possibly even those that overlap \(E\).

Step by step solution

01

Define the set \(F\)

Define a set \(F = E \cap I\), where \(I\) is any interval. By doing this, we ensure that \(F\) is bounded and any interval \(I'\) that contains \(F\) should also contain \(E\), as \(F\) is a subset of \(E\).
02

Evaluate the measure of \(I'\)

Consider the measure of any arbitrary interval \(I'\) such that \(I' \supset F\). If \(I' = I'' \cup (I'-I'')\), where \(I'' \supset E\), we can apply the given condition because \(I''\) contains \(E\). Thus, we get \(\mu^{*}(I') = \mu^{*}(I'') + \mu^{*}(I'-I'')\). Now, \(I'' \cap E = I'' \cap F\) and \(I''-F = I''-E\), so the equation can be rewritten as \(\mu^{*}(I') = \mu^{*}(I' \cap F) + \mu^{*}(I'-F)\).
03

Complete the proof

This equation shows that the property holds for any interval \(I'\) containing \(F\), which implies that the property holds for any interval that intersects \(E\), as \(F\) was defined to be a subset of \(E\). Therefore, we have shown that, under the given conditions, the measure property holds for all intervals, possibly even those overlapping \(E\).

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Most popular questions from this chapter

\(\operatorname{lf}(X, M)\) and \((Y, N)\) are measurable spaces, show that the projection maps \(\mathrm{pr}_{1}: X \times\) \(Y \rightarrow X\) and \(\mathrm{pr}_{2}: X \times Y \rightarrow Y\) defined by \(\mathrm{pr}_{1}(x, y)=x\) and \(\mathrm{pr}_{2}(x, y)=y\) are measurable functions,

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.

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

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

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.
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