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Chapter 11: Problem 6

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

Short Answer

Expert verified
Hence proved, the union of a sequence of sets of measure zero is a set of Lebesgue measure zero.

Step by step solution

01

Define a set of Measure Zero

A set \( S \) is said to have measure zero if, given any positive number \( \epsilon \), there exists a countable collection \( \{I_n\} \) of open intervals such that \( S \subseteq \bigcup_{n=1}^{\infty} I_n \) and the sum of the lengths of these intervals is less than \( \epsilon \), i.e. \( \sum_{n=1}^{\infty} \text{length}(I_n) < \epsilon \).
02

Assumption of the Problem

Assume that each set \( A_i \) in a sequence \( \{A_i\}_{i=1}^{\infty} \) is of measure zero. Hence for any given \( \epsilon \), there exists a collection of intervals \( \{I_{i, n}\}_{n=1}^{\infty} \) for each \( A_i \) such that \( A_i \subseteq \bigcup_{n=1}^{\infty} I_{i, n} \) and \( \sum_{n=1}^{\infty} \text{length}(I_{i, n}) < \frac{\epsilon}{2^{i}} \).
03

Prove the Conclusion

It's required to show that the union \( U = \bigcup_{i=1}^{\infty} A_i \) is also a measure zero set. Note \( U \subseteq \bigcup_{i=1}^{\infty} \bigcup_{n=1}^{\infty} I_{i, n} \). As each \( A_i \) is covered by the intervals \( I_{i, n} \) whose total length is less than \( \frac{\epsilon}{2^{i}} \), it follows that the total length of all intervals covering \( U \) is less than \( \sum_{i=1}^{\infty} \frac{\epsilon}{2^{i}} = \epsilon \). Thus, \( U \) also has measure zero according to the definition.

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Most popular questions from this chapter

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 \(E\).

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

\(\operatorname{lf}(X, M)\) 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,

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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