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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

Show that every countable subset of \(R\) is measurablc and has Lebesgue measure zero.

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 \(E\).

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 \(\mathbb{R}^{2}\).

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

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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