Exercises: Chapter 3, Section 2

  1. Prove that ch3b1.png is not of content 0 if ch3b2.png for ch3b3.png .

    Suppose ch3b4.png for ch3b5.png are closed rectangles which form a cover for ch3b6.png . By replacing the ch3b7.png with ch3b8.png , one can assume that ch3b9.png for all ch3b10.png . Let ch3b11.png . Choose a partition ch3b12.png which refines all of the partitions ch3b13.png where ch3b14.png Note that ch3b15.png is a rectangle of the cover ch3b16.png . Let ch3b17.png be any rectangle in ch3b18.png with non-empty interior. Since the intersection of any two rectangles of a partition is contained in their boundaries, if ch3b19.png contains an interior point ch3b20.png not in ch3b21.png for some ch3b22.png , then ch3b23.png contains only boundary points of ch3b24.png . So, if ch3b25.png has non-empty interior, then ch3b26.png is a subset of ch3b27.png for some ch3b28.png since the union of the ch3b29.png is ch3b30.png . The sum of the volumes of the rectangles of ch3b31.png is the volume of ch3b32.png , which is at most equal to the sum of the volumes of the ch3b33.png . So ch3b34.png is not of content 0 as it cannot be covered with rectangles of total area less than the volume of ch3b35.png .

    1. Show that an unbounded set ch3b36.png cannot have content 0.

      Suppose ch3b37.png where ch3b38.png are rectangles, say ch3b39.png . Let ch3b40.png where ch3b41.png and ch3b42.png . Then ch3b43.png contains all the ch3b44.png and hence also ch3b45.png . But then ch3b46.png is bounded, contrary to hypothesis.

    2. Give an example of a closed set of measure 0 which does not have content 0.

      The set of natural numbers is unbounded, and hence not of content 0 by part (a). On the other hand, it is of measure zero. Indeed, if ch3b47.png , then the union of the open intervals ch3b48.png for ch3b49.png . contains all the natural numbers and the total volume of all the intervals is ch3b50.png .

    1. If C is a set of content 0, show that the boundary of ch3b51.png also has content 0.

      Suppose a finite set of open rectangles ch3b52.png , ch3b53.png . cover of ch3b54.png and have total volume less than ch3b55.png where ch3b56.png . Let ch3b57.png where ch3b58.png . Then the union of the ch3b59.png cover the boundary of ch3b60.png and have total volume less than ch3b61.png . So the boundary of ch3b62.png is also of content 0.

    2. Give an example of a bounded set ch3b63.png of measure 0 such that the boundry of ch3b64.png does not have measure 0.

      The set of rational numbers in the interval ch3b65.png is of measure 0 (cf Proof of Problem 3-9 (b)), but its boundary ch3b66.png is not of measure 0 (by Theorem 3-6 and Problem 3-8).

  2. Let ch3b67.png be the set of Problem 1-18. If ch3b68.png , show that the boundary of ch3b69.png does not have measure 0.

    The set ch3b70.png closed and bounded, and hence compact. If it were also of measure 0, then it would be of content 0 by Theorem 3-6. But then there is a finite collection of open intervals which cover the set and have total volume less than ch3b71.png . Since the set these open intervals together with the set of ch3b72.png form an open cover of [0, 1], there is a finite subcover of ch3b73.png . But then the sum of the lengths of the intervals in this finite subcover would be less than 1, contrary to Theorem 3-5.

  3. Let ch3b74.png be an increasing function. Show that ch3b75.png is a set of measure 0.

    Using the hint, we know by Problem 1-30 that the set of ch3b76.png where ch3b77.png if finite for every ch3b78.png . Hence the set of discontinuities of ch3b79.png is a countable union of finite sets, and hence has measure 0 by Theorem 3-4.

    1. Show that the set of all rectangles ch3b80.png where each ch3b81.png and each ch3b82.png are rational can be arranged into a sequence (i.e. form a countable set).

      Since the set of rational numbers is countable, and cartesian products of countable sets are countable, so is the set of all ch3b83.png -tuples of rational numbers. Since the set of these intervals is just a subset of this set, it must be countable too.

    2. If ch3b84.png is any set and ch3b85.png is an open cover of ch3b86.png , show that there is a sequence ch3b87.png of members of ch3b88.png which also cover ch3b89.png .

      Following the hint, for each ch3b90.png in ch3b91.png , there is a rectangle B of the type in part (a) such that ch3b92.png has non-zero volume, contains ch3b93.png and is contained in some ch3b94.png in ch3b95.png . In fact, we can even assume that ch3b96.png is in the interior of the rectangle ch3b97.png . In particular, the union of the interiors of the rectangles ch3b98.png (where ch3b99.png is allowed to range throughout ch3b100.png ) is a cover of ch3b101.png . By part (a), the set of these ch3b102.png are countable, and hence so are the set of corresponding ch3b103.png 's; this set of corresponding ch3b104.png 's cover ch3b105.png .