Math Problems Vault

Problem 4.5.1

Let $E \notin \mathscr {L}^{k}$ denote the set constructed in this section, and let $A \subseteq E$ be such that $A \in \mathscr {L}^{k}$ . Then $\bar{\lambda }_{k}(A)=0$ .

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Remark.

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The construction of this section: for each $\boldsymbol {x}$ and $\boldsymbol {y}$ in $\mathbb {R}^{k}$ , write $\boldsymbol {x} \sim \boldsymbol {y}$ iff there is $\boldsymbol {r} \in \mathbb {Q}^{k}$ such that $\boldsymbol {x}+\boldsymbol {r}=\boldsymbol {y}$ ; then $\sim$ is an equivalence relation on $\mathbb {R}^{k}$ , determining an infinite number of equivalence classes, each itself infinite. Invoking the Axiom of Choice, form a set $E \subseteq \mathbb {R}^{k}$ consisting of exactly one element from each equivalence class, and consider the collection $\left\{ E \oplus \boldsymbol {r}: \boldsymbol {r} \in \mathbb {Q}^{k}\right\}$ of the translates of $E$ by the elements of $\mathbb {Q}^{k}$ . Claim 2: if $\boldsymbol {r}_{1}, \boldsymbol {r}_{2} \in \mathbb {Q}^{k}$ are distinct, then $E \oplus \boldsymbol {r}_{1}$ and $E \oplus \boldsymbol {r}_{2}$ are disjoint. Claim 3: $\mathbb {R}^{k}=\bigcup \left\{ (E \oplus \boldsymbol {r}): \boldsymbol {r} \in \mathbb {Q}^{k}\right\}$ , the union being disjoint. Claim 4: $\lambda_{k}^{*}(E)>0$ . Claim 5: $\sup \left\{ \lambda_{k}(K): K \text{ compact}, K \subseteq E\right\} =0$ . Claim 6: the set $E$ is not in $\mathscr {L}^{k}$ . This is a variant of the classical construction given by Vitali in 1905.

Problem 4.5.2

There exists no translation invariant measure $\mu$ defined for all subsets of $\mathbb {R}^{k}$ such that $\mu ((\boldsymbol {a}, \boldsymbol {b}])=\prod_{i=1}^{k}(b_{i}-a_{i})$ for all bounded $k$ -dimensional rsc rectangles $(\boldsymbol {a}, \boldsymbol {b}]$ .

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Problem 4.5.3

Here is an alternative but quite similar construction of a set outside $\mathscr {L}$ . We shall consider $(0,1]$ as our space. Let $\dot{+}$ denote addition modulo 1: for $x, y \in (0,1]$ ,

x \mathbin {\dot{+}} y= \begin{cases} x+y, & \text{if } x+y \leqslant 1, \\ x+y-1, & \text{if } x+y>1. \end{cases}

For each $A \subseteq (0,1]$ and $x \in (0,1]$ , we shall [for this exercise only] write $A \oplus x$ as translation modulo 1: $A \oplus x=\left\{ a \mathbin {\dot{+}} x: a \in A\right\}$ . Finally, [for this exercise only] we shall let $\mathbb {Q}$ denote the rational numbers of $(0,1]$ .

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(a)

For $x, y \in (0,1]$ write $x \sim y$ iff there is $r \in \mathbb {Q}$ such that $x \mathbin {\dot{+}} r=y$ . Then $\sim$ is an equivalence relation.

(b)

Let $F$ denote the set formed by choosing exactly one element from each equivalence class [use (AC)]. Show that $(F \oplus r_{1}) \cap (F \oplus r_{2})$ is empty if $r_{1}$ and $r_{2}$ are distinct elements of $\mathbb {Q}$ .

(c)

$(0,1]=\bigcup_{r \in \mathbb {Q}}(F \oplus r)$ , where the union is disjoint.

(d)

$F \notin \mathscr {L}$ and $0<\lambda^{*}(F) \leqslant 1$ .

(e)

If there exists a translation invariant measure $\mu$ defined on a $\sigma$ -field on $(0,1]$ that includes $F$ and is closed under translations, then $\mu ((0,1])$ is either $0$ or $+\infty$ .

(f)

Show that there can exist no translation invariant measure defined on all subsets of $(0,1]$ that agrees with $\bar{\lambda }$ on the Lebesgue sets of $(0,1]$ .

Remark.

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Argue by contradiction for (d): if $F \in \mathscr {L}$ , then

1=\bar{\lambda }((0,1])=\sum _{r \in \mathbb {Q}} \bar{\lambda }(F \oplus r)=\sum _{r \in \mathbb {Q}} \bar{\lambda }(F).

If $\bar{\lambda }(F)=0$ , we obtain $1=0$ . If $\bar{\lambda }(F)>0$ , then we obtain $1=+\infty$ . Conclude that $\bar{\lambda }(F)$ cannot exist, which means that $F \notin \mathscr {L}$ . For the other part, since $F \subseteq (0,1]$ , clearly $\lambda^{*}(F) \leqslant 1$ . To show that $\lambda^{*}(F)>0$ , write

1=\lambda ^{*}((0,1])=\lambda ^{*}\left(\bigcup _{r \in \mathbb {Q}}(F \oplus r)\right) \leqslant \sum _{r \in \mathbb {Q}} \lambda ^{*}(F \oplus r)=\sum _{r \in \mathbb {Q}} \lambda ^{*}(F).

Problem 4.5.4

(a)(b)

Prove the following assertions.

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(a)

There exists a sequence $\left\{ E_{n}\right\}_{n=1}^{\infty }$ of disjoint sets [not necessarily in $\mathscr {L}$ ] such that $\lambda^{*}\left(\bigcup_{n=1}^{\infty } E_{n}\right)<\sum_{n=1}^{\infty } \lambda^{*}(E_{n})$ .

(b)

There exists a nonincreasing sequence $\left\{ E_{n}\right\}_{n=1}^{\infty }$ of sets [not necessarily in $\mathscr {L}$ ] with $\lambda^{*}(E_{n})<+\infty$ for each $n \in \mathbb {N}$ such that $\lambda^{*}\left(\bigcap_{n=1}^{\infty } E_{n}\right)<\lim_{n} \lambda^{*}(E_{n})$ .

Remark.

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For (a): first, show that $\lambda^{*}$ is translation invariant. Let $\left\{ r_{n}\right\}_{n=1}^{\infty }$ denote the rationals in $(0,1]$ , and let $E_{n}=F \oplus r_{n}$ , where $F$ is the set from Exercise 4.5.3 [the set formed by choosing exactly one element from each equivalence class of $\sim$ on $(0,1]$ , where $x \sim y$ iff $x \mathbin {\dot{+}} r=y$ for some rational $r$ ] and $\oplus$ denotes translation modulo 1.

Problem 4.5.5

Let $A \in \mathscr {L}^{k}$ be such that $\bar{\lambda }_{k}(A)>0$ . There exist $\boldsymbol {x}, \boldsymbol {y} \in A$ so that $\boldsymbol {x}-\boldsymbol {y} \in \mathbb {Q}^{k}$ .

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Problem 4.5.6

(a)(b)

Let $T: \mathbb {R}^{2} \rightarrow \mathbb {R}$ be such that $T \boldsymbol {x}=x_{1}$ for all $\boldsymbol {x} \in \mathbb {R}^{2}$ . That is, $T$ projects points [perpendicularly] onto the $x_{1}$ -axis.

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(a)

Such a mapping is linear and singular.

(b)

There exists $A \in \mathscr {B}\left(\mathbb {R}^{2}\right)$ such that (i) $TA$ , viewed as a subset of $\mathbb {R}$ , is not in $\mathscr {B}$ , while (ii) $TA$ , viewed as a subset of $\mathbb {R}^{2}$ , is in $\mathscr {B}\left(\mathbb {R}^{2}\right)$ .

Remark.

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Consider the set $E \times \left\{ 0\right\}$ , where $E$ is the one-dimensional form of the set constructed in this section.

Problem 4.5.7

For any linear mapping $T$ [singular or nonsingular] defined on $\mathbb {R}^{k}$ , $A \in \mathscr {L}^{k}$ forces $TA \in \mathscr {L}^{k}$ .

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Remark.

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The result has been established for nonsingular $T$ (Claim 11(ii) of Section 4.4: if $T$ is a nonsingular transformation, then for every $A \in \mathscr {L}^{k}$ we have $TA \in \mathscr {L}^{k}$ and $\bar{\lambda }_{k}(TA)=\left|\operatorname {det}\left(T\right)\right| \bar{\lambda }_{k}(A)$ ). For singular $T$ , observe that $TA \subseteq T\mathbb {R}^{k}$ , which is a hyperplane, hence a Borel set of $\lambda_{k}$ -measure zero.