Lebesgue Measure

Let $x \in \mathbb {R}$ and $k \geqslant 2$. Then $\lambda (\left\{ x\right\} )=\lambda_{k}\left(\left\{ x\right\} \times \mathbb {R}^{k-1}\right)=0$. Next, for any $j \in \left\{ 1, \ldots , k-1\right\}$ and $\boldsymbol {x} \in \mathbb {R}^{j}$ we have $\lambda_{k}\left(\left\{ \boldsymbol {x}\right\} \times \mathbb {R}^{k-j}\right)=0$.

Enumerate the rationals in $(0,1]$ by $\left\{ q_{1}, q_{2}, \ldots \right\}$. Given arbitrarily small $\varepsilon >0$, remove the interval $A_{n}=\left(q_{n}-\varepsilon / 2^{n+1}, q_{n}+\varepsilon / 2^{n+1}\right) \cap (0,1]$. Let $A=\bigcup_{n=1}^{\infty } A_{n}$. Then $\bar{\lambda }(A) \leqslant \varepsilon$, despite the fact that $A$ is an open dense subset of $(0,1]$. Also, we have $\bar{\lambda }((0,1] \backslash A) \geqslant 1-\varepsilon$, even though $(0,1] \backslash A$ is a nowhere dense subset of $(0,1]$.

There cannot exist a closed subset of $(0,1]$ whose interior is empty, yet has $\bar{\lambda }$-measure of one.

$\lambda_{k}$ is nonatomic: any $A \in \mathscr {B}^{k}$ with $\lambda_{k}(A)>0$ has a proper subset $B \in \mathscr {B}^{k}$ with $0<\lambda_{k}(B)<\lambda_{k}(A)$. This forces $\bar{\lambda }_{k}$ to be nonatomic as well.

Let $k \geqslant 2$. There exists an uncountable set $U \in \mathscr {B}^{k}$ with $\lambda_{k}(U)=0$.

$\left|\mathscr {L}^{k}\right|=2^{\mathrm{c}}$ and $\left|\mathscr {B}^{k}\right|=\mathrm{c}$ for each $k \in \mathbb {N}$.

Assume that $\mathscr {L}=2^{\mathbb {R}}$; in particular every one of the $2^{\mathrm{c}}$ subsets of $\left[0,1\right]$ is a Lebesgue set. Let $B=\left\{ \bar{\lambda }(A): A \subseteq [0,1], \bar{\lambda }(A) \notin A\right\}$. Consideration of the set $B$ (which is in $\mathscr {L}$ be assumption) leads to contradiction.

Let $k \geqslant 2$. Every line is in $\mathscr {B}^{k}$, has $\lambda_{k}$-measure zero, and hence has $\bar{\lambda }_{k}$-measure.

Let $\mathscr {B}_{(0,1]}=\sigma (\left\{ (a, b]:(a, b] \subseteq (0,1]\right\} )$.

Let $\mu : \mathscr {B}^{k} \rightarrow \overline{\mathbb {R}}$ denote a measure that assigns finite measure to bounded $\mathscr {B}^{k}$-sets. Furthermore, suppose that $\mu$ is translation invariant: $\mu (B \oplus \boldsymbol {x})=\mu (B)$ for all $\boldsymbol {x} \in \mathbb {R}^{k}$. Then there is $c>0$ such that $\mu (B)=c \lambda_{k}(B)$ for all $B \in \mathscr {B}^{k}$.

Let $A \in \mathscr {L}$ be such that $\bar{\lambda }(A)>0$, and let $c \in [0,1)$. There exists an open interval $U$ such that $\bar{\lambda }(A \cap U) \geqslant c \lambda (U) .^{1}$

Let $A \in \mathscr {L}$ contain an open interval. Then there exists $a>0$ such that $(-a, a)$ is contained in $D(A)=\left\{ x-y: x, y \in A\right\}$.

Let $A \in \mathscr {L}$ be such that $\bar{\lambda }(A)>0$. Then there exists $a>0$ such that $(-a, a)$ is contained in $D(A)=\left\{ x-y: x, y \in A\right\}$.

Let $A$ be a dense subset of $\mathbb {R}$. Then $c A=\left\{ c a: a \in A\right\}$ is dense for any $c \neq 0$.

Let $\xi$ be an irrational number.

For $x, y \in \mathbb {R}$ write $x \sim y$ iff $x-y \in A=\left\{ n+m \xi : n, m \in \mathbb {Z}\right\}$, where $\xi$ is a fixed irrational number as in the previous exercise. Then $\sim$ is an equivalence relation, and hence $\mathbb {R}$ may be partitioned into disjoint equivalence classes.

We now invoke ( $A C$ ) to form a set $E_{0}$ consisting of exactly one element from each of the equivalence classes in the previous exercise. We will now show that $E_{0} \notin \mathscr {L}$.

In conjunction with the previous exercises, it is not possible to have Lebesgue measure defined for every subset of $\mathbb {R}$. That is, a measure extension $\mu$ of $\Delta_{F}$ [where $F \in \mathfrak {S}$ is the identity function] from the semiring $\mathcal{A}$ consisting of $\varnothing$ and the bounded rsc intervals $(a, b]$ to $2^{\mathbb {R}}$ fails to exist.

Let $B \in \mathscr {L}$ be such that $\bar{\lambda }(B)>0$. We will show that there is $E_{1} \subseteq B$ with $E_{1} \notin \mathscr {L}$ by following the outline below. The set $A$ appearing below is the set $A$ from Exercise 4.3.7, consisting of the numbers $n+m \xi$, where $n, m \in \mathbb {Z}$ and $\xi$ is a fixed irrational number.

Let $A \in \mathscr {L}$, let $B$ be a dense subset of $\mathbb {R}$, and suppose that $\bar{\lambda }(A \triangle (A \oplus x))=0$ for every $x \in B$. Then either $\bar{\lambda }(A)=0$ or $\bar{\lambda }(A^{c})=0$. This exercise is independent of the previous exercises.

Recall that a linear transformation $T: \mathbb {R}^{k} \rightarrow \mathbb {R}^{k}$ (for $k \geqslant 2$) is called special iff $T$ is a mapping of one of the following forms: a $k \times k$ permutation matrix (Claim 4); $T(\boldsymbol {x})=(p x_{1}, x_{2}, \ldots , x_{k})'$ for some $p \neq 0$ (Claim 5); $T(\boldsymbol {x})=(x_{1}+x_{2}, x_{2}, \ldots , x_{k})'$ (Claim 6); or $T(\boldsymbol {x})=(x_{1}-x_{2}, x_{2}, \ldots , x_{k})'$ (Claim 7). The matrices associated with such transformations are called special matrices. The inverse of a special matrix is a special matrix.

Any nonsingular $k \times k$ matrix $\mathbf{T}$ may be written as a finite product $\mathbf{T}_{N} \cdots \mathbf{T}_{2} \mathbf{T}_{1}$ of nonsingular $k \times k$ matrices, each matrix $\mathbf{T}_{j}$ being nonsingular and freely denoting the matrix corresponding to any one of the special linear transformations from $\mathbb {R}^{k}$ to $\mathbb {R}^{k}$ described earlier.

Suggestion: find a finite sequence $\mathbf{T}_{1}, \cdots , \mathbf{T}_{N_{1}}$ of special matrices such that

Then find another finite sequence of matrices $\mathbf{T}_{N_{1}+1}, \cdots , \mathbf{T}_{N_{1}+N_{2}}$ such that

Then, find another finite sequence of matrices $\mathbf{T}_{N_{2}+1}, \cdots , \mathbf{T}_{N_{2}+N_{3}}$ such that

Argue that the above matrix contains a $(k-1) \times (k-1)$ invertible submatrix [the blank spot in the above matrix] to which we can repeat this entire procedure without disturbing the 1's and 0's of the above matrix in order to obtain a matrix of the form

Argue that we may carry on this procedure until we have [written symbolically] a sequence $\mathbf{T}_{1}, \cdots , \mathbf{T}_{\alpha +\beta }$ of special matrices such that

Therefore,

Since the inverse of a special matrix is a special matrix by Exercise 4.4.1, the proof is complete. This gives an outline of the proof of Claim 10.

A mapping $F: \mathbb {R}^{k} \rightarrow \mathbb {R}^{k}$ is called a rigid motion [on $\mathbb {R}^{k}$] iff there exists $\boldsymbol {x}_{0} \in \mathbb {R}^{k}$ and an orthonormal matrix $\mathbf{O}$ such that $F(\boldsymbol {x})=\mathbf{O} \boldsymbol {x}+\boldsymbol {x}_{0}$ for all $\boldsymbol {x} \in \mathbb {R}^{k}$.

Let $\boldsymbol {a}_{1}, \cdots , \boldsymbol {a}_{k} \in \mathbb {R}^{k}$, and define

Use the main result of this section to show that $\lambda_{k}(A)$ is the absolute value of the determinant of the matrix whose columns are $\boldsymbol {a}_{1}, \cdots , \boldsymbol {a}_{k}$.

Consider a parallelogram with vertices $(0,0)$, $(a_{1}, 0)$, $(h, a_{2})$, and $(a_{1}+h, a_{2})$, where $a_{1}, a_{2}>0$ and $0 \leqslant h \leqslant a_{1}$. Find the matrix $\mathbf{T}$ that maps the unit square onto this parallelogram, and obtain the formula for the area of a parallelogram by using this section's main result.

Use Exercise 14 of Section 2.2 to give a quick proof that the Lebesgue measure of any hyperplane must be zero.

If $\mathbf{T}: \mathbb {R}^{k} \rightarrow \mathbb {R}^{k}$ is a singular linear transformation, then we do not necessarily have $\mathbf{T} A^{c}=(\mathbf{T} A)^{c}$ for every $A \subseteq \mathbb {R}^{k}$. Also, we do not necessarily have that $\left\{ \mathbf{T} A_{\gamma }: \gamma \in \Gamma \right\}$ is a disjoint collection of subsets of $\mathbb {R}^{k}$ whenever $\left\{ A_{\gamma }: \gamma \in \Gamma \right\}$ is a disjoint collection of subsets of $\mathbb {R}^{k}$.

Let $\mathbf{T}$ denote a nonsingular $k \times k$ matrix, and let $\gamma_{1} \geqslant \cdots \geqslant \gamma_{k}$ denote the [positive] eigenvalues of $\mathbf{T}' \mathbf{T}$. Then $\gamma_{k} \left\| \boldsymbol {x}\right\| \leqslant \left\| \mathbf{T} \boldsymbol {x}\right\| \leqslant \gamma_{1} \left\| \boldsymbol {x}\right\|$ for every $\boldsymbol {x} \in \mathbb {R}^{k}$. This generalizes Claim 1.

Prove Claim 7 by adapting the proof of Claim 6 accordingly.

Prove Claim 11(i) when $k=1$.

Let $T(x)=px$ for some $p \neq 0$. By Claim 3, $A \in \mathscr {B}$ forces $pA=\left\{ pa: a \in A\right\} \in \mathscr {B}$. Let $\nu (A)=\left|p\right| \lambda (A)$. Show that $\nu$ is a measure with domain $\mathscr {B}$, then show that $\nu (A)=\lambda (pA)$ whenever $A \in \mathcal{A}_{k}$. Appeal to the Uniqueness Theorem in Section 3.5 to finish things.

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 $\bar{\lambda }_{k}(A)=0$.

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 $(\boldsymbol {a}, \boldsymbol {b}]$.

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 $x, y \in (0,1]$,

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 $(0,1]$.

Prove the following assertions.

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

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.

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

If $x \in [0,1]$ has a finite ternary expansion that ends in two or more $1$'s, then $x \notin C$.

Let $x \in C$.

$1 / 4 \in C$, even though $1 / 4$ fails to be an endpoint of any removed interval $J_{d}$. Also, $0,1 \in C$.

There is a certain symmetry concerning $C$.

$\left\{ x-y: x, y \in C\right\} =[-1,1]$ and $\left\{ x+y: x, y \in C\right\} =[0,2]$, so that

That is, while $\lambda (C)=0$, both $C \oplus C$ and $C \ominus C$ have positive Lebesgue measure.

$C$ is perfect: each $x \in C$ is the limit of a sequence of distinct points of $C$.

Let $I_{C}$ be the indicator function for $C$: $I_{C}(x)$ is $1$ or $0$ as $x \in C$ or $x \notin C$. Then $I_{C}$ is continuous on $[0,1]-C$, and $I_{C}$ is discontinuous at each $x \in C$.