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A shear sends the unit cube to a set of Lebesgue measure one

Statement

Let $n \geq 2$ , assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), let $i \neq j$ be below $n$ and let $t$ be real. Let $T$ be the linear map with matrix the elementary matrix $T_{i j} (t)$ obtained from the identity by adding $t$ times row $j$ to row $i$ (Elementary matrices obtained by applying one elementary row operation to an identity matrix), so that

$T (x)_{i} = x_{i} + t x_{j}, T (x)_{l} = x_{l} (l \neq i) .$

Then $T [(0, 1]^{n}]$ is Lebesgue measurable and

$λ_{n} (T [(0, 1]^{n}]) = 1 = ∣ \det T_{i j} (t) ∣ .$

Facts & Assumptions

Given: A natural number $n \geq 2$ , the Axiom of Countable Choice, distinct indices $i, j < n$ , a real $t$ , and the shear $T$ with matrix $T_{i j} (t)$ .

[L1]

$λ_{n} (E + h) = λ_{n} (E)$ for every Lebesgue measurable $E$ and every $h$ (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation, Translation of a subset of $R^{n}$ ).

[L2]

An invertible linear map carries Borel sets to Borel sets and open sets to open sets (An invertible linear map of $R^{n}$ scales the Lebesgue measure of every Borel set by a positive constant depending only on the map, claim 1).

[L3]

Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable), and $λ_{n}$ is a complete measure on $L (R^{n})$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume).

[L4]

For real parameters $a_{l} \leq b_{l}$ , every set $R$ between the open box $R^{\circ}$ and closed box $\overline{R}$ is Lebesgue measurable with $λ_{n} (R) = \prod_{l < n} (b_{l} - a_{l})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included), and $(u, v]^{n} : = B (u, v)$ (Half-open boxes in $R^{n}$ and their volume).

[F1]

An elementary matrix is a matrix obtained by applying one elementary row operation to $I_{n}$ ; $T_{p q} (c)$ adds $c$ times row $q$ to the distinct row $p$ (Elementary matrices obtained by applying one elementary row operation to an identity matrix, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes), and for every linear map there is a unique such matrix acting by $(A x)_{i} = \sum_{l < n} a_{i l} x_{l}$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

[F2]

Adding $c$ times one row to a distinct row leaves the determinant equal to $\det (A)$ (For every square matrix, including singular ones, a row swap negates the determinant, scaling a row by any scalar scales it, and row addition leaves it unchanged, claim 3; For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix), and a triangular matrix has determinant the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[F3]

For every real $x$ there is exactly one integer $p$ with $p \leq x < p + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F4]

A measure is countably additive on pairwise disjoint measurable sequences, hence finitely additive after padding with empty sets (Measures on sigma-algebras).

[F5]

A subset $U$ is open in $(X, d)$ when every $x \in U$ has a ball $B (x, r) \subseteq U$ , a subset is closed when its complement is open, and a finite intersection of open sets is open (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space, Arbitrary unions and finite intersections of open sets are open, open balls are open and closed balls are closed, claim 3).

[F6]

For every $x \in R^{n}$ , $∥ x ∥_{\infty} \leq ∥ x ∥_{2}$ , and $∥ x - y ∥_{2} = d_{2} (x, y)$ , $∥ x - y ∥_{\infty} = d_{\infty} (x, y)$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ , claim 3; Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, claim 3; The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ ; $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[F7]

The Borel sigma-algebra is the sigma-algebra generated by the open sets, and a sigma-algebra is closed under complements and countable unions (The Borel sigma-algebra of a topological space, Sigma-algebras); in particular every open and every closed subset of $R^{n}$ is Borel.

Proof

technique · direct

1.1L2F1F2

The matrix $T_{i j} (t)$ is obtained from the identity by a row addition, so $\det T_{i j} (t) = \det I_{n} = 1$ and $T$ is invertible, with inverse the shear $T_{i j} (- t)$ ; consequently $T$ carries Borel sets to Borel sets.

1.2F3

For every real $s$ there is exactly one integer $k$ with $k < s \leq k + 1$ : applying the integer part to $- s$ gives the unique integer $p$ with $p \leq - s < p + 1$ , and $k : = - p - 1$ is the integer sought, uniqueness following the same way.

1.3F1F5F6F7

The linear functional $L (x) : = x_{i} + t x_{j}$ satisfies $∣ L (x) - L (y) ∣ \leq (1 + ∣ t ∣) d_{2} (x, y)$ , so for every real $s$ the set ${x : L (x) > s}$ is open and ${x : L (x) \leq s}$ is closed. Also $(0, 1]^{n} = [0, 1]^{n} \cap ⋂_{l < n} {x : x_{l} > 0}$ , with $[0, 1]^{n}$ closed and each ${x : x_{l} > 0}$ open, hence Borel by [F7]; therefore each $A_{k} : = (0, 1]^{n} \cap {x : L (x) > k} \cap {x : L (x) \leq k + 1}$ is a Borel set.

1.4F3

Only finitely many integers $k$ admit a point of $A_{k}$ : for $x \in (0, 1]^{n}$ one has $- ∣ t ∣ \leq L (x) \leq 1 + ∣ t ∣$ , so $k < 1 + ∣ t ∣$ and $k + 1 > - ∣ t ∣$ , and the integers satisfying both lie between the two integers supplied by the integer part of $- ∣ t ∣ - 1$ and of $1 + ∣ t ∣$ , hence form a finite consecutive list $K_{-}, \dots, K_{+}$ .

2.1step 1.2step 1.4

By step 1.2 every $x \in (0, 1]^{n}$ lies in exactly one $A_{k}$ , so the sets $A_{k}$ for $k$ in the list of step 1.4 are pairwise disjoint with union $(0, 1]^{n}$ .

3.1step 1.2step 2.1F1

Define $Φ : (0, 1]^{n} \to R^{n}$ by $Φ (x) : = T (x) - k e_{i}$ for the unique $k$ with $x \in A_{k}$ , where $e_{i}$ is the $i$ -th standard vector. Then $Φ$ takes values in $(0, 1]^{n}$ , since its $i$ -th coordinate is $L (x) - k \in (0, 1]$ and its other coordinates are those of $x$ .

4.1step 1.2step 3.1

$Φ$ is a bijection of $(0, 1]^{n}$ onto itself. It is injective: if $Φ (x) = Φ (y)$ then $x_{l} = y_{l}$ for every $l \neq i$ , so $x_{j} = y_{j}$ and $x_{i} - y_{i} = k (x) - k (y)$ is an integer of absolute value below $1$ , hence $0$ . It is surjective: given $z \in (0, 1]^{n}$ , step 1.2 supplies the unique integer $k$ with $z_{i} - t z_{j} + k \in (0, 1]$ ; setting $x_{l} : = z_{l}$ for $l \neq i$ and $x_{i} : = z_{i} - t z_{j} + k$ gives $x \in (0, 1]^{n}$ with $L (x) = z_{i} + k \in (k, k + 1]$ , so $x \in A_{k}$ and $Φ (x) = z$ .

5.1step 1.1step 1.3step 2.1step 3.1step 4.1L1L3

The sets $T [A_{k}]$ are pairwise disjoint, Borel and have union $T [(0, 1]^{n}]$ , because $T$ is an injective linear bijection; each $T [A_{k}] - k e_{i} = Φ [A_{k}]$ , so translation invariance gives $λ_{n} (T [A_{k}]) = λ_{n} (Φ [A_{k}])$ ; and by step 4.1 the sets $Φ [A_{k}]$ are pairwise disjoint with union $(0, 1]^{n}$ .

6.1step 1.1step 5.1L3L4F4∎

Finite additivity applied twice therefore gives $λ_{n} (T [(0, 1]^{n}]) = \sum_{k} λ_{n} (T [A_{k}]) = \sum_{k} λ_{n} (Φ [A_{k}]) = λ_{n} ((0, 1]^{n}) = 1$ , which with step 1.1 is the Statement.

Depends on

Used by

A linear map T of ℝⁿ sends Lebesgue measurable sets to Lebesgue measurable sets, with λₙ(T[E])=|det T| λₙ(E) when T is invertible and T[E] Lebesgue null when it is not Theorem

Dependency tree · two levels

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Sources

E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Section 3 (standard reference, not scraped)
John K. Hunter, Measure Theory (UC Davis lecture notes), Chapter 2 (standard reference, not scraped)