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The Lebesgue measure of the image of the unit cube under an explicit linear map of the plane and of three-space

Example

Assume the Axiom of Countable Choice. Let

$T (x, y) : = (2 x + y, x + 3 y), U (x, y, z) : = (2 x + y, 3 y - z, 4 z), S (x, y) : = (x, x) .$

Then the image of the unit square $(0, 1]^{2}$ under $T$ has Lebesgue measure $5$ , the image of the unit cube $(0, 1]^{3}$ under $U$ has Lebesgue measure $24$ , and the image of $(0, 1]^{2}$ under the singular map $S$ is Lebesgue null.

Facts & Assumptions

Given: The Axiom of Countable Choice and the three linear maps $T$ , $U$ and $S$ above.

[L1]

Assuming countable choice, a linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not).

[F1]

Its determinant is the Leibniz sum over permutations (For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F2]

If $A = (a_{i j}) \in M_{n} (R)$ is upper or lower triangular over a commutative ring, then $\det (A) = \prod_{i} a_{i i}$ (The determinant of a triangular matrix is the product of its diagonal entries).

[L2]

Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null (Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null).

Verification

technique · direct

1.1L1F1algebra

The matrix of $T$ is $(\begin{array}{cc} 2 & 1 \\ 1 & 3 \end{array})$ , so [F1] gives $\det T = 2 \cdot 3 - 1 \cdot 1 = 5$ ; therefore [L1] gives $λ_{2} (T [(0, 1]^{2}]) = 5 λ_{2} ((0, 1]^{2}) = 5$ .

1.2L1F1F2algebra

The matrix of $U$ is the upper triangular matrix $(\begin{array}{ccc} 2 & 1 & 0 \\ 0 & 3 & - 1 \\ 0 & 0 & 4 \end{array})$ , so [F2] gives $\det U = 2 \cdot 3 \cdot 4 = 24$ ; hence [L1] gives $λ_{3} (U [(0, 1]^{3}]) = 24 λ_{3} ((0, 1]^{3}) = 24$ .

2.1L1L2F1algebra∎

The matrix of $S$ is $(\begin{array}{cc} 1 & 0 \\ 1 & 0 \end{array})$ , so [F1] gives $\det S = 0$ , and $S [(0, 1]^{2}] = {(t, t) : 0 < t \leq 1}$ lies in the proper linear subspace ${(u, v) : u = v}$ ; therefore [L1] and [L2] give that $S [(0, 1]^{2}]$ is Lebesgue null.

Depends on

Used by

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Dependency tree · two levels

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