Linear Transformations and the Scaling of Lebesgue Measure

Step 1. If $|E| = 0$, then $|\Phi(E)| = 0$. Let $\varepsilon > 0$ be arbitrary, there exist cubes $\{C_i\}$ covering $E$ such that

where $\omega_n = \frac{\pi^{n/2}}{\Gamma(n/2+1)}$ is the volume of the unit ball $B(0,1) \subseteq \mathbb{R}^n$ and $C_i = [a_i, b_i]^n$. We have

From the definition of diameter,

Since $\Phi(C_i)$ is contained in a ball of radius $\frac{\operatorname{diam} C_i}{2}$,

Since $\{\Phi(C_i)\}$ covers $\Phi(E)$,

Since $\varepsilon$ is arbitrary, $|\Phi(E)| = 0$.

Step 2. If $F \subseteq \mathbb{R}^n$ is compact, then $\Phi(F)$ is compact.

Let $(x_n) \subseteq \Phi(F)$ such that $x_n \to x \in \mathbb{R}^n$. For each $n$, choose $y_n \in F$ such that $\Phi(y_n) = x_n$. Since $F$ is compact, there is a subsequence $(y_{n_k})$ converging to $y \in F$. Then $x_{n_k} \to \Phi(y) \in \Phi(F)$. As the limit is unique, $x = \Phi(y) \in \Phi(F)$. Hence $\Phi(F)$ is compact.

Step 3. Let $H$ be an $F_\sigma$ set in $\mathbb{R}^n$; then $\Phi(H)$ is an $F_\sigma$ set in $\mathbb{R}^n$.

Write $H = \bigcup_{i=1}^\infty F_i$, where $F_i$ are closed. For each $i$, define $K_{ij} = F_i \cap \overline{B(0,j)}$, so that

Since $F_i$ is closed, $K_{ij}$ is a closed subset of the compact set $\overline{B(0,j)}$, hence compact. It follows that $\Phi(K_{ij})$ is compact for all $i,j$, and one can write

Therefore $\Phi(H)$ is $F_\sigma$.

Step 4. Let $A$ be measurable. By inner regularity of Lebesgue measure,

For each $n$, choose a compact subset $K_n \subseteq A$ such that

Set $H = \bigcup_{n=1}^\infty K_n$; then $H$ is $F_\sigma$. Since $H \subseteq A$, one has $|H| \leq |A|$, and $|H| \geq |K_n| \geq |A| - \frac{1}{n}$ for all $n$, so $|H| = |A|$. Then $|A \setminus H| = 0$, so $N = A \setminus H$ is a null set.

Step 5. Write $E = H \cup N$, where $H$ is $F_\sigma$ and $N$ is a null set. Let $A \subseteq \mathbb{R}^n$ be any subset. Since $\Phi(H) \subseteq \Phi(E)$, we have $\Phi(E)^c \subseteq \Phi(H)^c$. Since $H$ is $F_\sigma$, it follows from Step 3 that $\Phi(H)$ is $F_\sigma$, hence Borel and therefore measurable. We have the estimate

By Step 1, $\Phi$ maps null sets to null sets, so $|\Phi(N)| = 0$. Then

Since $\Phi(H)$ is measurable, combining yields

The reverse inequality follows by monotonicity of outer measure. Hence $\Phi(E)$ satisfies the Carathéodory condition.

Let

be the smallest $\lambda$-system containing $\mathcal{P}$. Then $\mathcal{P} \subseteq \mathcal{S} \subseteq \mathcal{L}$, and $\mathcal{S}$ is itself a $\lambda$-system by construction.

Claim: $\mathcal{S}$ is a $\pi$-system.

Fix any $A \in \mathcal{S}$ and define $\mathcal{A} = \{C \subseteq X \mid A \cap C \in \mathcal{S}\}$. One can verify directly that $\mathcal{A}$ is a $\lambda$-system.

Step 1: First take $A \in \mathcal{P}$. For any $P \in \mathcal{P}$, since $\mathcal{P}$ is a $\pi$-system, $A \cap P \in \mathcal{P} \subseteq \mathcal{S}$, so $P \in \mathcal{A}$. Hence $\mathcal{P} \subseteq \mathcal{A}$, and since $\mathcal{S}$ is the smallest $\lambda$-system containing $\mathcal{P}$, we get $\mathcal{S} \subseteq \mathcal{A}$. This means $A \cap C \in \mathcal{S}$ for all $A \in \mathcal{P}$ and all $C \in \mathcal{S}$.

Step 2: Now take any $A \in \mathcal{S}$. By Step 1, for any $P \in \mathcal{P}$, $A \cap P \in \mathcal{S}$, so $P \in \mathcal{A}$. Hence $\mathcal{P} \subseteq \mathcal{A}$, and again $\mathcal{S} \subseteq \mathcal{A}$. In particular, for any $B \in \mathcal{S}$, $B \in \mathcal{A}$, which means $A \cap B \in \mathcal{S}$.

Hence $\mathcal{S}$ is closed under finite intersections, i.e., a $\pi$-system.

$\mathcal{S}$ is a $\sigma$-algebra. Since $\mathcal{S}$ is a $\lambda$-system, $X \in \mathcal{S}$ and $X \setminus X = \varnothing \in \mathcal{S}$. If $A \in \mathcal{S}$, then $A^c = X \setminus A \in \mathcal{S}$. For countable unions: given $\{A_k\} \subseteq \mathcal{S}$, set $B_k = A_1 \cup \cdots \cup A_k$. Since $\mathcal{S}$ is a $\pi$-system, it is closed under finite unions (by De Morgan and closure under complements and finite intersections), so $B_k \in \mathcal{S}$. Since $B_k \nearrow \bigcup_k A_k$ and $\mathcal{S}$ is a $\lambda$-system, $\bigcup_k A_k \in \mathcal{S}$. Hence $\mathcal{S}$ is a $\sigma$-algebra.

Therefore $\sigma(\mathcal{P}) \subseteq \sigma(\mathcal{S}) = \mathcal{S} \subseteq \mathcal{L}$.

Since $A$ is compact, there exists a closed interval $[a,b]$ such that $A \subseteq [a,b] \times \mathbb{R}^{n-1}$. For each $c \in [a,b]$, denote $A_c = \{x \in \mathbb{R}^{n-1} \mid (c,x) \in A\}$.

Let $\varepsilon > 0$ be arbitrary. Since $|A_c| = 0$ in $\mathbb{R}^{n-1}$, there exist finitely many $(n-1)$-dimensional open cubes $\{C_1, \dots, C_k\}$ covering $A_c$ with $\sum_i |C_i| < \varepsilon$. Set $U_c = C_1 \cup \cdots \cup C_k$.

Claim: There exists an open interval $J_c \ni c$ such that $A \cap (J_c \times \mathbb{R}^{n-1}) \subseteq J_c \times U_c$.

Suppose not. Then there exists a sequence $(c_i, x_i) \in A$ with $c_i \to c$ and $x_i \notin U_c$. Since $A$ is compact, passing to a subsequence, $(c_i, x_i) \to (c, x)$ for some $(c,x) \in A$. In particular $x \in A_c$. But $U_c$ is open and $x_i \notin U_c$ for all $i$, so $x \notin U_c$, contradicting $A_c \subseteq U_c$. This proves the claim.

Since $\{J_c\}_{c \in [a,b]}$ is an open cover of the compact set $[a,b]$, it admits a finite subcover $\{J_{c_1}, \dots, J_{c_m}\}$. If necessary, we can shrink overlapping parts so that $\sum_k |J_{c_k}| \leq 2(b-a)$. Then

Since $\varepsilon$ is arbitrary, $|A| = 0$.

Let $V$ be a proper affine subspace of $\mathbb{R}^n$. If $V$ is empty, it trivially has measure zero. Otherwise, there exist $a = (a_1, \dots, a_n) \neq 0$ and $b \in \mathbb{R}$ such that

Fix $1 \leq i \leq n$ such that $a_i \neq 0$. For any $x \in V$,

So $V$ is the graph of the continuous function $F: \mathbb{R}^{n-1} \to \mathbb{R}$. We apply the previous lemma: for any $c \in \mathbb{R}$, the slice $V \cap (\{c\} \times \mathbb{R}^{n-1})$ consists of at most one point (since $x_i$ is uniquely determined by the remaining coordinates), which has $(n-1)$-dimensional measure zero. Since $V$ is closed and every closed set is a countable union of compact sets, and each compact slice has $(n-1)$-measure zero, it follows that $V$ has Lebesgue measure zero.

Reduction. Since $\Phi(U) = A(U) + b$ and outer Lebesgue measure is translation invariant, $|A(U) + b| = |A(U)|$. So it suffices to prove $|A(U)| = |\det(A)|\,|U|$ for all $U \subseteq \mathbb{R}^n$.

Case 1: $\det(A) = 0$. Then $\operatorname{im}(A)$ is a proper affine subspace of $\mathbb{R}^n$, which has measure zero by the previous lemma. Since $A(U) \subseteq \operatorname{im}(A)$, monotonicity gives $|A(U)| = 0 = |\det(A)|\,|U|$.

Case 2: $\det(A) \neq 0$. Then $A$ is invertible. We proceed in five steps.

Step 1: Define a candidate measure $\nu$. Let $\nu: \mathcal{B}(\mathbb{R}^n) \to [0, +\infty]$ be defined by $\nu(U) = |A(U)|$. Clearly $\nu(\varnothing) = 0$. For any countable disjoint collection $\{U_i\} \subseteq \mathcal{B}(\mathbb{R}^n)$, since $A$ is a bijection,

Hence $\nu$ is a measure on $\mathcal{B}(\mathbb{R}^n)$.

Step 2: $\nu$ is translation invariant. For any $U \in \mathcal{B}(\mathbb{R}^n)$ and $x \in \mathbb{R}^n$, since $A$ is linear,

where the middle equality uses translation invariance of outer measure.

Step 3: $\nu$ agrees with $|\det A|\,|\cdot|$ on half-open cubes.

For any half-open cube $Q = \prod_{i=1}^n [a_i, b_i)$ of side length $s$, write $Q = a + s \cdot [0,1)^n$. By Step 2, $\nu(Q) = \nu(s \cdot [0,1)^n)$. Tile $[0,1)^n$ by $m^n$ disjoint half-open cubes $\{Q_j\}$ of side $\frac{1}{m}$. By translation invariance, all $\nu(Q_j)$ are equal, so $\nu([0,1)^n) = m^n \,\nu\!\left(\left[0,\tfrac{1}{m}\right)^n\right)$. A scaling argument gives $\nu([0,s)^n) = s^n\,\nu([0,1)^n)$ for rational $s$, and monotonicity extends this to all $s > 0$. The image $A([0,1)^n)$ is a parallelepiped whose volume is $|\det A|$ by the geometric interpretation of the determinant. Hence

Step 4: Conclude $\nu(U) = |\det A|\,|U|$ on $\mathcal{B}(\mathbb{R}^n)$. Both $\nu$ and $U\mapsto|\det A|\,|U|$ are $\sigma$-finite Borel measures that agree on all half-open cubes. Half-open cubes form a $\pi$-system generating $\mathcal{B}(\mathbb{R}^n)$. Define

One checks that $\mathcal{D}$ is a $\lambda$-system: $\mathbb{R}^n \in \mathcal{D}$ by Step 3; if $U \subseteq V$ are in $\mathcal{D}$ and both have finite measure then $V \setminus U \in \mathcal{D}$ by additivity; and $\mathcal{D}$ is closed under increasing unions by monotone convergence. Since $\mathcal{D}$ contains the $\pi$-system of half-open cubes, the $\pi$-$\lambda$ theorem gives $\mathcal{B}(\mathbb{R}^n) = \sigma(\text{half-open cubes}) \subseteq \mathcal{D}$. Hence

Step 5: Extend to all subsets. For arbitrary $U \subseteq \mathbb{R}^n$,

Since $A$ is a bijection, $A$ maps Borel sets to Borel sets and $U \subseteq V \Rightarrow A(U) \subseteq A(V)$. Therefore