Proof of Translation-Invariant Measures on ℝⁿ Are Multiples of Lebesgue Measure

Throughout, $\lambda_{n}$ is a measure with

$$\lambda_{n}(R(a,b))=\prod_{i=1}^{n}(b_{i}-a_{i})\qquad\text{whenever }a_{i}<b_{i}\text{ for every }i\in[n]$$

(Lebesgue Measure on $\mathbb{R}^n$ (i) and Existence and Uniqueness of Lebesgue Measure on ℝⁿ (iii)), open boxes are Borel sets (Existence and Uniqueness of Lebesgue Measure on ℝⁿ (i)), and $B+a\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(B+a)=\lambda_{n}(B)$ for $B\in\mathcal{B}(\mathbb{R}^{n})$ and $a\in\mathbb{R}^{n}$ (Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ (i)). For $t\in\mathbb{R}$, $t\mathbf{u}=(t,\dots,t)$. For $a,b\in\mathbb{R}^{n}$ let

$$H(a,b)=\{x\in\mathbb{R}^{n}:a_{i}\le x_{i}<b_{i}\text{ for every }i\in[n]\}$$

be the half-open box with corners $a$ and $b$, and let $Q=H(0,\mathbf{u})$ be the half-open unit cube.

Step 1: half-open boxes. For $a,b\in\mathbb{R}^{n}$,

$$H(a,b)=\bigcap_{m\in\mathbb{N}}R(a-m^{-1}\mathbf{u},\,b),$$

since a point with $x_{i}<a_{i}$ violates $a_{i}-m^{-1}<x_{i}$ as soon as $m^{-1}<a_{i}-x_{i}$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)). Hence $H(a,b)\in\mathcal{B}(\mathbb{R}^{n})$. Moreover $H(a,b)+t=H(a+t,b+t)$ for $t\in\mathbb{R}^{n}$, and $H(a,b)=\emptyset$ when $b_{i}\le a_{i}$ for some $i$.

Step 2: splitting a box into translates. Let $\rho$ be a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ with $\rho(B+t)=\rho(B)$ for all $B\in\mathcal{B}(\mathbb{R}^{n})$ and $t\in\mathbb{R}^{n}$; both $\mu$ and $\lambda_{n}$ are such measures. Let $h\in\mathbb{R}$ be positive, $a\in\mathbb{R}^{n}$, $m_{1},\dots,m_{n}\in\mathbb{N}$, and $b_{i}=a_{i}+m_{i}h$ for $i\in[n]$. We show

$$\rho(H(a,b))=m_{1}m_{2}\cdots m_{n}\,\rho(H(0,h\mathbf{u})),$$

the products in $[0,\infty]$. For $i\in\{0,\dots,n\}$ let $b^{[i]}\in\mathbb{R}^{n}$ have coordinates $a_{l}+h$ for $l\le i$ and $b_{l}$ for $l>i$, and let $P_{i}=H(a,b^{[i]})$; thus $P_{0}=H(a,b)$ and $P_{n}=H(0,h\mathbf{u})+a$. Fix $i\in[n]$ and let $e_{i}$ be the $i$th standard basis vector. For $r\in[m_{i}]$ the translate $P_{i}+(r-1)he_{i}$ consists of the points $x\in P_{i-1}$ with

$$a_{i}+(r-1)h\le x_{i}<a_{i}+rh .$$

Every $x\in P_{i-1}$ lies in exactly one of these translates: $r-1$ must be the integer part of $(x_{i}-a_{i})/h$, which lies in $\{0,\dots,m_{i}-1\}$ because $0\le(x_{i}-a_{i})/h<m_{i}$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iv) and The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (vi)). So $P_{i-1}$ is the disjoint union of the $m_{i}$ translates, and finite additivity and invariance give

$$\rho(P_{i-1})=\sum_{r=1}^{m_{i}}\rho\bigl(P_{i}+(r-1)he_{i}\bigr)=m_{i}\,\rho(P_{i}).$$

By induction on $i$, $\rho(P_{0})=m_{1}\cdots m_{n}\,\rho(P_{n})=m_{1}\cdots m_{n}\,\rho(H(0,h\mathbf{u}))$.

Step 3: the two constants. Put $q=\mu(Q)$ and $\ell=\lambda_{n}(Q)$. The cube $Q$ has all sides $2\cdot\frac12$, so Step 2 with $h=\frac12$ and $m_{i}=2$, together with invariance, gives

$$q=2^{n}\mu\bigl(H(0,\tfrac12\mathbf{u})\bigr)=2^{n}\mu\bigl(H(\tfrac12\mathbf{u},\mathbf{u})\bigr)\le2^{n}c<\infty,$$

because $H(\frac12\mathbf{u},\mathbf{u})\subseteq R(0,\mathbf{u})$. From $R(0,\mathbf{u})\subseteq Q\subseteq R(-\mathbf{u},\mathbf{u})$ we get $1\le\ell\le2^{n}$. Thus $q$ and $\ell$ are finite and $\ell>0$; put $\kappa=q/\ell\ge0$.

Step 4: boxes with dyadic sides. For $k\in\mathbb{N}_{0}$ put

$$w_{k}=\mu\bigl(H(0,2^{-k}\mathbf{u})\bigr),\qquad v_{k}=\lambda_{n}\bigl(H(0,2^{-k}\mathbf{u})\bigr);$$

as these boxes lie in $Q$, $w_{k}\le q$ and $v_{k}\le\ell$ are real. The box $H(0,2^{-k}\mathbf{u})$ has all sides $2\cdot2^{-(k+1)}$, so Step 2 gives $w_{k}=2^{n}w_{k+1}$ and $v_{k}=2^{n}v_{k+1}$. Since $w_{0}=q=\kappa\ell=\kappa v_{0}$, induction on $k$ yields $w_{k}=\kappa v_{k}$ for every $k\in\mathbb{N}_{0}$.

Now let $a,b\in\mathbb{R}^{n}$ and $k\in\mathbb{N}_{0}$ be such that $b_{i}-a_{i}=m_{i}2^{-k}$ with $m_{i}\in\mathbb{Z}$ for every $i\in[n]$. If $m_{i}\le0$ for some $i$, then $H(a,b)=\emptyset$. Otherwise Step 2 applies to $\mu$ and to $\lambda_{n}$ with $h=2^{-k}$, and

$$\mu(H(a,b))=m_{1}\cdots m_{n}\,w_{k}=\kappa\,m_{1}\cdots m_{n}\,v_{k}=\kappa\,\lambda_{n}(H(a,b)).$$

In either case

$$\mu(H(a,b))=\kappa\,\lambda_{n}(H(a,b)).\tag{$*$}$$

Step 5: open boxes. Let $a,b\in\mathbb{R}^{n}$. If $b_{i}\le a_{i}$ for some $i$, then $R(a,b)=\emptyset$ (Open Boxes: Emptiness, Intersections and Preimages under Concatenation (i)) and $\mu(R(a,b))=0=\kappa\lambda_{n}(R(a,b))$. Otherwise, with $\lfloor z\rfloor$ the integer part of $z\in\mathbb{R}$, define for $k\in\mathbb{N}$ points $a^{(k)},b^{(k)}\in\mathbb{R}^{n}$ by

$$a^{(k)}_{i}=2^{-k}\bigl(\lfloor2^{k}a_{i}\rfloor+1\bigr),\qquad b^{(k)}_{i}=2^{-k}\lfloor2^{k}b_{i}\rfloor\qquad(i\in[n]),$$

and put $H_{k}=H(a^{(k)},b^{(k)})$. The sides of $H_{k}$ are integer multiples of $2^{-k}$, so $(*)$ holds for $H_{k}$. Since $z-1<\lfloor z\rfloor\le z$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iv) and The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (v)),

$$a_{i}<a^{(k)}_{i}\le a_{i}+2^{-k},\qquad b_{i}-2^{-k}<b^{(k)}_{i}\le b_{i},$$

so $H_{k}\subseteq R(a,b)$. Since $2\lfloor z\rfloor\le\lfloor2z\rfloor\le2\lfloor z\rfloor+1$ (The Integer Part as the Greatest Integer Below, the Integer Part of a Double, and Powers of Two (ii)), applied to $z=2^{k}a_{i}$ and $z=2^{k}b_{i}$, we have $a^{(k+1)}_{i}\le a^{(k)}_{i}$ and $b^{(k)}_{i}\le b^{(k+1)}_{i}$, so $H_{k}\subseteq H_{k+1}$. Finally, let $x\in R(a,b)$ and let $\delta>0$ be the least of the numbers $x_{i}-a_{i}$ and $b_{i}-x_{i}$ ($i\in[n]$). Choose $k\in\mathbb{N}$ with $1/k<\delta$ (The Integers in the Real Numbers: Greatest Elements, the Integer Part, and Archimedean Consequences (iii)); then $2^{-k}<1/k<\delta$ (The Integer Part as the Greatest Integer Below, the Integer Part of a Double, and Powers of Two (iii)), and

$$a^{(k)}_{i}\le a_{i}+2^{-k}<x_{i}<b_{i}-2^{-k}<b^{(k)}_{i}\qquad(i\in[n]),$$

so $x\in H_{k}$. Thus $R(a,b)$ is the union of the increasing sequence $(H_{k})$, and continuity from below and $(*)$ give

$$\mu(R(a,b))=\sup_{k}\mu(H_{k})=\sup_{k}\kappa\,\lambda_{n}(H_{k})=\kappa\sup_{k}\lambda_{n}(H_{k})=\kappa\,\lambda_{n}(R(a,b))=\kappa\prod_{i=1}^{n}(b_{i}-a_{i})<\infty.$$

Step 6: conclusion. The function $\nu=\kappa\lambda_{n}$, $B\mapsto\kappa\,\lambda_{n}(B)$, is a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ (Integrals of Finite Sums of Nonnegative Functions, and Constant Multiples of a Measure (ii)). By Step 5, $\mu$ and $\nu$ agree on every open box and are finite there, so $\mu=\nu$ by Existence and Uniqueness of Lebesgue Measure on ℝⁿ (iv). In particular $c=\mu(R(0,\mathbf{u}))=\kappa\,\lambda_{n}(R(0,\mathbf{u}))=\kappa$, and therefore $\mu(B)=c\,\lambda_{n}(B)$ for every $B\in\mathcal{B}(\mathbb{R}^{n})$.