Uniqueness of Lebesgue Measure

Undergraduates with some exposure to measure theory may know that the Lebesgue measure is the unique, up to multiplication by a constant, translation-invariant Borel measure. If not, then no worries, as today’s post will cover the proof of this result. My motivation for selecting this topic for a blog post was I was trying to remember the proof that Lebesgue measure is invariant under transformation by an orthogonal matrix. The proof of this invariance result, along with a more general result about the behavior of the Lebesgue measure under linear transformations, will be proven in a later post.

Let $X$ be a set. A family $\mathcal{D}\subset\mathcal{P}(X)$ is a Dynkin system if

1. $X\in\mathcal{D}$
2. $D\in\mathcal{D}\Longleftrightarrow D^{c}\in\mathcal{D}$
3. $(D_{j})_{j \in \mathbb{N}}\subset D$ pairwise disjoint $\Rightarrow\bigcup_{j\in\mathbb{N}}D_{j}\in\mathcal{D}$.

It is immediate from the definition that $\emptyset\in\mathcal{D}$, and $\mathcal{D}$ is closed under finite disjoint union. It is evident that every $\sigma$-algebra is a Dynkin system, but the following example shows the converse is false:

Let $X=\left\{1,\cdots,2k\right\}\subset\mathbb{N}$ for some fixed $k\in\mathbb{N}$. Set $\mathcal{D}=\left\{A\subset X:\#{X} \mod{2}=0\right\}$. That $\mathcal{D}$ is a Dynkin system contains $X$ and is closed under countable disjoint union is obvious. For closure under relative complement, $\#{X}-\#{A}=\#{A}^{c}$ is the difference of two even numbers and therefore even. Evidently, $\mathcal{D}$ is not a $\sigma$-algebra, since $\left\{1,2\right\},\left\{1,3\right\}\in\mathcal{D}$, but $\left\{1,2\right\}\cup\left\{1,3\right\}=\left\{1,2,3\right\}\notin\mathcal{D}$.

Proposition 1. Let $\mathcal{G}\subset\mathcal{P}(X)$. Then there is a smallest (also minimal, coarset) Dynkin system $\delta(\mathcal{G})$ containing $\mathcal{G}$. $\delta(\mathcal{G})$ is called the Dynkin system generated by $\mathcal{G}$. Moreover, $\mathcal{G}\subset\delta(\mathcal{G})\subset \sigma\left(\mathcal{G}\right)$.

Proof. Define $\delta(\mathcal{G})$ to be the intersection of all Dynkin systems containing $G$. Then $\delta(\mathcal{G})$ exists and is non-empty since $\mathcal{P}(X)$ is a Dynkin system and $\mathcal{G}\subset\delta(\mathcal{G})$. That $\delta(G)$ is minimal is immediate from the definition. Since every $\sigma$-algebra is a Dynkin system and $\sigma(\mathcal{G})\supset\mathcal{G}$, we have that $\delta(\mathcal{G})\subset\sigma(\mathcal{G})$. $\Box$

Lemma 2. A Dynkin system $\mathcal{D}$ is a $\sigma$-algebra if and only if it is closed under finite intersection: $D,E \in\mathcal{D}\Rightarrow D\cap E\in\mathcal{D}$.

Proof. If $\mathcal{D}$ is a $\sigma$-algebra, then a fortiori it is closed under finite intersection. Suppose $\mathcal{D}$ is closed under finite intersection. Let $(D_{j})_{j\in\mathbb{N}}$ be an arbitrary sequence in $\mathcal{D}$. Define

$\displaystyle\tilde{D}_{j}=D_{j}\setminus\bigcup_{i < j}D_{i}$

By our $\bigcap$-stable hypothesis, $\tilde{D}_{j}\in\mathcal{D}$ for each $j$, and $(\tilde{D}_{j})_{j \in\mathbb{N}}\subset\mathcal{D}$ is pairwise disjoint. Hence, $\bigcup_{j\in\mathbb{N}}D_{j}=\bigcup_{j\in\mathbb{N}}\tilde{D}_{j}\in\mathcal{D}$. So $\mathcal{D}$ is closed under countable union, and therefore a $\sigma$-algebra. $\Box$

Proposition 3. If $\mathcal{G}\subset\mathcal{P}(X)$ is stable under finite intersections, then $\delta(\mathcal{G})=\sigma(\mathcal{G})$.

Proof. Tautologically, $\delta(\mathcal{G})\subset \sigma(\mathcal{G})$, so by definition of $\sigma(\mathcal{G})$, it suffices to show that $\delta(\mathcal{G})$ is a $\sigma$-algebra. By the preceding lemma, it suffices to show that $\delta(\mathcal{G})$ is stable under finite intersection. Fix $D\in\delta(\mathcal{G})$ and define the family

$\displaystyle\mathcal{D}_{D}=\left\{Q\in X:Q\cap D\in \delta(\mathcal{G})\right\}$

I claim that $\mathcal{D}_{D}$ is a Dynkin system. $X\cap D=D\in\delta(\mathcal{G})\Rightarrow X\in\mathcal{D}_{D}$. Observe that if $Q\cap D\in\delta(\mathcal{G})$, then

$\displaystyle Q^{c}\cap D=\left(Q^{c}\cup D^{c}\right)\cap D=\left(Q\cap D\right)^{c}\cap D=\left(\underbrace{(Q \cap D)}_{\in\delta(\mathcal{G})}\cup\underbrace{D^{c}}_{\in\delta(\mathcal{G})}\right)^{c}\in\delta(\mathcal{G})$

by the Dynkin axioms applied to $\delta(\mathcal{G})$. Closure under countable disjoint union is obvious. Observe that since $\mathcal{G}$ is $\bigcap$-stable, $\mathcal{G}\subset\mathcal{D}_{G}\forall G\in\mathcal{G}$. Since $\mathcal{D}_{G}$ is a Dynkin system, we have that $\delta(\mathcal{G})\subset\mathcal{D}_{G}$, when $G\in\mathcal{G}$. Hence,

$\displaystyle D\in\delta(\mathcal{G}),G\in\mathcal{G}\Rightarrow D\cap G\in\delta(\mathcal{G})\Rightarrow\mathcal{G}\subset\mathcal{D}_{D}\Rightarrow\delta(\mathcal{G})\subset\mathcal{D}_{D}$

which is equivalent to saying that $\delta(\mathcal{G})$ is closed under finite intersections. $\Box$

The proof of the following proposition is obvious from our preceding work.

Proposition 4.Let $\mathcal{A}$ be a $\sigma$-algebra, $\mathcal{D}$ be a Dynkin system, and $\mathcal{G}\subset\mathcal{H} \subset\mathcal{P}(X)$ be two collections of subsets of $X$. Then

1. $\delta(\mathcal{A})=\mathcal{A};\delta(\mathcal{D})=\mathcal{D}$
2. $\delta(\mathcal{G})\subset\delta(\mathcal{H})$
3. $\delta(\mathcal{G})\subset\sigma(\mathcal{G})$

The next proposition is a useful tool for checking the equivalence of two measures on a given measurable space.

Proposition 5. (Uniqueness of Measures) Assume that $(X,\mathcal{A})$ is a measurable space and that $\mathcal{A}=\sigma(\mathcal{G})$ is generated by a family $\mathcal{G}$ such that

1. $G,H\in\mathcal{G}\Rightarrow G\cap H\in\mathcal{G}$
2. There exists an exhausting sequence $(G_{j})_{j\in\mathbb{N}}\subset\mathcal{G}$ with $G_{j}\nearrow X$

Then any two measures $\mu,\nu$ that coincide on $\mathcal{G}$ and satisfy $\mu(G_{j})=\nu(G_{j})<\infty$ are equal on $\mathcal{A}$ (i.e. $\mu(A)=\nu(A)\forall A\in\mathcal{A}$).

Proof. Let $\mu,\nu$ be two measures that coincide on $\mathcal{G}$ and satisfy $\mu(G_{j})=\nu(G_{j})<\infty$ for each $j$. For each $j\in\mathbb{N}$, define

$\displaystyle\mathcal{D}_{j}=\left\{A\in\mathcal{A}:\mu(G_{j}\cap A)=\nu(G_{j}\cap A)<\infty\right\}$

I claim that $\mathcal{D}_{j}$ is a Dynkin system. Clearly, $X\in\mathcal{D}_{j}$, and if $A\in\mathcal{D}_{j}$, then by additivity,

$\displaystyle\mu(G_{j}\cap A^{c})=\mu(G_{j})-\mu(G_{j}\cap A)=\nu(G_{j})-\nu(G_{j}\cap A)=\nu(G_{j}\cap A^{c})\Rightarrow A^{c}\in\mathcal{D}_{j}$

If $(A_{k})_{k\in\mathbb{N}}$ is pairwise disjoint, then using that $\mu(G_{j})=\nu(G_{j})<\infty$,

$\begin{array}{lcl}\displaystyle\mu\left(G_{j}\cap\bigcup_{k=1}^{\infty}A_{k}\right)=\sum_{k=1}^{\infty}\mu(G_{j}\cap A_{k})&=&\displaystyle\sum_{k=1}^{\infty}\nu(G_{j}\cap A_{k})\\[.9 em]&=&\displaystyle\nu\left(G_{j}\cap\bigcup_{k=1}^{\infty}A_{k}\right)<\infty\\[.9 em]&\Rightarrow&\displaystyle\bigcup_{k=1}^{\infty}A_{k}\in\mathcal{D}_{j}\end{array}$

Obviously $\mathcal{D}_{j}\supset\mathcal{G}$ by hypothesis that $\mathcal{G}$ is $\bigcap$-stable and that $\mu,\nu$ agree on $\mathcal{G}$. Hence, $\delta(\mathcal{G})\subset\mathcal{D}_{j}$ for each $j$. Since $\mathcal{G}$ is $\bigcap$-stable, by Proposition 3, $\delta(\mathcal{G})=\sigma(\mathcal{G})=\mathcal{A}$. Hence, $\mathcal{D}_{j}=\mathcal{A}\forall j\in\mathbb{N}$. Since $G_{j}\nearrow X$, for every $A\in\mathcal{A}$,

$\displaystyle\mu(A)=\lim_{j\rightarrow\infty}\mu(G_{j}\cap A)=\lim_{j\rightarrow\infty}\nu(G_{j}\cap A)=\nu(A)$

We denote the Lebesgue measure on $\mathbb{R}^{n}$ by $\lambda_{n}$

Theorem 6. (Lebesgue Translation-Invariance)

The $n$-dimensional Lebesgue measure $\lambda_{n}$ is invariant under translations:

$\displaystyle\lambda_{n}(x+B)=\lambda_{n}(B), \indent\forall B\in\mathcal{B}(\mathbb{R}^{n})$

Every measure $\mu$ on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ which is invariant under translations and satisfies $\kappa=\mu\left([0,1)^{n}\right)<\infty$ is a real multiple of the Lebesgue measure: $\mu = \kappa\lambda_{n}$.

Proof. We first verify that $x+B$ is indeed a Borel set. Define

$\displaystyle\mathcal{A}_{x}=\left\{B\in\mathcal{B}(\mathbb{R}^{n}):x+B\in\mathcal{B}(\mathbb{R}^{n})\right\}\subset \mathcal{B}(\mathbb{R}^{n})$

I claim that $\mathcal{A}_{x}$ is a $\sigma$-algebra. It is clear that $\emptyset,X\in\mathcal{A}_{x}$, and $\mathcal{A}_{x}$ is closed under countable union. For closure under relative complement, I claim that $\left(x+B\right)^{c}=x+B^{c}$. Indeed,

$\displaystyle y\in(x+B)^{c}\Longleftrightarrow y\neq x+b,b\in B \Longleftrightarrow y-x\in B^{c}\Longleftrightarrow y=x+(y-x)\in x+B^{c}$

Since translation is a homeomorphism, we have that $\mathcal{O}\subset\mathcal{A}_{x}$, where $\mathcal{O}$ is the collection of open (in the Euclidean topology) subsets of $\mathbb{R}^{n}$. By definition, $\mathcal{A}_{x}\supset\mathcal{B}(\mathbb{R}^{n})$, which implies that $\mathcal{A}_{x}=\mathcal{B}(\mathbb{R}^{n})$.

For $B \in\mathcal{B}(\mathbb{R}^{n})$, set $\nu(B)=\lambda_{n}\left(x+B\right)$ for $x=(x_{1},\cdots,x_{n})\in\mathbb{R}^{n}$ fixed. It is clear that $\nu$ is a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$. Let $\mathcal{I}$ denote the collection of half-open rectangles in $\mathbb{R}^{n}$. Let $I=[a_{1},b_{1})\times\cdots\times[a_{n},b_{n})$ and observe that

$\displaystyle x+I=[a_{1}+x_{1},b_{1}+x_{1})\times\cdots\times [a_{n}+x_{n},b_{n}+x_{n})\in\mathcal{I}$

Hence,

$\displaystyle\nu(I)=\lambda_{n}(x+I)=\prod_{j=1}^{n}\left((b_{j}+x_{j})-(a_{j}+x_{j})\right)=\prod_{j=1}^{n}\left(b_{j}-a_{j}\right)=\lambda_{n}(I)$

which shows that $\nu|_{\mathcal{I}}=\lambda_{n}|_{\mathcal{I}}$. Since $\mathcal{I}$ is $\bigcap$-stable, generates $\mathcal{B}(\mathbb{R}^{n})$, and has an exhausting sequence $\left([-k,k]^{j}\right)_{j\in\mathbb{N}}$, we have by Proposition 5 that $\nu=\lambda_{n}\Longleftrightarrow \lambda_{n}(x + B)=\lambda_{n}(B)\forall B\in\mathcal{B}(\mathbb{R}^{n})$.

Let $\mu$ be a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ which satisfies $\kappa=\mu\left([0,1)^{n}\right)<\infty$ and is translation invariant. Observe that, if $q$ is a positive integer, then by additivity of $\mu$, we see that

$\begin{array}{lcl}\displaystyle\kappa=\sum_{j_{1}=1}^{q}\cdots\sum_{j_{n}}^{q}\mu\left(\prod_{i=1}^{n}\left[\frac{j_{i}-1}{q},\frac{j_{i}}{q}\right)\right)&=&\displaystyle\sum_{j_{1}=1}^{q}\cdots\sum_{j_{n}=1}^{q}\mu\left(\left(\frac{j_{1}-1}{q},\ldots,\frac{j_{n}-1}{q}\right)+\left[0,\frac{1}{q}\right)^{n}\right)\\[.9 em]&=&\displaystyle q^{n}\mu\left(\left[0,\frac{1}{q}\right)^{n}\right)\end{array}$,

which implies that $\mu([0,\frac{1}{q})^{n})=\frac{\kappa}{q^{n}}$. If $p_{1},\ldots,p_{n}\geq0$ are positive integers, then

$\begin{array}{lcl}\displaystyle\mu\left(\prod_{i=1}^{n}\left[0,\frac{p_{i}}{q}\right)\right)=p_{1}\cdots p_{n}\cdot\mu\left(\left[0,\frac{1}{q}\right)^{n}\right)=\dfrac{p_{1}\cdots p_{n}}{q^{n}}\cdot\kappa&=&\displaystyle\dfrac{p_{1}\cdots p_{n}}{q^{n}}\cdot\kappa\cdot\lambda_{n}\left(\left[0,1\right)^{n}\right)\\[.9 em]&=&\displaystyle p_{1}\cdots p_{n}\cdot\kappa\cdot\lambda_{n}\left(\left[0,\frac{1}{q}\right)^{n}\right)\\[.9 em]&=&\displaystyle\kappa\cdot\lambda_{n}\left(\prod_{i=1}^{n}\left[0,\frac{p_{i}}{q}\right)\right),\end{array}$

where we use the additivity of measures. It follows from the translation-invariance of $\mu$ that, for half-open rectangle $I\in\mathcal{I}$ as above, but with $a_{j}, b_{j}\in\mathbb{Q}$ for $1\leq j\leq n$, the formula

$\displaystyle\mu(I)=\kappa\lambda_{n}(I)$

holds. By Proposition 5, $\mu=\kappa\lambda_{n}$. $\Box$