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}



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


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

This entry was posted in math.CA, math.FA and tagged , , , , , . Bookmark the permalink.

One Response to Uniqueness of Lebesgue Measure

  1. Pingback: Lebesgue Measure and Linear Transformations | Math by Matt

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