I am looking for a unique characterization of the Borel measure $\mu$ on $(\mathbb{R}, \mathcal{B})$, the measurable space $\mathbb{R}$ equipped with the Borel $\sigma$-algebra.

Based on this note on wikipedia , I know that $\mu$ must map intervals to their length: $\mu((a,b])=b-a$. Is this sufficient to uniquely characterize $\mu$?

I am unsure because for the Lebesgue measure $\lambda$ on $(\mathbb{R},\Sigma)$ (where $\Sigma$ contains the Lebesgue measurable sets), sufficient criterions for uniqueness given here are translation-invariance and $\lambda([0,1])=1$.


2 Answers 2


To understand your confusion (and the Wikipedia article you are referring to) one should recall why the theory of measures exists at all: After giving exact formulas for the volume of a lot of geometric objects (Archimedes calculated the volume of a sphere) people (namely Cantor, Borel and Lebesgue as far as I know) started asking how to calculate the volume of arbitrary subsets of $\mathbb R^d$.

Borel and Lebesgue wanted to find a function $\mu: \mathcal P(R^d) \to [0, \infty ]$ that satisfies the following conditions:

  • $\sigma-$additive
  • translation-invariant
  • normalized, i.e., $\mu([0,1]^d) =1 $

It turned out (Vitali, 1905) that such a general function does not exist and Banach and Tarski (1924) showed the strange things that can happen if you consider arbitrary subsets of $\mathbb R^d$ (see Banach-Tarski paradoxon for further information).

As we all know it turned out that we can define certain functions, called measures, on $\sigma-$algebras. They satisfy the following conditions

  • non-negative
  • $\sigma-$additive
  • $\mu(\emptyset)=0$

Now what about Borel and Lebesgue and their idea of finding volumes for subsets of $\mathbb R^d$. They constructed a measure called Lebesgue-(Borel-)-measure that yields the following theorem

Let $\mathcal B(\mathbb R^d)$ be the Borel-$\sigma-$algebra on $\mathbb R^d$. Then there exists exactly one measure $\lambda^d$ called the Lebesgue-(Borel-)-measure such that for all $Q=\times^d_{\alpha=1}[a_\alpha, b_ \alpha[$ with $-\infty < a_\alpha \le b_\alpha < \infty$ for all $\alpha=1, \ldots, d$ the following holds $$\lambda^d(Q) = \prod_{\alpha=1}^d(b_\alpha-a_\alpha).$$

Thats the milestone in measure-theory: Borel and Lebesgue were able to abstract the concept of volumes and created (among others) the modern measure-theory. This theory yields to a measure that is very natural in the sense that it does what we expect a volume function to do.

Since the Lebesgue-(Borel-)-measure is natural it would be nice if Lebesgue and Borle could have achieved their goal of finding a translation-invariant and normed volume function wouldn't it? And these guys were clearly happy: Their measure $\lambda^d$ is indeed translation-invariant and normalized. But even more holds:

Let $\mu$ be a measure of $\mathcal B(\mathbb R^d)$. Then the following are equivalent:

  • $\mu= \lambda^d$
  • $\mu$ is translation-invariant, i.e., $\mu(B)=\mu(x+B)$ for all $B \in \mathcal B(\mathbb R^d)$ and $x \in \mathbb R^d$ and normalized, i.e., $\mu([0,1[^d)=1$ where $[0,1[^d = \times^d_{\alpha=1}[0,1[$.

To conclude: The fact that the Lebesgue-(Borel-)-measure for hyperrectangle is just the product of their lenghts is sufficient for its uniqueness. It turns out that this measure is also translation-invariant and normalized, in fact both characterizations are equivalent.

  • $\begingroup$ I'm also trying to understand basic measure theory. Vitali showed that there were subsets of $\mathbb R^d$ that would not comply with $\sigma$-additivity, translation invariance, and normalization for any function chosen as a measure. So Borel solved this by simply excluding such subsets, or to put it the other way round, he defined the Borel sigma algebra on the topology of the reals and showed that it could have a measure function with all the desired properties. Lebesgue extended the scope of measurable sets by including sets of measure zero in the basis of the sigma algebra. Am I close ? $\endgroup$ Jun 29, 2017 at 10:38

There is actually a correspondence between Borel Measures and increasing right continuous functions $F:\mathbf{R}\to \mathbf{R}$. That is, we can associate to each increasing right continuous function $F$ a Borel Measure $\mu_F$ such that $\mu_F((a,b])=F(b)-F(a)$. This Borel Measure is unique. If $G:\mathbf{R}\to \mathbf{R}$ is another right continuous function such that $F-G$ is a constant, then $\mu_F=\mu_G$.

The converse statement says that if $\mu$ is a Borel Measure which is finite on all bounded Borel Sets, then defining $$ F(x)=\begin{cases} \mu((0,x])&x>0\\ 0 &x=0\\ \mu((x,0])&x<0 \end{cases} $$ we can see that $\mu_F=\mu$. So, if you define your function $F$ to be $$F(x)=x $$ then you recover your "usual" Borel Measure. So, yes. Knowing $\mu((a,b])=b-a$ should be enough data.


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