Uniqueness of Borel measure

Question

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 $\lambda([0,1])=1$.

Answer 1

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.

Answer 2

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.