# Lebesgue Stieltjes measure unique for invariance on $\mathscr{B}_\mathbb{R}$

Exercsise: Let $$\mu$$ be a Lebesgue-Stieltjes measure on $$\mathscr{B}_{\mathbb{R}}$$ invariant for the class of right half-closed intervals of $$\mathbb{R}$$, so that, $$\mu(a+I)=\mu(I)$$, for all $$a\in\mathbb{R}$$ and $$I=(x,y]$$. Show that, in $$\mathscr{B}_\mathbb{R}$$, $$\mu=c.Leb$$ where c\in$$\mathbb{R}$$ and Leb denotes the Lebesgue measure.

Attempted resolution:

Lets assume that $$\lambda$$ is the Lebesgue measure and $$\nu$$ is a measure defined on the same space such that $$\nu(a+I)=\nu(I)$$

If $$\mathscr{I}_n=(0,0+\frac{1}{n}]$$

Then $$\bigcap_{n\geqslant 1}\mathscr{I}_n=\{0\}$$

As $$nu$$ is a measure then $$\nu(\mathscr{I}_n+a)=\nu(\mathscr{I}_n)=\lim_{n\to\infty}\nu(\mathscr{I}_n+a)=\lim_{n\to\infty}\nu(\mathscr{I}_n)\implies \nu(a)=\nu(0)$$ and $$a\in\mathbb{R}$$ is an arbitrary point. In the Lebesgue measure it happens the same by definition of length $$\lambda(a)=\lambda(0)=0$$. But I do not think that this resemblance proves the measures to be equal.

Question:

What should I do to prove the statement?

• I am not certain, but I think Carathéodory's theorem implies that if you show $\mu = c \lambda$ for the class of right half-closed intervals, then the measures are equal on $\mathscr{B}_{\mathbb{R}}$. Dec 22, 2018 at 19:46
• @angryavian Carathéodory's theorem assures the extension is unique, so I do not know at what extent linear transformations would be tolerated. But the measure $\mu$ is Lebesgue-Stieltjes and not just Lebesgue. Extension is understood: when you have two algebraic structure $S$ you have an extension to $R$ if the measures in those different $\mu(E)=\mu'(E)\forall E in S$. Dec 22, 2018 at 20:00
If $$\nu (A)=\infty$$ for every non-empty Borel set then $$\nu$$ has the invariance property but it is not constant times Lebesgue measure. If you assume that $$\nu (E) <\infty$$ for bounded Borel sets then the fact that $$\nu (a)=\nu (0)$$ for all $$a$$ implies that $$\nu (a)=0$$ for all $$a$$, so $$\nu$$ is actually translation invariant on the class of all intervals from which we can deduce that it is translation invariant.
• How would I conclude the argument? I need to prove the constant times Lebesgue is equal to $\mu$. Should I use Radon-Nikodym? How did you deduce $\nu(a)=0$? Thanks for your answer! Dec 23, 2018 at 11:28
• @PedroGomes Suppose $\nu (a) >0$ .If $\nu (A)<\infty$ then you can have atmost $[\frac {\nu (A)} {\nu(0)}]+1$ points in $A$ because if $a_1,a_2,..,a_k$ are distinct points of $A$ then $\mu (A) \geq \nu(a_1)+\nu(a_2)+...+\nu(a_k) =k\nu(0)$. In particular any open interval has a finite number of points. This contradcition shows that $\nu(a)\nu (0)=0$ for all $a$. The last part is a well known theorem Lebesgue measure is, up to a constant factor, the only transaltion invariant Borel measure finite on compact sets. Dec 23, 2018 at 11:41