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?
Thanks in advance!
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.