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.


What should I do to prove the statement?

Thanks in advance!

  • $\begingroup$ 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 $\mathscr{B}_{\mathbb{R}}$. $\endgroup$
    – angryavian
    Dec 22, 2018 at 19:46
  • $\begingroup$ @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 $\mu(E)=\mu'(E)\forall E in S$. $\endgroup$ Dec 22, 2018 at 20:00
  • $\begingroup$ I deleted my answer since it was flawed a little bit. $\endgroup$
    – Shashi
    Dec 22, 2018 at 22:20
  • $\begingroup$ @Shashi Are you going to upload it again? $\endgroup$ Dec 22, 2018 at 22:21
  • $\begingroup$ @Shashi I need all the hints I can get. $\endgroup$ Dec 22, 2018 at 22:21

1 Answer 1


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.

  • $\begingroup$ 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! $\endgroup$ Dec 23, 2018 at 11:28
  • $\begingroup$ @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. $\endgroup$ Dec 23, 2018 at 11:41

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