# Lebesgue measure problem

• Sep 5th 2011, 09:42 AM
TheApp
Lebesgue measure problem
Hi,

I have the following problem:

Let m be the Lebesgue measure on R and let v be a measure on the Lebesgue sigma-algebra such that:
(i) v is absolutely cont. with respect to m.
(ii) v({x}+A)=v(A) for all x in R and A in the Lebesgue sigma-algebra.
Show: v=c*m for some constant c in R.

Really all I can see is that (and this is given by assumption (i)) the statement holds for all E such that m(E)=0. Other then that, nada. Any hints?
• Sep 5th 2011, 10:45 AM
girdav
Re: Lebesgue measure problem
Apply Radon-Nikodym theorem: it will give you an integrable function $\displaystyle f$ such that for all $\displaystyle A$ Lebesgue-measurable, $\displaystyle \nu(A)=\int_A f(t)dm(t)$. Now, fix $\displaystyle x\in\mathbb R$, and put $\displaystyle h(u):=f(u+x)-f(u)$. h is integrable and for all $\displaystyle A$ Lebesgue-measurable, $\displaystyle \int_A h(t)dm(t)=0$. It shows that $\displaystyle f$ is constant.