Assume $A\in\mathbb{R}^n$ is a Lebesgue null set and $\mu$ is a positive $\sigma$-finite measure living on $A$ (i.e. $\mu(A^c)=0$) such that $$ \mu(A+r)=\mu(A),\forall r\in\mathbb{R}^n $$ does this imply that $\mu=0$?
I think of this when I am learning the Lebesgue-Radon-Nikodym theorem and the invariant property under translation of Lebesgue measure. I think the answer is yes but still need more justification. Any hint would be appreciated!
Thanks to Fubini's theorem, we have $$ \mu(A)=\int_0^1 \mu(A+r)\,dr =\int_0^1\int 1_{A+r}(y)\,d\mu(y)\,dr=\int \int_0^1 1_{A+r}(y)\,dr\,d\mu(y)\leq \int \lambda(y-A)\,d\mu(y)=0, $$ where $\lambda$ is the Lebesgue measure. Here, I used that $\lambda(y-A)=0$ since $\lambda(A)=0$ (and by elementary properties of the Lebesgue measure).