Doubts about Radon-Nikodym Theorem

Question

I'm learning Radon-Nikodym theorem and I have some doubts now. Let $\mu(B(x,r))=r^2$ for each $x\in\mathbb{R}$ and $r>0$, then $\mu$ is a radon measure on $\mathbb{R}$ and $\mu<<\mathscr{H}^1$. By Radon-Nikodym theorem, we have $\mu(E)=\int_E f d\mathscr{H}^1$ for some $f$. We can calculate that $f(x)=\lim_{r\to 0}\frac{\mu(x,r)}{\mathscr{H}^1(B(x,r))}=\lim_{r\to 0}\frac{r^2}{2r}=0$. That is, $\mu(B(x,1))=\int_{B(x,1)}0d\mathscr{H}^1=0$, which contradicts the fact that $\mu(B(x,1))=1$. What's wrong? I couldn't understand.

Answer 1

The problem is that $\mu$ is not additive: $\mu([0, 2]) = \mu(B(1, 1)) = 1^2 = 1$, but $\mu([0, 1]) + \mu([1, 2]) = \mu(B(\frac 1 2, \frac 1 2)) + \mu(B(1 + \frac 1 2, \frac 1 2)) = (\frac 1 2)^2 + (\frac 1 2)^2 = \frac 1 4 + \frac 1 4 = \frac 1 2$. So $\mu$ is not even a measure.