# A counter-example to Radon-Nikodym Theorem?

I am confused about the Radon-Nikodym Theorem, because of the following example which came to my mind. What is wrong in my argument?

Consider $X=[0,1]$ with the $\sigma$-algebra of Borel measurable sets. Let $m$ be the Lebesgue measure on $[0,1]$, and let $\mu$ be defined by $$\mu(E)=\text{counting measure}(E\cap\mathbb{Q}),$$ i.e. $\mu$ counts the number of rational points in $E$. Then, $\mu$ is a positive $\sigma$-finite measure because by letting $\mathbb{Q}=\{r_n\}_{n=1}^\infty$ we get $$[0,1]=([0,1]\cap\mathbb{Q}^c)\cup\bigcup_{n=1}^\infty\{r_n\}.$$ Now, $m\ll \mu$, so by the Radon-Nikodym theorem there exists $h\in L^1(\mu)$ such that $$m(E)=\int_E hd\mu,\tag{1}$$ for all measurable sets $E\subseteq[0,1]$.

But then we have problem: For all $x\in\mathbb{Q}$, we get $$0=m(\{x\})=\int_{\{x\}}hd\mu=h(x)\mu(\{x\})=h(x),$$ so $h=0$ on $\mathbb{Q}$. But $\mu$ is concentrated on $\mathbb{Q}$, so (1) shows that $m=0$.

What's wrong?

• $m$ isn't absolutely continuous w.r.t. $\mu$. For example, irrationals are $\mu$-null but not $m$-null. – hot_queen Mar 12 '15 at 19:18

We have $$\mu([0,1]-\mathbb{Q})=0$$ but $$m([0,1]-\mathbb{Q})=1,$$ so $m\not\ll \mu$.