Let $m$ be the Lebesgue measure on $(0,1)$, and $\lambda$ the counting measure on $(0,1)$. I am trying to prove that there is no decomposition $$\lambda=\lambda_a+\lambda_s$$ where $\lambda_a$ is absolutely continuous with respect to $m$, and $\lambda_s$ is singular with respect to $m$.
I am trying to show that the decomposition would give a contradiction, but I am stuck. Any idea?
My attempt: We get $\lambda(E)=\lambda_s(E)$ for all countable sets $E$, so (maybe) this implies that $\lambda=\lambda_s$ and hence $\lambda\perp m$, which is false.