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.
Best Answer
Here is my attempt. For each $x \in (0,1)$, we have $1 = \lambda (x) = \lambda_a (x) + \lambda_s(x) = 0 + \lambda_s(x)$. So, $\lambda_s(x) = 1 \ \forall x \in (0, 1)$. Since $\lambda_s$ and $m $ are mutually singular, there exist $A \sqcup B = (0,1)$ s.t. $m(A)= \lambda_s(B) = 0$. Now, if $B$ is nonempty, take $x \in B$, and note that $\lambda_s(B) \geq \lambda_s(x) =1 $, a contradiction. So $B = \emptyset.$ And since $A, B$ disjoint, we have $A= (0,1)$, leading to $0 = m(A)=m((0,1))=1$. A contradiction. So no lebesgue decomposition exists.