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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.

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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.

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Suppose there exist such a decomposition.

As $\lambda_a$ is absolutely continuous, $\lambda_a(\{x\}) = 0 \forall x$. Hence $\lambda_s(\{x\}) = 1 \forall x$

But as $\lambda_s$ is singular, there exist a set $S\in \mathcal{B}(\mathbb{R})$ such that $\forall B\in \mathcal{B}(S), \lambda_s(B) = 0$.

But this set contain a $x$, so $\{ x \} \in \mathcal{B}(S)$. Then $\lambda_s(\{x\}) = 0 $: contradiction

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    $\begingroup$ $\lambda_a$ being absolutely continuous means that if $\lambda(S)=0$ then $\lambda_a(S)=0$. Why can you assume that $\lambda_a(\{x\})=0$? $\endgroup$ – Squirtle Mar 6 '15 at 0:10
  • $\begingroup$ @Squirtle I upvoted your comment, but you are actually wrong. $\lambda_{a}$ is absolutely continuous with respect to Lebesgue measure $m$. So since $m( \{ x \}) = 0$, then $\lambda_{a}( \{ x \} ) = 0$. It's important to specify which measure you are absolutely continuous with respect to. $\endgroup$ – layman Mar 6 '15 at 0:54
  • $\begingroup$ Yeah.... I agree... i made a typo, my $\lambda$ above should be $m$. $\endgroup$ – Squirtle Mar 6 '15 at 0:58
  • $\begingroup$ @Squirtle : for all x, $m(\{x\}) = 0$, then, as $\lambda_a$ is absolutly continuous with respect to $m$,for all x, $\lambda_a(\{x\}) = 0$ $\endgroup$ – Tryss Mar 6 '15 at 8:12
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If you are familiar with the purely atomic and nonatomic decomposition of any measure then my answer will work wonderfully... otherwise, you may consider studying the great classic and very readable Atomic and Nonatomic Measures by Roy A. Johnson.

Proof: Suppose such a decomposition exists. As $\lambda$ is atomic (the singletons are its atoms), and $m$ is nonatomic such that $m \ll \lambda$ then $\lambda \perp m$ by Theorem 2.3 of the paper. Hence $m\perp (\lambda_a + \lambda_s)$, but as $m\perp \lambda_s$ this implies $m\perp \lambda_a$ but we also assumed that $\lambda_a\ll m$. Because $m\perp \lambda_a \ll m$ we conclude that $m=0$ (the zero measure), a contradiction.

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