# prove that $\mu$ is a measure $\sigma$-finite.

Let $A \subseteq \mathbb{R}$ a countable subset. Let $a: A \rightarrow \mathbb{R}^{+}$ any function. For $E \subseteq \mathbb{R}$ define $\mu(E)=\sum_{x\in E\cap A} a(x)$.

(a) Prove that $\mu$ is a measure $σ$-finite on $(\mathbb{R},P(\mathbb{R}))$

(B) find Lebesgue decomposition $\mu= \mu_{a}+ \mu_{s}$ of $\mu$ with respect to the Lebesgue measure $\lambda$. ($\mu_{a} \ll \lambda$ and $\mu_{s}\perp \lambda$)

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Is this a homework problem? –  George V. Williams Feb 2 '13 at 0:48

Hints:

a) $A$ is countable and $a(x)$ is finite.

b) $\mu$ is concentrated on $A$ and $\lambda(A) = 0$.

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but not necessarily E is countable. –  Alexander Osorio Feb 2 '13 at 0:59
@AlexanderOsorio Ah, I meant $A$ (and therefore $A \cap E \subset A$ too). Fixed now. –  Ayman Hourieh Feb 2 '13 at 1:01
But E ∩ A can have infinite elements. thus µ(E) is infinite –  Alexander Osorio Feb 2 '13 at 1:06
That's OK. $\mu$ is required to be $\sigma$-finite, not finite. $\sigma$-finite means any $E$ is a countable union of measurable sets of finite measure. What's the measure of each $\{x\}$ where $x \in E$? –  Ayman Hourieh Feb 2 '13 at 1:19