Let $\nu$ be a finite measure on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$. Let $C=\{x\in\mathbb{R}:\nu(\{x\})\neq0\}$. I want to show that $\nu$ can be decomposed into the sum of a discrete meausre, a continuous but singular measure, and an absolutely continuous measure.
I have proved that $C$ is countable. So if I define $\nu_1:\mathscr{B}(\mathbb{R})\to\mathbb{R}_+$ by $\nu_1(A)=\nu(A\bigcap C)$, then I could show that $\nu_1$ is a discrete measure. And what remains is to apply the Lebesgue Decomposition Theorem.
However, I got stuck on proving $A\bigcap C$ belongs to $\mathscr{B}(\mathbb{R})$. Since $\nu$ is a measure, if $A\bigcap C$ did not belong to $\mathscr{B}(\mathbb{R})$, then we cannot write $\nu(A\bigcap C)$. So I tried to show that $C\in\mathscr{B}(\mathbb{R})$, but I failed. Could someone please help me with this? Thanks a lot in advance!
- $C$ is countable:
Let $n$ be a positive integer. Assume to the contrary that there are infinitely many points $x$ such that $\nu(\{x\})\geq\frac{1}{n}$. Define $A_n=\{x\in\mathbb{R}:\nu(\{x\})\geq\frac{1}{n}\}\subseteq C$. Then $\nu(A_n)=\nu\left(\bigcup_{x\in A_n}\{x\}\right)=\sum_{x\in A_n}\nu(\{x\})\geq\sum_{x\in A_n}\frac{1}{n}=+\infty$, and so $\nu(A_n)=+\infty$, contradicting $\nu$ being a finite measure. Thus for each positive integer $n$, there are only finitely many $x$ such that $\nu(\{x\})\geq\frac{1}{n}$. So $C=\left(\bigcup_{n\in\mathbb{N}}A_n\right)$, as a countable union of countable set, must be countable.
Thanks to @Mark's answer, $C$ is Borel-measurable.
$\nu_1$ defined above is a discrete measure:
First, $\nu_1$ is nonnegative-valued. Next, $\nu_1(\emptyset)=\nu(\emptyset\bigcap C)=\nu(\emptyset)=0$. Moreover, let $\{A_j\}$ be a sequence of disjoint sets that belong to $\mathscr{A}$. Then $\nu_1\left(\bigcup_{i=1}^{\infty}A_i\right) = \nu\left(\left(\bigcup_{i=1}^{\infty}A_i\right)\bigcap C\right) = \nu\left(\bigcup_{i=1}^{\infty}\left(A_i\bigcap C\right)\right) = \sum_{i=1}^{\infty}\nu\left(A_i\bigcap C^c\right) = \sum_{i=1}^{\infty}\nu_1(A_i)$. So $\nu_1$ is a measure. Now, clearly, $\nu_1$ is finite. Then since $\nu_1\left(C^c\right)=\nu\left(C^c\bigcap C\right)=\nu(\emptyset)=0$, it follows that $\nu$ is a discrete measure.
- Lebesgue decomposition
Define $\phi:\mathscr{B}(\mathbb{R})\to\mathbb{R}_+$ by $\phi(A)=\nu\left(A\bigcap C^c\right)$. Then $\phi$ is a finite positive measure. Now we construct the Lebesgue decomposition $\nu_2$ and $\nu_3$ of $\phi$ with respect to the Lebesgue measure $\lambda$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$, where $\nu_2$ is continuous but singular with respect to $\lambda$ and $\nu_3$ is absolutely continuous with respect to $\lambda$. Define $\mathscr{N}_{\lambda}$ by $\mathscr{N}_{\lambda}=\{B\in\mathscr{B}(\mathbb{R}):\lambda(B)=0\}$, and choose a sequence $\{B_j\}$ of sets in $\mathscr{N}_{\lambda}$ such that $\lim_{j\to\infty}\phi(B_j)=\sup\{\phi(B):B\in\mathscr{N}_{\lambda}\}$. Let $N=\bigcup_jB_j$, and define measures $\nu_2$ and $\nu_3$ on $(\mathbb{R},\mathscr{B}(\mathbb{R}))$ by $\nu_2(A)=\phi\left(A\bigcap N^c\right)=\nu\left(A\bigcap N^c\bigcap C^c\right)$ and $\nu_3(A)=\phi(A\bigcap N)=\nu\left(A\bigcap N\bigcap C^c\right)$. (Note that $\nu_2$ is continuous. To see this, let $x\in\mathbb{R}$. Then $\{x\}\bigcap N^c\bigcap C^c$ is either $\emptyset$ or a singleton subset $\{x\}$ of $C^c$ so that $x\in C^c$. In either case, $\nu_2(\{x\})=\nu\left(\{x\}\bigcap N^c\bigcap C^c\right)=0$.)