=== Existence === For each $n \in \N$, define $\Omega_n$ by: :$\Omega_n := \set {\omega \in \Omega: \map P {\set \omega} \ge \dfrac 1 n}$ Suppose that $\Omega_n$ has more than $n$ elements, and define $\Omega'_n$ to be a finite subset of $\Omega_n$ with $n + 1$ elements. From Measure is Monotone and Measure is Finitely Additive Function, it follows that: :$\displaystyle 1 = \map P \Omega \ge \map P {\Omega'_n} = \sum_{\omega \mathop \in \Omega'_n} \map P {\set \omega} \ge \frac {\card {\Omega'_n} } n = \frac {n + 1} n$ This is an obvious contradiction, whence $\Omega_n$ has at most $n$ elements, and in particular, $\Omega_n$ is finite. It follows from Sequence of Reciprocals is Null Sequence that: :$\forall p > 0: \exists n \in \N: \dfrac 1 n < p$ Therefore, defining $\Omega_\infty$ by: :$\Omega_\infty := \set {\omega \in \Omega: \map P {\set \omega} > 0}$ we have: :$\displaystyle \Omega_\infty = \bigcup_{n \mathop \in \N} \Omega_n$ whence by Countable Union of Countable Sets is Countable, $\Omega_\infty$ is countable. Also, being the countable union of elements in $\Sigma$, $\Omega_\infty \in \Sigma$. Thus, let $\paren {\omega_n}_{n \in \N}$ be an enumeration of $\Omega_\infty$. Now define the discrete measure $\nu$ by: :$\displaystyle \nu := \sum_{n \mathop \in \N} \map P {\omega_n} \delta_{\omega_n}$ Next, we define $\mu$ in the only possible way: :$\mu: \Sigma \to \overline \R: \map \mu E := \map P E - \map \nu E$ It remains to verify that $\mu$ is a measure, and diffuse. So let $E \in \Sigma$, and write, by Set Difference and Intersection form Partition: :$E = \paren {E \cap \Omega_\infty} \sqcup \paren {E \setminus \Omega_\infty}$ where $\sqcup$ signifies disjoint union. Then we also have, trivially, the decomposition: :$\displaystyle E \cap \Omega_\infty = \bigsqcup_{\omega_n \mathop \in E} \set {\omega_n}$ Now, we can compute: {{begin-eqn}} {{eqn | l = \map \nu E       | r = \sum_{n \mathop \in \N} \map P {\set {\omega_n} } \map {\delta_{\omega_n} } E       | c = Definition of $\nu$ }} {{eqn | r = \sum_{\omega_n \mathop \in E} \map P {\set {\omega_n} }       | c = {{Defof|Dirac Measure}} }} {{eqn | r = \map P {E \cap \Omega_\infty}       | c = $P$ is a Measure }} {{eqn | ll= \leadsto       | l = \map P E - \map \nu E       | r = \map P {E \cap \Omega_\infty} + \map P {E \setminus \Omega_\infty} - \map P {E \cap \Omega_\infty}       | c = Measure is Finitely Additive Function }} {{eqn | r = \map P {E \setminus \Omega_\infty} }} {{end-eqn}} From Set Difference as Intersection with Complement, we may write: :$E \setminus \Omega_\infty = E \cap \paren {\Omega \setminus \Omega_\infty}$ So for every $E \in \Sigma$, we have: :$\map \mu E = \map P {E \cap \paren {\Omega \setminus \Omega_\infty} }$ whence by $\mu$ is an intersection measure, and by Intersection Measure is Measure, a measure. To show that $\mu$ is diffuse, let $\omega \in \Omega$. Then: :$\map \mu {\set \omega} = \map P {\set \omega \cap \paren {\Omega \setminus \Omega_\infty} }$ Now, by definition of $\Omega_\infty$, we have: :$\set \omega \setminus \Omega_\infty = \begin{cases} \set \omega & \text {if $\map P {\set \omega} = 0$} \\ \O & \text{otherwise} \end{cases}$ and so in either case, it follows that: :$\map P {\set \omega \cap \paren {\Omega \setminus \Omega_\infty} } = 0$ that is to say, $\mu$ is diffuse. {{qed|lemma}} === Uniqueness === Let $P = \mu_1 + \nu_1 = \mu_2 + \nu_2$ be two decompositions. Since $\mu_1$ and $\mu_2$ are diffuse, it follows that, for any $\omega \in \Omega$: :$\map P {\set \omega} = \nu_1 \paren {\set \omega} = \map {\nu_2} {\set \omega}$ Suppose that we have the discrete measures $\nu_1$ and $\nu_2$ defined as: :$\displaystyle \nu_1 = \sum_{n \mathop \in \N} \kappa_n \delta_{x_n}$ :$\displaystyle \nu_2 = \sum_{m \mathop \in \N} \lambda_m \delta_{y_m}$ where we take $\sequence {x_n}_{n \mathop \in \N}$ and $\sequence {y_n}_{n \mathop \in \N}$ to be sequences of distinct terms. Then the equality derived for arbitrary $\omega$ above yields, taking $\omega = x_n$: :$\displaystyle \kappa_n = \sum_{m \mathop \in \N} \lambda_m \map {\delta_{y_m} } {\set {x_n} }$ which by definition of Dirac measure implies the existence of a unique $m \in \N$ such that: :$x_n = y_m, \, \kappa_n = \lambda_m$ and reversing the argument, also an $n \in \N$ with this equality for every $y_m$. That is, it must be that $\nu_1 = \nu_2$. Whence $\mu_1 = P - \nu_1 = P - \nu_2 = \mu_2$, and uniqueness follows. {{qed}} {{ACC|Countable Union of Countable Sets is Countable}}