Decomposition of Probability Measures

Theorem
Let $\left({\Omega, \Sigma, P}\right)$ be a probability space.

Suppose that for every $\omega \in \Omega$, it holds that:


 * $\left\{{\omega}\right\} \in \Sigma$

i.e., that $\Sigma$ contains all singletons.

Then there exist a unique diffuse measure $\mu$ and a unique discrete measure $\nu$ such that:


 * $P = \mu + \nu$

Existence
For each $n \in \N$, define $\Omega_n$ by:


 * $\Omega_n := \left\{{\omega \in \Omega: P \left({\left\{{\omega}\right\}}\right) \ge \dfrac 1 n}\right\}$

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:


 * $\displaystyle 1 = P \left({\Omega}\right) \ge P \left({\Omega'_n}\right) = \sum_{\omega \mathop \in \Omega'_n} P \left({\left\{{\omega}\right\}}\right) \ge \frac {\left\vert{\Omega'_n}\right\vert} {n} = \frac {n + 1} {n}$

This is an obvious contradiction, whence $\Omega_n$ has at most $n$ elements, and in particular, $\Omega_n$ is finite.

It follows from Power of Reciprocal: Corollary that:


 * $\forall p > 0: \exists n \in \N: \dfrac 1 n < p$

Therefore, defining $\Omega_\infty$ by:


 * $\Omega_\infty := \left\{{\omega \in \Omega: P \left({\left\{{\omega}\right\}}\right) > 0}\right\}$

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.

Thus, let $\left({\omega_n}\right)_{n \in \N}$ be an enumeration of $\Omega_\infty$.

Now define the discrete measure $\nu$ by:


 * $\displaystyle \nu := \sum_{n \mathop \in \N} P \left({\omega_n}\right) \delta_{\omega_n}$

Next, we define $\mu$ in the only possible way:


 * $\mu: \Sigma \to \overline{\R}: \mu \left({E}\right) := P \left({E}\right) - \nu \left({E}\right)$

It remains to verify that $\mu$ is a measure, and diffuse.