Decomposition of Probability Measures

== Theorem ==
Let $\struct {\Omega, \Sigma, P}$ be a probability space.
Suppose that for every $\omega \in \Omega$, it holds that:
:$\set \omega \in \Sigma$
that is, that $\Sigma$ contains all singletons.
Then there exist a unique diffuse measure $\mu$ and a unique discrete measure $\nu$ such that:
:$P = \mu + \nu$
== Proof ==
=== Existence ===
For each $n \in \N$, define $\Omega_n$ by:
:$\Omega_n := \set {\omega \in \Omega: \map P {\set \omega} \ge \dfrac 1 n}$
 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:
:$\ds 1 = \map P \Omega \ge \map P {\Omega'_n} = \sum_{\omega \mathop \in \Omega'_n} \map P {\set \omega} \ge \frac {\card {\Omega'_n} } n = \frac {n + 1} n$
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:
:$\forall p \in \R_{>0}: \exists n \in \N: \dfrac 1 n < p$
Therefore, defining $\Omega_\infty$ by:
:$\Omega_\infty := \set {\omega \in \Omega: \map P {\set \omega} > 0}$
we have:
:$\ds \Omega_\infty = \bigcup_{n \mathop \in \N} \Omega_n$
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 $\Omega_\infty = \set {\omega_0, \omega_1, \omega_2, \ldots}$ be a countably infinite enumeration.
Now define the discrete measure $\nu$ by:
:$\ds \nu := \sum_{n \mathop \in \N} \map P {\set {\omega_n} } \delta_{\omega_n}$
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:
:$\ds E \cap \Omega_\infty = \bigsqcup_{\substack n \mathop \in \N \\ \omega_n \mathop \in E } \set {\omega_n}$
Now, we can compute:
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.
=== 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:
:$\ds \nu_1 = \sum_{n \mathop \in \N} \kappa_n \delta_{x_n}$
:$\ds \nu_2 = \sum_{m \mathop \in \N} \lambda_m \delta_{y_m}$
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 $\omega = x_n$:
:$\ds \kappa_n = \sum_{m \mathop \in \N} \lambda_m \map {\delta_{y_m} } {\set {x_n} }$
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.
== Sources ==
* : $\S 6$: Problem $5$
Category:Probability Measures