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:


 * $P = \mu + \nu$

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:


 * $\ds 1 = \map P \Omega \ge \map P {\Omega'_n} = \sum_{\omega \mathop \in \Omega'_n} \map P {\set \omega} \ge \frac {\card {\Omega'_n} } n = \frac {n + 1} n$

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:


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

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 $\paren {\omega_n}_{n \in \N}$ be an enumeration of $\Omega_\infty$.

Now define the discrete measure $\nu$ by:


 * $\ds \nu := \sum_{n \mathop \in \N} \map P {\omega_n} \delta_{\omega_n}$

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:


 * $\ds E \cap \Omega_\infty = \bigsqcup_{\omega_n \mathop \in E} \set {\omega_n}$

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.