Probability Measure on Single-Subset Event Space

Theorem
Let $\EE$ be an experiment whose sample space is $\Omega$.

Let $\O \subsetneqq A \subsetneqq \Omega$.

Let $\Sigma := \set {\O, A, \Omega \setminus A, \Omega}$ be the event space of $\EE$.

Let $\Pr: \Sigma \to \R$ be a probability measure on $\struct {\Omega, \Sigma}$.

Then $\Pr$ has the form:

for some $p \in \R$ satisfying $0 \le p \le 1$.

Proof
From Event Space from Single Subset of Sample Space, we have that $\Sigma$ is an event space.

Recall the Kolmogorov axioms:

First we determine that $\Pr$ as defined is actually a probability measure.

Axiom $(1)$ and axiom $(2)$ are fulfilled trivially by definition.

Then we note that, apart from $\O$, $\set {A, \Omega \setminus A}$ are the only pairwise disjoint events whose union is $\Omega$.

Hence by definition of $\Pr$ we see that axiom $(3)$ degenerates to:

and thus axiom $(3)$ is fulfilled.

Thus $\Pr$ is indeed a probability measure.

Now it is established that a probability measure on $\Sigma$ is of the form of $\Pr$ as defined.

First we note that from Probability of Empty Event is Zero:
 * $\map \Pr \O = 0$

Hence $\text {Pr 1}$ is satisfied.

From Axiom $(2)$ we have that $\map \Pr \Omega = 1$

Hence $\text {Pr 4}$ is satisfied.

By definition of probability measure, $\map \Pr A$ is between $0$ and $1$ inclusive.

That is:
 * $\map \Pr A = p$

for some $p \in \R$ such that $0 \le p \le 1$.

This is exactly $\text {Pr 2}$, which is hence seen to be satisfied.

Then we note that from Probability of Event not Occurring:
 * $\map \Pr {\Omega \setminus A} = 1 - \map \Pr A$

That is:
 * $\map \Pr {\Omega \setminus A} = 1 - p$

Hence $\text {Pr 3}$ is satisfied.

Thus it has been proved that a probability measure on such an an event space $\Sigma$ has to be of the form $\Pr$ as defined.

Proof

 * : $1$: Events and probabilities: $1.3$: Probabilities: Example $10$