Completeness

Consider a set which is even smaller (subset) than a null set. A null set is a set $S$ such that $\Pr(S) = 0$.

This subset of the null set also intuitively has $\Pr(\cdot) = 0$. However, this small intuitive idea cannot be derived from the definition of a probability measure.

Why? Because there is no guarantee that if $A \in \cF$ and $B \subseteq A$, then $B \in \cF$ for a $\sigma$-field $\cF$.

If $B \in \cF$ (i.e. if we knew it was measurable), then by monotonicity the idea above is true. But we don’t know if the subset is even measurable (in $\cF$).

Definition: Complete Probability Space

A probability space $(\Omega, \cF, \Pr)$ is complete if

$$A \subseteq B \text{ and } \Pr(B) = 0 \implies A \in \cF$$

(and consequently $\Pr(A) = 0$).

Consequences of completeness

If $(\Omega, \cF, \Pr)$ is complete, then for a set $A'$, if there exists $A \in \cF$ such that $A \Delta A' \subseteq B$ for some $B$ with $\Pr(B) = 0$, then $A' \in \cF$ and $\Pr(A') = \Pr(A)$.

Note: $A \Delta A'$ is the symmetric difference, defined as $A \Delta A' = (A \cup A') - (A \cap A')$.

Since $A \Delta A' \subseteq B$ and $\Pr(B) = 0$, the subsets $A \cap A'^c$ and $A^c \cap A'$ are subsets of a null set in a complete space, so they are null sets. This implies $\Pr(A) = \Pr(A') = \Pr(A \cap A')$.

Existence of complete extension

A probability space always has a complete extension.

Theorem

Statement

Let $(\Omega, \cF, \Pr)$ be a probability space. Then there exists a complete probability space $(\Omega, \cF', \Pr')$ such that $\cF \subseteq \cF'$ and $\Pr(A) = \Pr'(A)$ for all $A \in \cF$.

Proof

1. Recall: The outer measure $\Pr^*$ defined in the extension result is a probability measure on the $\sigma$-field $\mathcal{M}$ where $\mathcal{M}$ is the class of all $\Pr^*$-measurable sets.

   $$A \in \mathcal{M} \iff \Pr^*(E) = \Pr^*(A \cap E) + \Pr^*(A^c \cap E) \quad \forall E \subseteq \Omega$$

   We are going to show that $(\Omega, \mathcal{M}, \Pr^*)$ is a complete probability space and it will be the required extension.

2. Let $\Pr^*(B) = 0$ and $A \subseteq B$. Our goal is to show $\Pr^*(A) = 0$, to show that we need to just show $A \in \mathcal{M}$.

   For any $E \subseteq \Omega$, consider:

   $$\Pr^*(A \cap E) + \Pr^*(A^c \cap E)$$

   Note that:

   • $A \cap E \subseteq A \subseteq B \implies \Pr^*(A \cap E) \le \Pr^*(B) = 0$. So $\Pr^*(A \cap E) = 0$.
   • $A^c \cap E \subseteq E \implies \Pr^*(A^c \cap E) \le \Pr^*(E)$.

   Therefore, by monotonicity of $\Pr^*$:

   $$\Pr^*(A \cap E) + \Pr^*(A^c \cap E) \le 0 + \Pr^*(E) = \Pr^*(E)$$

3. The other side of the inequality is trivial, as:

   $$\Pr^*(E) \le \Pr^*(A \cap E) + \Pr^*(A^c \cap E)$$

   by the definition of outer measure (subadditivity).

   Thus, we have equality:

   $$\Pr^*(A \cap E) + \Pr^*(A^c \cap E) = \Pr^*(E)$$

   This implies that $A$ is $\Pr^*$-measurable, so $A \in \mathcal{M}$.

   Also, $\Pr^*(A) = 0$ follows from the monotonicity of $\Pr^*$ (since $A \subseteq B$ and $\Pr^*(B)=0$).