3
$\begingroup$

Given a probability space $(\Omega,\mathcal{F},\mathcal{P})$, we can construct a complete probability space $(\Omega,\bar{\mathcal{F}},\bar{\mathcal{P}})$ such that $\mathcal{F}\subseteq \bar{\mathcal{F}}$ and $\left.\bar{\mathcal{P}}\right|_\mathcal{F}=\mathcal{P}$.

For the proof, we define $\bar{\mathcal{F}}=\{A\subseteq \Omega:A \triangle B \in \mathcal{N},$ for some $B \in \mathcal{F}\}$ where $A\triangle B$ is the symmetric difference i.e. $A\triangle B= \left(A \setminus B\right) \cup \left(B \setminus A\right)$ and $\bar{\mathcal{P}}(A)=\mathcal{P}(B)$. Also $\mathcal{N}$ is the set of null events of $(\Omega,\mathcal{F},\mathcal{P})$.

I've already shown most of what I need to show, in particular, $\mathcal{F}\subseteq \bar{\mathcal{F}}$, that $\bar{\mathcal{P}}$ is well-defined, $\bar{\mathcal{P}}(\Omega)=1 $, $\left.\bar{\mathcal{P}}\right|_\mathcal{F}=\mathcal{P}$, and $\bar{\mathcal{F}}$ is indeed a $\sigma$-field. The only thing that I have left, and what is giving me trouble, is that $\bar{\mathcal{P}}$ is $\sigma$-additive.

So I start with $A_i \in \bar{\mathcal{F}}$, $i=1,2,...$ such that we have pairwise disjointedness. I want to prove $\bar{\mathcal{P}}(\bigcup\limits_{i=1}^{\infty} A_{i})=\sum_{i=1}^{\infty} \bar{\mathcal{P}}(A_i)$. Now I've shown that $(\bigcup\limits_{i=1}^{\infty} A_{i})\triangle (\bigcup\limits_{i=1}^{\infty} B_{i})\in \mathcal{N}$, where for each $A_i$, there is a $B_i \in \mathcal{N}$ such that $A_i \triangle B_i \in \mathcal{N}$

This problem becomes trivial if I can show the collection of $B_i$ is pairwise disjoint (because then I can just use $\bar{\mathcal{P}}(A)=\mathcal{P}(B)$ and the $\sigma$-additivity of $\mathcal{P}$), however I don't believe this can be the case.

I've thought of using subadditivity, but showing the opposite inequality is also giving me trouble. Any help is very much appreciated.

$\endgroup$
1
$\begingroup$

Hints: let $B_1'=B_1,B_2'=B_2\setminus B_1, B_3'=B_3\setminus (B_1\cup B_2),...$. Then $B_n$'s are disjoint and $\cup_n B_n'=\cup_n B_n$. Now use the fact that $P(\cup_n B_n)=P(\cup_n B_n')=\sum_n P(B_n')$. Now you will be able to complete the proof provided you know that $P(B_n')=P(B_n)$ for all $n$. To prove this observe that $B_n\cap B_i \subset (A_n\cap A_i) \cup (A_n\Delta B_n) \cup (A_i\Delta B_i)$ for any $i <n$. From this it should be clear that $P(B_n')=P(B_n \setminus \cup_{i<n} B_i) =P(B_n)$.

$\endgroup$
2
  • $\begingroup$ To create a disjoint collection is a good idea. I have a question about the indices you use at the end. Do you mean $P(B_n')=P(B_n \setminus \cup_{i<n} B_i) =P(B_n)$. $\endgroup$ – HCS Sep 11 '19 at 22:56
  • $\begingroup$ @HCS Yes, there was a typo. $\endgroup$ – Kavi Rama Murthy Sep 11 '19 at 23:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.