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.


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 $P(B_n')=P(B_n \setminus \cup_{i<n} B_i) =P(B_n)$.

  • $\begingroup$ 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

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