Completion of a Probability Space using Symmetric Difference

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 $$i . From this it should be clear that $$P(B_n')=P(B_n \setminus \cup_{i.
• To create a disjoint collection is a good idea. I have a question about the indices you use at the end. Do you mean $P(B_n')=P(B_n \setminus \cup_{i<n} B_i) =P(B_n)$. – HCS Sep 11 '19 at 22:56