# Augmented sigma-algebra and conditional expectation.

I want to prove this:

You have the probability space $(\Omega, \mathcal{F},P)$, and a random variable Y on this space. You have a sub-sigma algebra $\mathcal{A}$ on this space, you take the augmented sigma algebra for $\mathcal{A}$ like this: Let $\mathcal{B}$ be the sigma algebra generated by $\mathcal{A}\cup\mathcal{N}$, where $\mathcal{N}$ is the collection of all sets of measure 0. We then have 3 sigma algebras: $\mathcal{A}\subset \mathcal{B} \subset\mathcal{F}$.

I want to prove that $\mathbb{E}(Y|\mathcal{A})=\mathbb{E}(Y|\mathcal{B})$ a.s.

By the uniquness in the Radon-Nikodym theorem, this amounts to showing that for any $B \in \mathcal{B}$ we have $\int_B\mathbb{E}(Y|\mathcal{A})dP=\int_BYdP$.

But how do I get this? The problem is that I do not have much control over the B sets? If we had the completion instead it would be easier, because then we know that each B set is of the form of a set in $\mathcal{A}$ union with a subset of a set of measure 0, but in this case, I am not sure how the sets look. I tried proving that all the sets are of the form $A\cup N$, but wasn't able to do it, and is it the case that they are?

This may help: $B\subset\Omega$ is an element of $\mathcal B$ if and only if there exists $A\in\mathcal A$ such that $A\setminus B$ and $B\setminus A$ are both elements of $\mathcal N$.