Sigma Algebra Generated by Almost Sure Zero Random Variable Let $(\Omega, \mathcal{F}, P)$ be a probability space and $X: \Omega \rightarrow \mathbb{R}$ be a random variable with $P(X = 0) =1.$ Let $\mathcal{N}$ be the class of sets $A \subset \Omega$ such that there exists a set $N \in \mathcal{F}$ with $P(N)=0$ and $A \subset N.$ Let $\mathcal{G}$ be the $\sigma$-algebra generated by $X$ and $\mathcal{N}.$
Let $Y:\Omega \rightarrow \mathbb{R}$ be $\mathcal{G}$-measurable. Is it true that $P(Y = c) =1 $ for some $c \in \mathbb{R}?$
I know this result is true if $\mathcal{G}$ were the trivial $\sigma$-algebra.  
 A: A first problem is that $\{Y=c\}$ may not be $\mathcal F$-measurable, hence in order to apply the measure $\mathbb P$, we have to rather consider its completion. This means that instead of $\mathcal F$, we work with the $\sigma$-algebra of the subsets of   $\Omega$ of the form $F\cup N$, where $F\in\mathcal F$ and $N\in\mathcal N$ (then $\mathbb P\left(F\cup N\right)$ is defined as $\mathbb P\left(F\right)$).  
We need the following intermediate lemma:

Let $\mathcal A$ and $\mathcal B$ be two sub-$\sigma$-algebras of a probability space $\left(\Omega,\mathcal F,\mathbb P\right)$ such that for each $A\in\mathbb A$, $\mathbb P\left(A\right)\in\{0,1\}$ and for each $B\in\mathbb B$, $\mathbb P\left(B\right)\in\{0,1\}$. Then each element of the $\sigma$-algebra generated by $\mathcal A$ and $\mathcal B$ has probability zero or one.

To this this, let 
$$
\mathcal C:=\left\{C\in\mathcal F\mid \mathbb P\left(C\right)\in\{0,1\}\right\}.
$$
Then $\mathcal C$ is a $\sigma$-algebra containing $\mathcal A$ and $\mathcal B$.
